Re: [R] Archive format
Hi Joe, I have read your question with great interest. I am a little bit astonished to read about your project. There is a big national institute in Germany called GESIS (https://de.wikipedia.org/wiki/GESIS_%E2%80%93_Leibniz-Institut_f%C3%BCr_Sozialwissenschaften) which does the same job you are trying to set-up since 1986 now. You could try to exchange ideas with them. Your subject is very complex with regard to reproducible research. You might want to have a look at (1) https://cran.r-project.org/web/views/ReproducibleResearch.html (2) Gandrud, Christopher: Reproducible Research with R and R Studio (https://www.amazon.com/Reproducible-Research-Studio-Second-Chapman/dp/1498715370) Kind regards Georg > Gesendet: Mittwoch, 29. März 2017 um 10:44 Uhr > Von: "Joe Gain" > An: R-help@r-project.org > Cc: bwfdm-i...@lists.kit.edu > Betreff: [R] Archive format > > Hello, > > we are collecting information on the subject of research data management > in German on the webplatform: > > www.forschungsdaten.info > > One of the topics, which we are writing about, is how to *archive* data. > Unfortunately, none of us in the project is an expert with respect to R > and so I would like to ask the list, what they recommend? A related > question is to do with the sharing of data. We have already asked some > academics, who have basically replied that they don't really know other > than to strongly recommend a plain text format. > > We would also like to know, if members of the list recommend converting > formats from commercial software such as S-Plus, Terr, SPSS etc. to an > R-compatible format for long term archivation? Are there any general > rules and best practices, when it comes to archiving (and sharing) > statistical data and statistical programs? > > Any comments would be much appreciated! > Joe > > -- > B 1003 > Kommunikations-, Informations-, Medienzentrum (KIM) > Universitaet Konstanz > > t: ++49-7531-883234 > e: joe.g...@uni-konstanz.de > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Antwort: Re: Way to Plot Multiple Variables and Change Color
Hi Ulrik,
many thanks for your reply. I had to take an unplanned break and was not
in the office during the last two weeks. Thus my late reply.
I followed your advice and converted the variable in argument "fill" to
factor. Now the color change works:
-- cut --
d_result <- structure(list("variable" = c("Item 1 (ø = 3.3) ", "Item 1 (ø
= 3.3) ",
"Item 1 (ø = 3.3) ", "Item 1 (ø =
3.3) ", "Item 1 (ø = 3.3) ",
"Item 1 (ø = 3.3) ", "Item 2 (ø =
3.8) ", "Item 2 (ø = 3.8) ",
"Item 2 (ø = 3.8) ", "Item 2 (ø =
3.8) ", "Item 2 (ø = 3.8) ",
"Item 2 (ø = 3.8) ", "Item 3 (ø =
3.4) ", "Item 3 (ø = 3.4) ",
"Item 3 (ø = 3.4) ", "Item 3 (ø =
3.4) ", "Item 3 (ø = 3.4) ",
"Item 3 (ø = 3.4) ", "Item 4 (ø =
3.4) ", "Item 4 (ø = 3.4) ",
"Item 4 (ø = 3.4) ", "Item 4 (ø =
3.4) ", "Item 4 (ø = 3.4) ",
"Item 4 (ø = 3.4) ", "Item 5 (ø =
3.5) ", "Item 5 (ø = 3.5) ",
"Item 5 (ø = 3.5) ", "Item 5 (ø =
3.5) ", "Item 5 (ø = 3.5) ",
"Item 5 (ø = 3.5) ", "Item 6 (ø =
3.5) ", "Item 6 (ø = 3.5) ",
"Item 6 (ø = 3.5) ", "Item 6 (ø =
3.5) ", "Item 6 (ø = 3.5) ",
"Item 6 (ø = 3.5) ", "Item 7 (ø =
3.4) ", "Item 7 (ø = 3.4) ",
"Item 7 (ø = 3.4) ", "Item 7 (ø =
3.4) ", "Item 7 (ø = 3.4) ",
"Item 7 (ø = 3.4) ", "Item 8 (ø =
3.3) ", "Item 8 (ø = 3.3) ",
"Item 8 (ø = 3.3) ", "Item 8 (ø =
3.3) ", "Item 8 (ø = 3.3) ",
"Item 8 (ø = 3.3) "), value =
structure(c(1L, 2L, 3L, 4L, 5L,
6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L,
4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L,
2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L), .Label = c("1 = very
satisfied",
"2", "3",
"4", "5", "6 = very dissatified"), class = "factor"),
n = c(14L, 20L, 24L, 14L, 16L, 14L, 9L, 15L,
21L, 20L, 14L,
23L, 19L, 17L, 16L, 14L, 16L, 20L, 22L,
17L, 15L, 16L, 20L,
12L, 19L, 15L, 16L, 15L, 18L, 19L, 18L,
15L, 18L, 18L, 16L,
17L, 17L, 20L, 17L, 17L, 14L, 16L, 16L,
25L, 16L, 17L, 8L,
20L)), .Names = c("variable", "value",
"n"), row.names =
c(NA,
-48L), vars = list("variable"), drop = TRUE,
indices =
list(0:5,
6:11, 12:17, 18:23, 24:29, 30:35, 36:41,
42:47),
group_sizes = c(6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L),
biggest_group_size = 6L,
labels = structure(list(
"variable" = structure(1:8, .Label = c("Item 1 (ø
= 3.3) ",
"Item 2 (ø =
3.8) ", "Item 3 (ø = 3.4) ", "Item 4 (ø = 3.4) ",
"Item 5 (ø =
3.5) ", "Item 6 (ø = 3.5) ", "Item 7 (ø = 3.4) ",
"Item 8 (ø =
3.3) "), class = "factor")),
row.names = c(NA,
-8L), class = "data.frame", vars =
list("variable"),
drop = TRUE, .Names = "variable"),
class = c("grouped_df",
"tbl_df", "tbl", "data.frame"))
ggplot(
d_result,
aes(x = variable, y = n, fill = rev(factor(value +
geom_bar(
stat = "identity") +
coord_cartesian(ylim = c(0,100)) +
coord_flip() +
scale_y_continuous(name = "Percent") +
scale_fill_manual(
values = rev(
c(
"forestgreen", "limegreen",
"gold", "orange1",
"tomato3", "darkred"))) +
ggtitle(
paste(
"Question 8: Satisfaction?")) +
labs(fill = "Rating") +
scale_x_discrete(
name = element_blank()) +
# scale_color_manual(
# values = rev(
# c(
# "forestgreen", "limegreen",
# "gold", "orange1",
# "tomato3", "darkred"))) +
geom_text(
aes(label = n),
color = "white",
position = position_stack(vjust = 0.5)) +
theme_minimal() +
theme(
legend.position = "right")
-- cut --
I tried to change the order of the items on the y-axis, e.g. Item 8
should be last and Item 1 first. I tried to reverse the order of the items
within ggplot using rev()
[R] Antwort: Re: Antwort: Re: Antwort: Re: Way to Plot Multiple Variables and Change Color (SOLVED)
Hi David,
many thanks for your answer.
I followed your suggesting and came up with the following code:
-- cut --
ggplot(
d_result,
aes(x = variable, y = n, fill = value)) +
geom_bar(
stat = "identity") +
coord_cartesian(ylim = c(0,100)) +
coord_flip() +
scale_y_continuous(name = "Percent") +
scale_fill_manual(
values = rev(
c(
"forestgreen", "limegreen",
"gold", "orange1",
"tomato3", "darkred"))) +
ggtitle(
paste(
"Question 8: Some Text")) +
labs(fill = "Rating") +
scale_x_discrete(
name = element_blank(),
drop = FALSE) + # keep factor levels if no value exists
geom_text(
aes(label = n),
color = "white",
position = position_stack(vjust = 0.5)) +
theme_minimal() +
theme(
legend.position = "right") +
guides(fill = guide_legend(reverse = TRUE))
-- cut --
In addition to your suggestion I changed "fill = rev(factor(value))" to
"fill = value" and I added
guides(fill = guide_legend(reverse = TRUE))
to get the legend in the order from 1 .. 6 instead of 6 .. 1.
In my data I added the counts (n) before the mean value in the labels of
the left hand side. Now it looks to me as a version conforming to the
ESOMAR and BVM standards.
Many thanks again for your help.
Kind regards
Georg
Von:David Winsemius
An: g.maub...@weinwolf.de,
Kopie: r-help@r-project.org
Datum: 10.04.2017 22:21
Betreff:Re: [R] Antwort: Re: Antwort: Re: Way to Plot Multiple
Variables and Change Color
> On Apr 10, 2017, at 1:06 PM, David Winsemius
wrote:
>
>
>> On Apr 10, 2017, at 7:45 AM, g.maub...@weinwolf.de wrote:
>>
>> Hi Ulrik,
>>
>> many thanks for your reply. I had to take an unplanned break and was
not
>> in the office during the last two weeks. Thus my late reply.
>>
>> I followed your advice and converted the variable in argument "fill" to
>> factor. Now the color change works:
>>
>> -- cut --
>>
>> d_result <- structure(list("variable" = c("Item 1 (ø = 3.3) ", "Item 1
(ø = 3.3) ",
>> "Item 1 (ø = 3.3) ", "Item 1 (ø =
3.3) ", "Item 1 (ø = 3.3) ",
>> "Item 1 (ø = 3.3) ", "Item 2 (ø =
3.8) ", "Item 2 (ø = 3.8) ",
>> "Item 2 (ø = 3.8) ", "Item 2 (ø =
3.8) ", "Item 2 (ø = 3.8) ",
>> "Item 2 (ø = 3.8) ", "Item 3 (ø =
3.4) ", "Item 3 (ø = 3.4) ",
>> "Item 3 (ø = 3.4) ", "Item 3 (ø =
3.4) ", "Item 3 (ø = 3.4) ",
>> "Item 3 (ø = 3.4) ", "Item 4 (ø =
3.4) ", "Item 4 (ø = 3.4) ",
>> "Item 4 (ø = 3.4) ", "Item 4 (ø =
3.4) ", "Item 4 (ø = 3.4) ",
>> "Item 4 (ø = 3.4) ", "Item 5 (ø =
3.5) ", "Item 5 (ø = 3.5) ",
>> "Item 5 (ø = 3.5) ", "Item 5 (ø =
3.5) ", "Item 5 (ø = 3.5) ",
>> "Item 5 (ø = 3.5) ", "Item 6 (ø =
3.5) ", "Item 6 (ø = 3.5) ",
>> "Item 6 (ø = 3.5) ", "Item 6 (ø =
3.5) ", "Item 6 (ø = 3.5) ",
>> "Item 6 (ø = 3.5) ", "Item 7 (ø =
3.4) ", "Item 7 (ø = 3.4) ",
>> "Item 7 (ø = 3.4) ", "Item 7 (ø =
3.4) ", "Item 7 (ø = 3.4) ",
>> "Item 7 (ø = 3.4) ", "Item 8 (ø =
3.3) ", "Item 8 (ø = 3.3) ",
>> "Item 8 (ø = 3.3) ", "Item 8 (ø =
3.3) ", "Item 8 (ø = 3.3) ",
>> "Item 8 (ø = 3.3) "), value =
>> structure(c(1L, 2L, 3L, 4L, 5L,
>> 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L,
>> 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L,
>> 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L), .Label = c("1 = very
>> satisfied",
>> "2", "3",
>> "4", "5", "6 = very dissatified"), class = "factor"),
>> n = c(14L, 20L, 24L, 14L, 16L, 14L, 9L, 15L,
>> 21L, 20L, 14L,
>>23L, 19L, 17L, 16L, 14L, 16L, 20L, 22L,
>> 17L, 15L, 16L, 20L,
>>12L, 19L, 15L, 16L, 15L, 18L, 19L, 18L,
>> 15L, 18L, 18L, 16L,
>>17L, 17L, 20L, 17L, 17L, 14L, 16L, 16L,
>> 25L, 16L, 17L, 8L,
>>20L)), .Names = c("variable", "value",
>> "n"), row.names =
>> c(NA,
>> -48L), vars = list("variable"), drop = TRUE,
>> indices =
>> list(0:5,
>>6:11, 12:17, 18:23, 24:29, 30:35, 36:41,
>> 42:47),
>> group_sizes = c(6L,
>> 6L, 6L, 6L, 6L, 6L, 6L, 6L),
>> biggest_group_size = 6L,
>> labels = st
[R] ggplot2: ..n.. and ..count.. in geom_text
Hi All,
I have the following code:
-- cut
(g03_02_p02 <- ggplot(data = d_kzb_input) +
geom_bar(
mapping = aes(x = v03_02_r01, y = round(..prop.. * 100, 0)),
fill = c_ww_palette["blue"]) +
scale_y_continuous(limits = c(0, c_y_limit)) +
theme_classic() +
ggtitle(paste0("Question 3",
"(n = ", <>, ")")) + # How can I refer to the number of cases
for this plot? Is there something like "..n.."?
xlab("Orders") +
ylab("Percent") +
geom_text(
aes(label = ..count..), # How can I refer to the counts for the
labels of the columns?
color = "white",
position = position_stack(vjust = 0.5)))
-- cut --
I would like to refer to the internal statistics of the geom_bar():
How can I refer to the number of cases for this plot? Is there something
like "..n.."?
How can I refer to the counts for the labels of the columns?
Kind regards
Georg
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Follow-up: RStudio: Place for Storing Options (as plain text)
Hi All, some time ago I asded a question about the places where RStudio stores it configuration information. I came across this posting https://support.rstudio.com/hc/en-us/articles/206382178?version=1.0.136&mode=desktop explaining RStudio keybindings (predefined and customized). At the end of the article is the information that RStudio stores keybindings in ~/.R/rstudio/keybindings/rstudio_commands.json ~/.R/rstudio/keybindings/editor_commands.json I want to share this with you. Kind regards Georg - Weitergeleitet von Georg Maubach/WWBO/WW/HAW am 19.04.2017 10:10 - Von:Georg Maubach/WWBO/WW/HAW An: R-help mailing list , Kopie: Martin Maechler , Jeff Newmiller Datum: 08.03.2017 08:59 Betreff:Follow-up: [R] RStudio: Place for Storing Options (as plain text) Hi All, I got a late reply from RStudio Support concerning the question where RStudio store options and configurations: -- cut -- The post RStudio Config Files has a new comment. . . . Unfortunately, it's unlikely that we'll be able to provide a programmatic R interface in the near future -- the way we lay out and store RStudio's client state does not make it as amenable to public consumption as we might hope. That said, you can generally copy everything within that folder to a new machine (at the same relative path from the user home directory), and expect preferences to be respected + restored as you might expect. . . . --cut -- The result of the discussion is: We can copy the complete RStudio directory for storing options and configurations under %localappdata%\RStudio-Desktop or C:\Users\\AppData\Local\RStudio-Desktop and copy it completely to a new installation of RStudio. A programmatic approach to edit RStudio options and configurations is not possible due to design decisions. The purpose of the initial question was to find a way to save RStudio options and configurations, e g. on git/github or similar. This is possible by initialising the above given directory with git or similar. An open question is what happens if a new RStudio release makes changes to the options and configurations. If the stored directory can be completely used would need additional clearification, i.e. for each new version. Kind regards Georg Von:Martin Maechler An: Kopie: , Datum: 23.02.2017 08:37 Betreff:Re: [R] RStudio: Place for Storing Options > Jeff Newmiller > on Sat, 11 Feb 2017 08:09:36 -0800 writes: > For the record, then, Google listened to my incantation of > "rstudio configuration file" and the second result was: > https://support.rstudio.com/hc/en-us/articles/200534577-Resetting-RStudio-Desktop-s-State > RStudio Desktop is also open source, so you can download > the source code and look at the operating-system-specific > bits (for "where") if the above link goes out of date or > disappears. Thanks a lot, Jeff! And for the archives: On reasonable OS's, the hidden directory/folder containing all the info is ~/.rstudio-desktop/ and if "things are broken" the recommendation is to rename that mv ~/.rstudio-desktop ~/backup-rstudio-desktop and (zip and) send along with your e-mail to the experts for diagnosis. > On Thu, 9 Feb 2017, Martin Maechler wrote: >> >>> Ulrik Stervbo on Thu, 9 >>> Feb 2017 14:37:57 + writes: >> >> > Hi Georg, > maybe someone here knows, but I think you >> are more likely to get answers to > Rstudio related >> questions with RStudio support: > >> https://support.rstudio.com/hc/en-us >> >> > Best, > Ulrik >> >> Indeed, thank you, Ulrik. >> >> In this special case, however, I'm quite sure many >> readers of R-help would be interested in the answer; so >> once you receive an answer, please post it (or a link to >> a public URL with it) here on R-help, thank you in >> advance. >> >> We would like to be able to *save*, or sometimes *set* / >> *reset* such options "in a scripted manner", e.g. for >> controlled exam sessions. >> >> Martin Maechler, ETH Zurich >> >> > On Thu, 9 Feb 2017 at 12:35 >> wrote: >> >> >> Hi All, >> I would like to make a backup of my RStudio >> IDE options I configure using >> "Tools/Global Options" >> from the menu bar. Searching the >> web did not reveal >> anything. >> >> >> Can you tell me where RStudio IDE does store its >> configuration? >> >> >> Kind regards >> Georg >> >> __ >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and >> more, see https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html and provide >> commented, minimal, self-contained, reproducible code. >> > -
[R] Multiple-Response Analysis: Cleaning of Duplicate Codes
Hi All,
in my current project I am working with multiple-response questions
(MRSets):
-- Coding --
100 Main Code 1
110 Sub Code 1.1
120 Sub Code 1.2
130 Sub Code 1.3
200 Main Code 2
210 Sub Code 2.1
220 Sub Code 2.2
230 Sub Code 2.3
300 Main Code 3
310 Sub Code 3.1
320 Sub Code 3.2
The coding for the variables is to detailed. Therefore I have recoded all
sub codes to the respective main code, e.g. all 110, 120 and 130 to 100,
all 210, 220 and 230 to 200 and all 310, 320 and 330 to 300.
Now it happens that some respondents get several times the same main code.
If the coding was done for respondent 1 with 120 and 130 after recoding
the values are 100 and 100. If I count this, it would mean that I weight
the multiple values of this respondent by factor 2. This is not my aim. I
would like to count the 100 for the respective respondent only once.
Here is my script so far:
# -- cut --
library(expss)
d_sample <-
structure(
list(
c05_01 = c(
110,
110,
130,
110,
110,
110,
110,
110,
110,
110,
110,
999,
110,
495,
160,
110,
410
),
c05_02 = c(NA,
NA, 120, NA, NA, 150, NA, NA, 170, 160, NA, NA, NA, NA,
170,
NA, 130),
c05_03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 410,
NA, NA, NA, NA, NA, NA, NA),
c05_04 = c(
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_
),
c05_05 = c(
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_
)
),
.Names = c("c05_01",
"c05_02", "c05_03", "c05_04", "c05_05"),
row.names = c(
"1",
"2",
"3",
"4",
"5",
"10",
"11",
"12",
"13",
"14",
"15",
"20",
"21",
"22",
"23",
"24",
"25"
),
class = "data.frame"
)
c05_xx_r01 <- d_sample %>%
select(starts_with("c05_")) %>%
recode(c(
110 %thru% 195 ~ 100,
210 %thru% 295 ~ 200,
310 %thru% 395 ~ 300,
410 %thru% 495 ~ 400,
510 %thru% 595 ~ 500,
810 %thru% 895 ~ 800,
910 %thru% 999 ~ 900))
names(c05_xx_r01) <- paste0("c05_0", 1:5, "_r01")
d_sample <- cbind(d_sample, c05_xx_r01)
# -- cut --
I would like to eliminate all duplicates codes, e. g. 100 and 100 for
respondents in row 3, 6, 13, 14 and 15 to 100 only once:
# -- cut --
d_sample_1 <-
structure(
list(
c05_01 = c(
110,
110,
130,
110,
110,
110,
110,
110,
110,
110,
110,
999,
110,
495,
160,
110,
410
),
c05_02 = c(NA,
NA, 120, NA, NA, 150, NA, NA, 170, 160, NA, NA, NA, NA,
170,
NA, 130),
c05_03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 410,
NA, NA, NA, NA, NA, NA, NA),
c05_04 = c(
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_
),
c05_05 = c(
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_,
NA_real_
),
c05_01_r01 = c(
100,
100,
100,
100,
100,
100,
100,
100,
100,
100,
100,
900,
100,
400,
100,
100,
400
),
c05_02_r01 = c(NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, 100),
c05_03_r01 = c(NA, NA,
NA, NA, NA, NA, NA, NA, NA, 400, NA, NA, NA, NA, NA,
NA, NA),
c05_04_r01 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA),
c05_05_r01 = c(NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)
),
.Names = c(
"c05_01",
"c05_02",
"c05_03",
"c05_04",
"c05_05",
"c05_01_r01",
"c05_02_r01",
"c05_03_r01",
"c05_04_r01
[R] Antwort: Re: Multiple-Response Analysis: Cleaning of Duplicate Codes (SOLVED)
Hi Bert,
many thanks for your reply. I appreciate your help a lot.
I would like to do the operation (= finding the duplicates) row-wise.
During this night a solution showed up in my dreams :) Instead of using
duplicates() to flag and filter the values I could use unique instead with
the same result. I tested:
# -- cut --
apply(X = c05_xx_r01, MARGIN = 1, unique)
# -- cut --
This finds the unique values for each row. That is nice but lacks the
requirement that I need a dataframe with a set of variables back that is
as long as the total amount of unique values for the complete
data.frame/matrix or the amount of variable of the original data.frame
respectively.
The result of the above operation gives a list instead of a data.frame due
to the fact that the amount of resulting values vary from 1 to 7.
Therefore no data.frame but a list is returned.
I search the web for a solution and found:
http://stackoverflow.com/questions/15753091/convert-mixed-length-named-list-to-data-frame
The complete solution would then look like:
# -- cut --
library(stringi)
library(tidyverse)
my_list <- apply(c05_xx_r01, MARGIN = 1, unique)
my_tibble <- as_tibble(stringi::stri_list2matrix(my_list, byrow = TRUE)
# DONE !
# -- cut --
All-in-all thanks again for your help.
Kind regards
Georg
P.S: I had a look into ?unique. The statement "unique(c05_xx_r01, MARGIN =
1) does not do the job, cause this looks for unique combinations of values
on all columns. But that is not the desired outcome.
Von:Bert Gunter
An: g.maub...@weinwolf.de,
Kopie: R-help
Datum: 25.04.2017 19:10
Betreff:Re: [R] Multiple-Response Analysis: Cleaning of Duplicate
Codes
If I understand you correctly, one way is:
> z <- rep(LETTERS[1:3],4)
> z
[1] "A" "B" "C" "A" "B" "C" "A" "B" "C" "A" "B" "C"
> z[!duplicated(z)]
[1] "A" "B" "C"
?duplicated
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue, Apr 25, 2017 at 9:36 AM, wrote:
> Hi All,
>
> in my current project I am working with multiple-response questions
> (MRSets):
>
> -- Coding --
> 100 Main Code 1
> 110 Sub Code 1.1
> 120 Sub Code 1.2
> 130 Sub Code 1.3
>
> 200 Main Code 2
> 210 Sub Code 2.1
> 220 Sub Code 2.2
> 230 Sub Code 2.3
>
> 300 Main Code 3
> 310 Sub Code 3.1
> 320 Sub Code 3.2
>
> The coding for the variables is to detailed. Therefore I have recoded
all
> sub codes to the respective main code, e.g. all 110, 120 and 130 to 100,
> all 210, 220 and 230 to 200 and all 310, 320 and 330 to 300.
>
> Now it happens that some respondents get several times the same main
code.
> If the coding was done for respondent 1 with 120 and 130 after recoding
> the values are 100 and 100. If I count this, it would mean that I weight
> the multiple values of this respondent by factor 2. This is not my aim.
I
> would like to count the 100 for the respective respondent only once.
>
> Here is my script so far:
>
> # -- cut --
>
> library(expss)
>
> d_sample <-
> structure(
> list(
> c05_01 = c(
> 110,
> 110,
> 130,
> 110,
> 110,
> 110,
> 110,
> 110,
> 110,
> 110,
> 110,
> 999,
> 110,
> 495,
> 160,
> 110,
> 410
> ),
> c05_02 = c(NA,
> NA, 120, NA, NA, 150, NA, NA, 170, 160, NA, NA, NA, NA,
> 170,
> NA, 130),
> c05_03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 410,
> NA, NA, NA, NA, NA, NA, NA),
> c05_04 = c(
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_
> ),
> c05_05 = c(
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_
> )
> ),
> .Names = c("c05_01",
>"c05_02", "c05_03", "c05_04", "c05_05"),
> row.names = c(
> "1",
> "2",
> "3",
> "4",
> "5",
> "10",
> "11",
> "12",
> "13",
> "14",
> "15",
> "20",
> "21",
> "22",
> "23",
> "24",
> "25"
> ),
> class = "data.frame"
> )
>
> c05_xx_r01 <- d_sample %>%
> select(starts_with("c05_")) %>%
> recode(c(
> 110 %thru% 195 ~ 100,
> 210 %thru% 295 ~ 200,
> 310 %thru% 395 ~ 300,
> 410 %thru% 495 ~ 400,
> 510 %thru% 595
[R] Antwort: Re: Multiple-Response Analysis: Cleaning of Duplicate Codes (SOLVED)
Hi Bert,
many thanks for your reply. I appreciate your help a lot.
I would like to do the operation (= finding the duplicates) row-wise.
During this night a solution showed up in my dreams :) Instead of using
duplicates() to flag and filter the values I could use unique instead with
the same result. I tested:
# -- cut --
apply(X = c05_xx_r01, MARGIN = 1, unique)
# -- cut --
This finds the unique values for each row. That is nice but lacks the
requirement that I need a dataframe with a set of variables back that is
as long as the total amount of unique values for the complete
data.frame/matrix or the amount of variable of the original data.frame
respectively.
The result of the above operation gives a list instead of a data.frame due
to the fact that the amount of resulting values vary from 1 to 7.
Therefore no data.frame but a list is returned.
I search the web for a solution and found:
http://stackoverflow.com/questions/15753091/convert-mixed-length-named-list-to-data-frame
The complete solution would then look like:
# -- cut --
library(stringi)
library(tidyverse)
my_list <- apply(c05_xx_r01, MARGIN = 1, unique)
my_tibble <- as_tibble(stringi::stri_list2matrix(my_list, byrow = TRUE)
# DONE !
# -- cut --
All-in-all thanks again for your help.
Kind regards
Georg
P.S: I had a look into ?unique. The statement "unique(c05_xx_r01, MARGIN =
1) does not do the job, cause this looks for unique combinations of values
on all columns. But that is not the desired outcome.
Von:Bert Gunter
An: g.maub...@weinwolf.de,
Kopie: R-help
Datum: 25.04.2017 19:10
Betreff:Re: [R] Multiple-Response Analysis: Cleaning of Duplicate
Codes
If I understand you correctly, one way is:
> z <- rep(LETTERS[1:3],4)
> z
[1] "A" "B" "C" "A" "B" "C" "A" "B" "C" "A" "B" "C"
> z[!duplicated(z)]
[1] "A" "B" "C"
?duplicated
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue, Apr 25, 2017 at 9:36 AM, wrote:
> Hi All,
>
> in my current project I am working with multiple-response questions
> (MRSets):
>
> -- Coding --
> 100 Main Code 1
> 110 Sub Code 1.1
> 120 Sub Code 1.2
> 130 Sub Code 1.3
>
> 200 Main Code 2
> 210 Sub Code 2.1
> 220 Sub Code 2.2
> 230 Sub Code 2.3
>
> 300 Main Code 3
> 310 Sub Code 3.1
> 320 Sub Code 3.2
>
> The coding for the variables is to detailed. Therefore I have recoded
all
> sub codes to the respective main code, e.g. all 110, 120 and 130 to 100,
> all 210, 220 and 230 to 200 and all 310, 320 and 330 to 300.
>
> Now it happens that some respondents get several times the same main
code.
> If the coding was done for respondent 1 with 120 and 130 after recoding
> the values are 100 and 100. If I count this, it would mean that I weight
> the multiple values of this respondent by factor 2. This is not my aim.
I
> would like to count the 100 for the respective respondent only once.
>
> Here is my script so far:
>
> # -- cut --
>
> library(expss)
>
> d_sample <-
> structure(
> list(
> c05_01 = c(
> 110,
> 110,
> 130,
> 110,
> 110,
> 110,
> 110,
> 110,
> 110,
> 110,
> 110,
> 999,
> 110,
> 495,
> 160,
> 110,
> 410
> ),
> c05_02 = c(NA,
> NA, 120, NA, NA, 150, NA, NA, 170, 160, NA, NA, NA, NA,
> 170,
> NA, 130),
> c05_03 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 410,
> NA, NA, NA, NA, NA, NA, NA),
> c05_04 = c(
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_
> ),
> c05_05 = c(
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_,
> NA_real_
> )
> ),
> .Names = c("c05_01",
>"c05_02", "c05_03", "c05_04", "c05_05"),
> row.names = c(
> "1",
> "2",
> "3",
> "4",
> "5",
> "10",
> "11",
> "12",
> "13",
> "14",
> "15",
> "20",
> "21",
> "22",
> "23",
> "24",
> "25"
> ),
> class = "data.frame"
> )
>
> c05_xx_r01 <- d_sample %>%
> select(starts_with("c05_")) %>%
> recode(c(
> 110 %thru% 195 ~ 100,
> 210 %thru% 295 ~ 200,
> 310 %thru% 395 ~ 300,
> 410 %thru% 495 ~ 400,
> 510 %thru% 595
[R] Factors and Alternatives
Hi All,
I am using factors in a study for the social sciences.
I discovered the following:
-- cut --
library(dplyr)
test1 <- c(rep(1, 4), rep(0, 6))
d_test1 <- data.frame(test)
test2 <- factor(test1)
d_test2 <- data.frame(test2)
test3 <- factor(test1,
levels = c(0, 1),
labels = c("WITHOUT Contact", "WITH Contact"))
d_test3 <- data.frame(test3)
d_test1 %>% filter(test1 == 0) # works OK
d_test2 %>% filter(test2 == 0) # works OK
d_test3 %>% filter(test3 == 0) # does not work, why?
myf <- function(ds) {
print(levels(ds$test3))
print(labels(ds$test3))
print(as.numeric(ds$test3))
print(as.character(ds$test3))
}
# This showsthat it is not possible to access the original
# values which were the basis to build the factor:
myf(d_test3)
-- cut --
Why is it not possible to use a factor with labels for filtering with the
original values?
Is there a data structure that works like a factor but gives also access
to the original values?
Kind regards
Georg
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Factors and Alternatives
Hi Bob,
many thanks for your reply.
I have read the documentation. In my current project I use "item
batteries" for dimensions of touchpoints which are rated by our customers.
I wrote functions to analyse them. If I create a factor before filtering
and analysing I lose the original values of the variable. If I use the
original variable for filtering and analysis I might happen that for some
dimensions values were not selected. This means they are not NA but none
of the respondents chose "4" for instance on a scale from 1 to 6. That
means that creating a factor from the analysed data with the complete
scale (1:6) fails due the different vector length (amount of remaining
unique values in the analysis vs values in the scale). As I have a
function doing the analysis I am looking for a way to make my function
robust to such circumstances and be able to use it to analyse all "item
batteries". Thus my question. I believe my findings are not odd. Maybe
there is a way dealing with that kind of problems in R and I am eager to
learn how it can be solved using R.
What would you suggest?
Kind regards
Georg
Von:"Bob O'Hara"
An: g.maub...@weinwolf.de,
Kopie: r-help
Datum: 09.05.2017 12:26
Betreff:Re: [R] Factors and Alternatives
That's easy! First
> str(test3)
Factor w/ 2 levels "WITHOUT Contact",..: 2 2 2 2 1 1 1 1 1 1
tells you that the internal values are 1 and 2, and the labels are
"WITHOUT Contact" and "WITH Contact". If you read the help page for
factor() you'll see this:
levels: an optional vector of the values (as character strings) that
‘x’ might have taken. The default is the unique set of
values taken by ‘as.character(x)’, sorted into increasing
order _of ‘x’_. Note that this set can be specified as
smaller than ‘sort(unique(x))’.
labels: _either_ an optional character vector of (unique) labels for
the levels (in the same order as ‘levels’ after removing
those in ‘exclude’), _or_ a character string of length 1.
So, when you create test3 you say that test can take values 0 and 1,
and these should be labelled as "WITHOUT Contact" and "WITH Contact".
So R internally codes "1" as 1 and "0" as 2 (internally R codes
factors as integers, which can be both useful and dangerous), and then
gives them labels "WITHOUT Contact" and "WITH Contact". It now doesn't
care that they were 1 and 0, because you've told it to change the
labels.
If you want to filter by the original values, then don't change the
labels (or at least not until after you've filtered by the original
labels), or convert the filter to the new labels. You're asking for a
data structure with two sets of labels, which sounds odd in general.
Bob
On 9 May 2017 at 12:12, wrote:
> Hi All,
>
> I am using factors in a study for the social sciences.
>
> I discovered the following:
>
> -- cut --
>
> library(dplyr)
>
> test1 <- c(rep(1, 4), rep(0, 6))
> d_test1 <- data.frame(test)
>
> test2 <- factor(test1)
> d_test2 <- data.frame(test2)
>
> test3 <- factor(test1,
> levels = c(0, 1),
> labels = c("WITHOUT Contact", "WITH Contact"))
> d_test3 <- data.frame(test3)
>
> d_test1 %>% filter(test1 == 0) # works OK
> d_test2 %>% filter(test2 == 0) # works OK
> d_test3 %>% filter(test3 == 0) # does not work, why?
>
> myf <- function(ds) {
> print(levels(ds$test3))
> print(labels(ds$test3))
> print(as.numeric(ds$test3))
> print(as.character(ds$test3))
> }
>
> # This showsthat it is not possible to access the original
> # values which were the basis to build the factor:
> myf(d_test3)
>
> -- cut --
>
> Why is it not possible to use a factor with labels for filtering with
the
> original values?
> Is there a data structure that works like a factor but gives also access
> to the original values?
>
> Kind regards
>
> Georg
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Bob O'Hara
NOTE NEW ADDRESS!!!
Institutt for matematiske fag
NTNU
7491 Trondheim
Norway
Mobile: +49 1515 888 5440
Journal of Negative Results - EEB: www.jnr-eeb.org
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: RE: Antwort: Re: Factors and Alternatives (SOLVED)
Hi David,
Hi Bob,
many thanks for your help.
Your solution - just to use all levels instead of just the one's found in
the data - helped.
The original code looked like this:
-- cut --
c_v10_val_labs <- c(
"1 = sehr gut",
"2", "3", "4", "5",
"6 = sehr schlecht"
)
# where c_v10_val_labs is handed over to my function as "val_labs".
ds_results$value <- factor(ds_results$value,
levels = sort(unique(ds_results$value)), #
old code
labels = sort(unique(val_labs)))
-- cut --
If I write instead
-- cut --
ds_results$value <- factor(ds_results$value,
levels = seq_along(val_labs), # new code 1st
version
labels = sort(unique(val_labs)))
-- cut --
Your solution builds a factor with all factor levels even if a value for
factor is not present (not NA, but does just not occur in the data, i.e.
not stated by any respondent).
In Zumel's book "Practical Data Science with R" (
https://www.amazon.de/Practical-Data-Science-Nina-Zumel/dp/1617291560),
Shelter Island: Manning, 2014, p. 23-24, Listing 2-5, a mapping using
subscripts is described:
-- cut --
mapping <- list(
'A40'='car (new)',
'A41'='car (used)',
'A42'='furniture/equipment',
'A43'='radio/television',
'A44'='domestic appliances',
...
)
for(i in 1:(dim(d))[2]) {
if(class(d[,i])=='character') {
d[,i] <- as.factor(as.character(mapping[d[,i]]))
}
}
-- cut -
Simple stated this would mean:
-- cut --
val_labs <- list(
"1" = "1 = sehr gut",
"2" = "2",
"3" = "3",
"4" = "4",
"5" = "5",
"6" = "6 = sehr schlecht"
)
set.seed(12345)
answers = c(sample(1:5, 10, replace = TRUE))
test <- factor(unlist(val_labs[answers]))
# or just
val_labs <- c(
"1 = sehr gut",
"2",
"3",
"4",
"5",
"6 = sehr schlecht"
)
set.seed(12345)
answers = c(sample(1:5, 10, replace = TRUE))
test <- val_labs[answers]
-- cut --
Adapting this to my code would give:
-- cut --
ds_results$value <- factor(ds_results$value,
levels = sort(unique(ds_results$value)),
labels =
val_labs[sort(unique(ds_results$value))]) # new code 2nd version
-- cut --
This results in a factor just as long as the vector of unique resulting
values.
Both solutions work. Which version is best depends on the overall process
and the purpose of the code. I document all this for use by readers who
refer later to the list archives.
Using your version and running my code reveals that ggplot runs into
difficulties cause the legend lacks values and the sequence and coloring
of the legend is wrong. But that's another story.
Many thanks again for your help.
Kind regards
Georg
Von:David L Carlson
An: "g.maub...@weinwolf.de" , "Bob O'Hara"
,
Kopie: r-help
Datum: 09.05.2017 14:37
Betreff:RE: [R] Antwort: Re: Factors and Alternatives
I'm not sure I understand your question, but you can easily include all
possible answers when you create the factor by using the levels= argument
as Bob pointed out. Here is an example of values that range from 1 to 6,
but value 3 is not represented. Notice that a factor level 3 is created
even though it does not appear in the data:
> set.seed(42)
> x <- sample.int(6, 10, replace=TRUE)
> table(x)
x
1 2 4 5 6
1 1 3 3 2
> y <- factor(x, levels=1:6)
> y
[1] 6 6 2 5 4 4 5 1 4 5
Levels: 1 2 3 4 5 6
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
Von:"Bob O'Hara"
An: g.maub...@weinwolf.de,
Kopie: r-help
Datum: 09.05.2017 13:58
Betreff:Re: Re: [R] Factors and Alternatives
For the problem you state, would it be enough to explicitly define your
levels?
fac <- rep(c("a", "b", "d"), each=4)
fac.f <- factor(fac, levels=c("a", "b", "c", "d"))
table(fac.f)
# but be warned...
fac.f2 <- factor(fac.f)
table(fac.f2)
This has the advantage that the code explicitly documents what the
possible values are, so if something goes wrong down-stream, you know
it is a real problem (well, unless you have some type conversions
screwing things up). You might also want to do some defensive
programming, and put some checks in the code, to make sure your
factors have the right number of levels.
Bob
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
g.maub...@weinwolf.de
Sent: Tuesday, May 9, 2017 6:37 AM
To: Bob O'Hara
Cc: r-help
Subject: [R] Antwort: Re: Factors and Alternatives
Hi Bob,
many thanks for your reply.
I have read the documentation. In my current project I use "item
batteries" for dimensions of touchpoints which are rated by our customers.
I wrote functions to analyse them. If I create a factor before filtering
and analysing I lose the original values of the variable. If I use the
original variable for filtering and analysis I might happen that for some
dimensions values were not selected. T
[R] Off-Topic: Project Organisation
Hi All, this post is somewhat off-topic cause it deals with a meta issue related to project organisation instead of real R code. I have updated my blog concerning a possible directory and file structure for marketing research projects and data mining projects alike: https://github.com/gmaubach/R-Know-How/wiki/R-Blog There I condensed best practices already communicated in articels, books, packages and guidelines into a new universial structure. It shall serve as a template and construction kit which you can use to create a structure that suits your project best. Comments and suggestions are welcome. Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ggplot: Pie Chart with correct labels
Hi All,
I would like to do the following pie chart using ggplot from an official
data source (
http://www.deutscheweine.de/fileadmin/user_upload/Website/Service/Downloads/Statistik_2016-2017-neu.pdf
, Tab 8, Page 14):
-- cut --
cat("# weinimport_piechart.R\n")
# -- Input
d_wine_import_DE <- structure(list(Land = structure(1:24, .Label =
c("Italien", "Frankreich",
"Spanien", "USA",
"Südafrika", "Chile", "Österreich", "Australien",
"Portugal",
"Griechenland", "Argentinien", "Neuseeland", "Ungarn",
"Mazedonien", "Schweiz",
"Dänemark", "Moldawien", "Türkei", "Belgien/Luxemburg",
"Rumänien", "Ukraine",
"Kroatien", "Israel", "Georgien"), class = "factor"),
Menge_hl_2015 = c(5481000, 2248000, 3824000, 493000,
845000,
539000, 308000, 446000, 153000, 99000,
64000, 43000, 123000,
186000, 5000, 9000, 28000, 7000, 1,
15000, 4000, 4000,
2000, 2000)), .Names = c("Land",
"Menge_hl_2015"), class = "data.frame", row.names = c(NA,
-24L))
names(d_wine_import_DE)
# -- Data -
d_result <- data.frame(
country = d_wine_import_DE$Land,
abs = d_wine_import_DE$Menge_hl_2015) %>%
mutate(rel = round(abs / sum(abs) * 100, 1)) %>%
dplyr::arrange(desc(abs)) %>%
dplyr::mutate(rel_labs = paste(rel, "%")) %>% # rev() does not work
dplyr::mutate(breaks = cumsum(abs) - (abs / 2)) # rev() does not work
# -- Plot -
d_result %>%
ggplot() +
geom_bar(
aes(x = "", y = abs, fill = country),
stat = "identity") +
# %SOURCE%
# coord_polar(): Wickham: ggplot2, Springer, 2nd Ed., p. 166
coord_polar(theta = "y", start = 0) +
guides(
fill = guide_legend(
title = "Länder",
reverse = FALSE)
) +
scale_y_continuous(
breaks = d_result$breaks, # simply "breaks" does not work
labels = d_result$rel_labs, # simply "breaks" does not work
trans = "reverse"
) +
# %SOURCE%
# Kassambra: Guide to Create Beautiful Graphics
# in R, sthda.com, 2nd Ed., 2013, p. 136ff
theme_minimal() +
theme(
panel.border = element_blank(),
panel.grid = element_blank(),
axis.title.x = element_blank(),
axis.title.y = element_blank()
# axis.text.x = element_text(size = 15)
) +
labs(
title = paste0("Weinimport nach Deutschland 2015"))
-- cut --
I can not figure out how to align the labels (values in %) with the
reverse printed countries. Also the breaks and labels do need the dataset
name although I thought "breaks" and "rel_labs" is sufficient due to the
piping operator.
Can you help me by telling how to
1. get the order of the labels right
2. Why I need to reference "breaks" and "labels" completely?
Kind regards
Georg
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[R] purrr::pmap does not work
Hi All,
I try to do a scatterplot for a bunch of variables. I plot a dependent
variable against a bunch of independent variables:
-- cut --
graphics::plot(
v01_r01 ~ v08_01_up11,
data = dataset,
xlab = "Dependent",
ylab = "Independent #1"
)
-- cut --
It is tedious to repeat the statement for all independent variables. Found
an alternative, i.e. :
-- cut --
mu <- list(5, 10, -3)
sigma <- list(1, 5, 10)
n <- list(1, 3, 5)
fargs <- list(mean = mu, sd = sigma, n = n)
fargs %>%
purrr::pmap(rnorm) %>%
str()
-- cut --
I tried to use this for may scatterplot task:
-- cut --
var_battery$v08 <- paste0("v08_", formatC(1:8, width = 2, format = "d",
flag = "0"))
v08_var_labs <- paste0("Label_", 1:8)
dataset <- as.data.frame(
matrix(
data = sample(
x = 1:11,
size = 90,
replace = TRUE),
nrow = 10,
ncol = 9))
names(dataset) <- c("v01_r01", var_battery$v08)
independent <- as.list(dataset$v01_r01)
dependent <- as.list(dataset[var_battery$v08])
fargs <- list(
x = independent,
y = dependent,
ylab = v08_var_labs)
fargs %>%
purrr::pmap(
function(d = dataset, xvalue = x, yvalue = y,
xlab = "Label for x variable",
ylab = ylab) {
graphics::plot(
xvalue ~ yvalue,
data = d,
xlab = xlab,
ylab = ylab)
}
)
-- cut --
The last statement comes back with
Error: Element 2 has length 8, not 1 or 10.
How can I get it up n running? Do you suggest a better solution for the
task described?
Kind regards
Georg
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[R] Paths in knitr
Hi All,
I have to compile a report for the management and decided to use RMarkdown
and knitr. I compiled all needed plots (using separate R scripts) before
compiling the report, thus all plots reside in my graphics directory. The
RMarkdown report needs to access these files. I have defined
```{r setup, include = FALSE}
knitr::opts_knit$set(
echo = FALSE,
xtable.type = "html",
base.dir = "H:/2017/Analysen/Kundenzufriedenheit/Auswertung",
root_dir = "H:/2017/Analysen/Kundenzufriedenheit/Auswertung",
fig.path = "results/graphics") # relative path required, see
http://yihui.name/knitr/options
```
and then referenced my plot using
because I want to be able to customize the plotting attributes.
But that fails with the message "pandoc.exe: Could not fetch
email_distribution_pie.png".
If I give it the absolute path
"H:/2017/Analysen/Kundenzufriedenheit/Auswertung/results/graphics/email_distribution_pie.png"
it works fine as well if I copy the plot into the directory where the
report.RMD file resides.
How can I tell knitr to fetch the ready-made plots from the graphics
directory?
Kind regards
Georg
__
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Re: [R] Paths in knitr
Hi Yihui,
Hi Duncan,
I corrected my typo. Unfortunately knitr did not find my plots in the
directory where they reside which is different from the Rmd document.
The documentation of knitr says:
base.dir: (NULL) an absolute directory under which the plots are generate
root.dir: (NULL) the root directory when evaluating code chunks; if NULL,
the directory of the input document will be used
>From that description I thought, if the base.dir can be used for writng
plots, it is then also used for reading plots if set? No, it is not.
If I set the root directory to the plots/graphics directory will knitr
then find my plots? No, it does not.
Reading blog posts my thoughts looked not so strange to me, e.g.
https://philmikejones.wordpress.com/2015/05/20/set-root-directory-knitr/.
Unfortunately, it does not work for me.
I am using a RStudio project file. Could it be that this interferes which
the knitr options?
I tried the solution that Duncan suggested:
c_path_plots <-
"H:/2017/Analysen/Kundenzufriedenheit/Auswertung/results/graphics
`r knitr::include_graphics(file.path(c_path_plots,
"email_distribution_pie.png"))`
This solution works fine. I will go with it for this project as I have to
finish my report soon.
I read Hadley's book on bulding R Packages (
https://www.amazon.de/R-Packages-Hadley-Wickham/dp/1491910593) and found
it quite complicated and time consuming to build one. Thus I did not try
yet to build my own packages. At the end of last week I heard from another
library (http://reaktanz.de/R/pckg/roxyPackage/) which shall make building
packages much easier. I plan to try that shortly.
On my path to become better in analytics using R, I will try to use
modules of Rmd files which can then easily be integrated into a Rmd
report. I have yet to see how I can include these file into a complete
report.
Kind regards
Georg
- Weitergeleitet von Georg Maubach/WWBO/WW/HAW am 12.06.2017 08:47
-
Von:Yihui Xie
An: g.maub...@gmx.de,
Kopie: R Help
Datum: 09.06.2017 20:53
Betreff:Re: [R] Paths in knitr
Gesendet von: "R-help"
I'd say it is an expert-only option. If you do not understand what it
means, I strongly recommend you not to set it.
Similarly, you set the root_dir option and I don't know why you did it,
but
it is a typo anyway (should be root.dir).
Regards,
Yihui
--
https://yihui.name
On Fri, Jun 9, 2017 at 4:50 AM, wrote:
> Hi Yi,
>
> many thanks for your reply.
>
> Why I do have to se the base.dir option? Cause, to me it is not clear
from
> the documentation, where knitr looks for data files and how I can adjust
> knitr to tell it where to look. base.dir was a try, but did not work.
>
> Can you give me a hint where I can find information/documentation on
this
> path issue?
>
> Kind regards
>
> Georg
>
>
> > Gesendet: Donnerstag, 08. Juni 2017 um 15:05 Uhr
> > Von: "Yihui Xie"
> > An: g.maub...@weinwolf.de
> > Cc: "R Help"
> > Betreff: Re: [R] Paths in knitr
> >
> > Why do you have to set the base.dir option?
> >
> > Regards,
> > Yihui
> > --
> > https://yihui.name
> >
> >
> > On Thu, Jun 8, 2017 at 6:15 AM, wrote:
> > > Hi All,
> > >
> > > I have to compile a report for the management and decided to use
> RMarkdown
> > > and knitr. I compiled all needed plots (using separate R scripts)
> before
> > > compiling the report, thus all plots reside in my graphics
directory.
> The
> > > RMarkdown report needs to access these files. I have defined
> > >
> > > ```{r setup, include = FALSE}
> > > knitr::opts_knit$set(
> > > echo = FALSE,
> > > xtable.type = "html",
> > > base.dir = "H:/2017/Analysen/Kundenzufriedenheit/Auswertung",
> > > root_dir = "H:/2017/Analysen/Kundenzufriedenheit/Auswertung",
> > > fig.path = "results/graphics") # relative path required, see
> > > http://yihui.name/knitr/options
> > > ```
> > >
> > > and then referenced my plot using
> > >
> > >
> > >
> > > because I want to be able to customize the plotting attributes.
> > >
> > > But that fails with the message "pandoc.exe: Could not fetch
> > > email_distribution_pie.png".
> > >
> > > If I give it the absolute path
> > > "H:/2017/Analysen/Kundenzufriedenheit/Auswertung/results/
> graphics/email_distribution_pie.png"
> > > it works fine as well if I copy the plot into the directory where
the
> > > report.RMD file resides.
> > >
> > > How can I tell knitr to fetch the ready-made plots from the graphics
> > > directory?
> > >
> > > Kind regards
> > >
> > > Georg
>
[[alternative HTML version deleted]]
__
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[R] Antwort: Re: Re: Paths in knitr
Hi Yihui,
I took root.dir and base.dir out. Everything works fine despite the
change.
I have implemented the solution Duncun suggested. I have difficulties with
the scaling / image size in my report. Some plots are too big, some are
too small. I need to adjust any plot. Steep learning curve :)
Kind regards
Georg
Von:Yihui Xie
An: g.maub...@weinwolf.de,
Kopie: Duncan Murdoch , R Help
Datum: 12.06.2017 18:29
Betreff:Re: Re: [R] Paths in knitr
Gesendet von: xieyi...@gmail.com
Will there be anything wrong if you do not set these options?
Regards,
Yihui
--
https://yihui.name
On Mon, Jun 12, 2017 at 2:24 AM, wrote:
> Hi Yihui,
> Hi Duncan,
>
> I corrected my typo. Unfortunately knitr did not find my plots in the
> directory where they reside which is different from the Rmd document.
>
> The documentation of knitr says:
>
> base.dir: (NULL) an absolute directory under which the plots are
generate
> root.dir: (NULL) the root directory when evaluating code chunks; if
NULL,
> the directory of the input document will be used
>
> From that description I thought, if the base.dir can be used for writng
> plots, it is then also used for reading plots if set? No, it is not.
> If I set the root directory to the plots/graphics directory will knitr
> then find my plots? No, it does not.
>
> Reading blog posts my thoughts looked not so strange to me, e.g.
> https://philmikejones.wordpress.com/2015/05/20/set-root-directory-knitr/
.
> Unfortunately, it does not work for me.
>
> I am using a RStudio project file. Could it be that this interferes
which
> the knitr options?
>
> I tried the solution that Duncan suggested:
>
> c_path_plots <-
> "H:/2017/Analysen/Kundenzufriedenheit/Auswertung/results/graphics
>
> `r knitr::include_graphics(file.path(c_path_plots,
> "email_distribution_pie.png"))`
>
> This solution works fine. I will go with it for this project as I have
to
> finish my report soon.
>
> I read Hadley's book on bulding R Packages (
> https://www.amazon.de/R-Packages-Hadley-Wickham/dp/1491910593) and found
> it quite complicated and time consuming to build one. Thus I did not try
> yet to build my own packages. At the end of last week I heard from
another
> library (http://reaktanz.de/R/pckg/roxyPackage/) which shall make
building
> packages much easier. I plan to try that shortly.
>
> On my path to become better in analytics using R, I will try to use
> modules of Rmd files which can then easily be integrated into a Rmd
> report. I have yet to see how I can include these file into a complete
> report.
>
> Kind regards
>
> Georg
>
>
> - Weitergeleitet von Georg Maubach/WWBO/WW/HAW am 12.06.2017 08:47
> -
>
> Von:Yihui Xie
> An: g.maub...@gmx.de,
> Kopie: R Help
> Datum: 09.06.2017 20:53
> Betreff:Re: [R] Paths in knitr
> Gesendet von: "R-help"
>
>
>
> I'd say it is an expert-only option. If you do not understand what it
> means, I strongly recommend you not to set it.
>
> Similarly, you set the root_dir option and I don't know why you did it,
> but
> it is a typo anyway (should be root.dir).
>
> Regards,
> Yihui
> --
> https://yihui.name
>
> On Fri, Jun 9, 2017 at 4:50 AM, wrote:
>
>> Hi Yi,
>>
>> many thanks for your reply.
>>
>> Why I do have to se the base.dir option? Cause, to me it is not clear
> from
>> the documentation, where knitr looks for data files and how I can
adjust
>> knitr to tell it where to look. base.dir was a try, but did not work.
>>
>> Can you give me a hint where I can find information/documentation on
> this
>> path issue?
>>
>> Kind regards
>>
>> Georg
>>
>>
>> > Gesendet: Donnerstag, 08. Juni 2017 um 15:05 Uhr
>> > Von: "Yihui Xie"
>> > An: g.maub...@weinwolf.de
>> > Cc: "R Help"
>> > Betreff: Re: [R] Paths in knitr
>> >
>> > Why do you have to set the base.dir option?
>> >
>> > Regards,
>> > Yihui
>> > --
>> > https://yihui.name
>> >
>> >
>> > On Thu, Jun 8, 2017 at 6:15 AM, wrote:
>> > > Hi All,
>> > >
>> > > I have to compile a report for the management and decided to use
>> RMarkdown
>> > > and knitr. I compiled all needed plots (using separate R scripts)
>> before
>> > > compiling the report, thus all plots reside in my graphics
> directory.
>> The
>> > > RMarkdown report needs to access these files. I have defined
>> > >
>> > > ```{r setup, include = FALSE}
>> > > knitr::opts_knit$set(
>> > > echo = FALSE,
>> > > xtable.type = "html",
>> > > base.dir = "H:/2017/Analysen/Kundenzufriedenheit/Auswertung",
>> > > root_dir = "H:/2017/Analysen/Kundenzufriedenheit/Auswertung",
>> > > fig.path = "results/graphics") # relative path required, see
>> > > http://yihui.name/knitr/options
>> > > ```
>> > >
>> > > and then referenced my plot using
>> > >
>> > >
>> > >
>> > > because I want to be able to customize the plotting attributes.
>> > >
>> > > But that fails with the message "pandoc.exe: Could not fetch
>> > > email_distribution_pie.png".
>> > >
>> > > If I give it the absolute path
>> > > "H:/2017
[R] WG: Fw: Re: rmarkdown and font size
Hi Dan,
Hi All,
I read the below post. I am wondering how do I know which "keys" are
available, e.g. "core.r" and "pre". Where kind I find the definition of
what can be adjusted and which "words" to use?
Kind regards
Georg
> Gesendet: Donnerstag, 08. Juni 2017 um 16:16 Uhr
> Von: "Nordlund, Dan (DSHS/RDA)"
> An: "MacQueen, Don" , "r-help@r-project.org"
> Betreff: Re: [R] rmarkdown and font size
>
> You can change the style, modifying a variety of things. E.g,
>
> ---
> title: Test
> ---
>
>
>
> body{ /* Normal */
> font-size: 12px;
> }
> td { /* Table */
> font-size: 8px;
> }
> h1.title {
> font-size: 38px;
> color: DarkRed;
> }
> h1 { /* Header 1 */
> font-size: 28px;
> color: DarkBlue;
> }
> h2 { /* Header 2 */
> font-size: 22px;
> color: DarkBlue;
> }
> h3 { /* Header 3 */
> font-size: 18px;
> font-family: "Times New Roman", Times, serif;
> color: DarkBlue;
> }
> code.r{ /* Code block */
> font-size: 12px;
> }
> pre { /* Code block - determines code spacing between lines */
> font-size: 14px;
> }
>
>
> Here is some normal text. It is a 12-point font. The table is in
8-point .
>
> ```{r example, echo=FALSE, results='asis'}
> tmp <- data.frame(a=1:5, b=letters[1:5])
> print( knitr::kable(tmp, row.names=FALSE))
> ```
>
>
> Hope this is helpful,
>
> Dan
>
> Daniel Nordlund, PhD
> Research and Data Analysis Division
> Services & Enterprise Support Administration
> Washington State Department of Social and Health Services
>
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> > MacQueen, Don
> > Sent: Wednesday, June 07, 2017 4:58 PM
> > To: r-help@r-project.org
> > Subject: [R] rmarkdown and font size
> >
> > Suppose I have a file (named "tmp.rmd") containing:
> >
> >
> > ---
> > title: Test
> > ---
> >
> > ```{r example, echo=FALSE, results='asis'}
> > tmp <- data.frame(a=1:5, b=letters[1:5])
> > print( knitr::kable(tmp, row.names=FALSE))
> > ```
> >
> >
> >
> > And I render it with:
> >
> > rmarkdown::render('tmp.rmd',
> > output_format=c('html_document','pdf_document'))
> >
> > I get two files:
> > tmp.pdf
> > tmp.html
> >
> > Is there a way to control (change or specify) the font size of the
table in the
> > pdf output?
> > (or of the entire document, if it can't be changed for just the table)
> >
> > With my actual data, the table is too wide to fit on a page in the pdf
output;
> > perhaps if I reduce the font size I can get it to fit.
> >
> > I would like the html version to still look decent, but I don't care
very much
> > what happens to its font size.
> >
> > Thanks!
> > -Don
> >
> > --
> > Don MacQueen
> >
> > Lawrence Livermore National Laboratory
> > 7000 East Ave., L-627
> > Livermore, CA 94550
> > 925-423-1062
> >
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-
> > guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Filtering String Variables
# Hi All,
#
# I have the following data frame (example):
Debitor <- c("968691", "968691", "968691",
"A04046", "A04046",
"L0006", "L0006", "L0006",
"L0023", "L0023",
"L0056", "L0056",
"L0094", "L0094", "L0094",
"L0124", "L0124",
"L0143",
"L0170",
"13459",
"473908",
"394704",
"4711",
"4712",
"4713")
Debitor <- as.character(Debitor)
var1 <- c(11, 12, 13,
14, 14,
12, 13, 14,
10, 11,
12, 12,
12, 12, 12,
15, 17,
11,
14,
12,
17,
13,
15,
16,
11)
ds_example <- data.frame(Debitor, var1)
ds_example$case_id <- 1:nrow(ds_example)
ds_example <- ds_example[, sort(colnames(ds_example))]
ds_example
# I would like to generate a data frame that contains the duplicates AND
the
# corresponding non-duplicates to the duplicates.
# For example, finding the duplicates with deliver case 2 and 3 but the
list
# should also contain case 1 because case 1 is the corresponding case to
the
# duplicate cases 2 and 3.
# For the whole example dataset that would be:
needed <- c(1, 1, 1,
1, 1,
1, 1, 1,
1, 1,
1, 1,
1, 1, 1,
1, 1,
0, 0, 0, 0, 0, 0, 0, 0)
needed <- as.logical(needed)
ds_example <- data.frame(ds_example, needed)
ds_example
# To find the duplicates and the corresponding non-duplicates
duplicates <- duplicated(ds_example$Debitor)
list_of_duplicated_debitors <- as.character(ds_example[duplicates,
"Debitor"])
filter_variable <- unique(list_of_duplicated_debitors)
ds_duplicates <- ds_example["Debitor" == filter_variable] # Result:
dataset with 0 columns
ds_duplicates <- ds_example["Debitor"] %in% filter_variable # Result:
FALSE
# How can I create a dataset like this
ds_example <- ds_example[needed, ]
ds_example
# using the Debitor IDs?
Kind regards
Georg Maubach
__
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and provide commented, minimal, self-contained, reproducible code.
[R] WG: Filtering String Variables (SOLVED)
Hi All,
the solution for my question is as follows
## Filter duplicates and correpsonding non-duplicates
### To filter duplicates and their corresponding non-duplicates use the
### following code snippet:
Debitor <- c("968691", "968691", "968691",
"A04046", "A04046",
"L0006", "L0006", "L0006",
"L0023", "L0023",
"L0056", "L0056",
"L0094", "L0094", "L0094",
"L0124", "L0124",
"L0143",
"L0170",
"13459",
"473908",
"394704",
"4711",
"4712",
"4713")
Debitor <- as.character(Debitor)
var1 <- c(11, 12, 13,
14, 14,
12, 13, 14,
10, 11,
12, 12,
12, 12, 12,
15, 17,
11,
14,
12,
17,
13,
15,
16,
11)
ds_example <- data.frame(Debitor, var1)
ds_example$case_id <- 1:nrow(ds_example)
ds_example <- ds_example[, sort(colnames(ds_example))]
ds_example
# This task is to generate a data frame that contains the duplicates AND
the
# corresponding non-duplicates to the duplicates.
# For example, finding the duplicates will deliver case 2 and 3 but the
list
# should also contain case 1 because case 1 is the corresponding case to
the
# duplicate cases 2 and 3.
# For the whole example dataset that would be:
needed <- c(1, 1, 1,
1, 1,
1, 1, 1,
1, 1,
1, 1,
1, 1, 1,
1, 1,
0, 0, 0, 0, 0, 0, 0, 0)
needed <- as.logical(needed)
ds_example <- data.frame(ds_example, needed)
ds_example
# To find the duplicates and the corresponding non-duplicates
duplicates <- duplicated(ds_example$Debitor)
list_of_duplicated_debitors <- as.character(ds_example[duplicates,
"Debitor"])
filter_variable <- unique(list_of_duplicated_debitors)
### Wrong code. Do not run.
### ds_duplicates <- ds_example["Debitor" == filter_variable] # Result:
dataset with 0 columns
### duplicates_and_correponding_non_duplicates <- ds_example["Debitor"]
%in% filter_variable # Result: FALSE
duplicates_and_correponding_non_duplicates <- ds_example$Debitor %in%
filter_variable # Result: OK
duplicates_and_correponding_non_duplicates <- ds_example[, "Debitor"] %in%
filter_variable # Result: OK
### Create the dataset with duplicates and corresponding non-duplicates
ds_example <- ds_example[duplicates_and_correponding_non_duplicates, ]
ds_example
It was a simple mistake when subscripting.
Kind regards
Georg Maubach
- Weitergeleitet von Georg Maubach/WWBO/WW/HAW am 23.05.2016 15:54
-
Von:Georg Maubach/WWBO/WW/HAW
An: r-help@r-project.org,
Datum: 23.05.2016 15:28
Betreff:Filtering String Variables
# Hi All,
#
# I have the following data frame (example):
Debitor <- c("968691", "968691", "968691",
"A04046", "A04046",
"L0006", "L0006", "L0006",
"L0023", "L0023",
"L0056", "L0056",
"L0094", "L0094", "L0094",
"L0124", "L0124",
"L0143",
"L0170",
"13459",
"473908",
"394704",
"4711",
"4712",
"4713")
Debitor <- as.character(Debitor)
var1 <- c(11, 12, 13,
14, 14,
12, 13, 14,
10, 11,
12, 12,
12, 12, 12,
15, 17,
11,
14,
12,
17,
13,
15,
16,
11)
ds_example <- data.frame(Debitor, var1)
ds_example$case_id <- 1:nrow(ds_example)
ds_example <- ds_example[, sort(colnames(ds_example))]
ds_example
# I would like to generate a data frame that contains the duplicates AND
the
# corresponding non-duplicates to the duplicates.
# For example, finding the duplicates with deliver case 2 and 3 but the
list
# should also contain case 1 because case 1 is the corresponding case to
the
# duplicate cases 2 and 3.
# For the whole example dataset that would be:
needed <- c(1, 1, 1,
1, 1,
1, 1, 1,
1, 1,
1, 1,
1, 1, 1,
1, 1,
0, 0, 0, 0, 0, 0, 0, 0)
needed <- as.logical(needed)
ds_example <- data.frame(ds_example, needed)
ds_example
# To find the duplicates and the corresponding non-duplicates
duplicates <- duplicated(ds_example$Debitor)
list_of_duplicated_debitors <- as.character(ds_example[duplicates,
"Debitor"])
filter_variable <- unique(list_of_duplicated_debitors)
ds_duplicates <- ds_example["Debitor" == filter_variable] # Result:
dataset with 0 columns
ds_duplicates <- ds_example["Debitor"] %in% filter_variable # Result:
FALSE
# How can I create a dataset like this
ds_example <- ds_example[needed, ]
ds_example
# using the Debitor IDs?
Kind regards
Georg Maubach
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[R] Creating a data frame from scratch
Hi All,
I need to create a data frame from scratch and fill variables created on the
fly with values. What I have so far:
-- schnipp --
# Example dataset
gene <-
c("ENSG0208234","ENSG0199674","ENSG0221622","ENSG0207604",
"ENSG0207431","ENSG0221312","ENSG00134940305","ENSG00394039490",
"ENSG09943004048")
hsap <- c(0,0,0, 0, 0, 0, 1,1, 1)
mmul <- c(NA,2 ,3, NA, 2, 1 , NA,2, NA)
mmus <- c(NA,2 ,NA, NA, NA, 2 , NA,3, 1)
rnor <- c(NA,2 ,NA, 1 , NA, 3 , NA,NA, 2)
cfam <- c(NA,2,NA, 2, 1, 2, 2,NA, NA)
ds_example <- data.frame(gene, hsap, mmul, mmus, rnor, cfam)
ds_example$gene <- as.character(ds_example$gene)
t_count_na <- function(dataset,
variables = "all")
# credit:
http://stackoverflow.com/questions/4862178/remove-rows-with-nas-in-data-frame
{
ds_na <- data.frame()
# if variables = "all" create character vector of variable names
if (variables == "all") {
variable_list <- dimnames(dataset)[[ 2 ]]
}
# if a character vector with variable names is given
# to run the function on a defined set of selected variables
else {
variable_list <- variables
}
for (var in variable_list) {
new_name <- paste0("na_", var)
ds_na[[ new_name ]] <- as.data.frame(is.na(dataset[[ var ]]))
}
ds_na[[ "na_count" ]] <- rowSums(ds_na)
return(ds_na)
}
test <- t_count_na(dataset = ds_example, variables = c("mmul", "mmus"))
-- schnipp --
gives:
Error in `[[<-.data.frame`(`*tmp*`, new_name, value =
list(`is.na(dataset[[var]])` = c(TRUE, :
replacement has 9 rows, data has 0 In addition: Warning message:
In if (variables == "all") { :
the condition has length > 1 and only the first element will be used
My goal is to create a dataset from scratch on the fly which has the same
amount of variables as the dataset ds_example plus a single variable storing
the amount of NA's in a row for the given variables. This is the basis for a
decious which cases to keep and which to drop.
I do not want to alter the base dataset like ds_example in the first place nor
do I want to make a copy of the existing dataset due to memory allocation. The
function shall also work with big data, e. g. datasets with more than 1 GB
memory consumption.
I also do not want the newly created variables to be stored in the original
data frame. They shall be separate.
A former similar solution worked:
http://r.789695.n4.nabble.com/Creating-variables-on-the-fly-td4720034.html
Why doesn't this one?
How do I create the variables within the data frame if the data frame is empty?
Kind regards
Georg Maubach
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Creating a data frame from scratch (SOLVED)
Hi Dan,
Hi All,
many thanks for your help.
Please find enclosed my little function for your use:
-- cut --
#---
# Module: t_count_na.R
# Author: Georg Maubach
# Date : 2016-05-24
# Update: 2016-05-25
# Description : Count NA's
# Source System : R 3.2.2 (64 Bit)
# Target System : R 3.2.2 (64 Bit)
# License : CC-BY-SA-NC
#1-2-3-4-5-6-7-8
test <- FALSE
t_count_na <- function(dataset,
variables = "all") {
# Counts the number of NA within given set of veriables
#
# Args:
# dataset : Object with dimnames, e.g. data frame, data table.
# variables: Character vector with variable names.
#
# Operation:
# Adds the variable "na_count" to the given dataset containing the
count of
# NA's within the given variables
#
# Returns:
# Original dataset with variable "na_count" added.
#
# Error handling:
# None.
#
# Credits:
#
http://stackoverflow.com/questions/4862178/remove-rows-with-nas-in-data-frame
#
http://r.789695.n4.nabble.com/Creating-variables-on-the-fly-td4720034.html
version <- "2016-05-25"
if (identical(variables, "all")) {
variable_list <- names(dataset)
} else {
variable_list <- variables
}
dataset[["na_count"]] <- apply(dataset[,variable_list],
1,
function(x) sum(is.na(x)))
return(dataset)
}
#---
test <- function(do_test = FALSE) {
cat("\n", "\n", "Test function t_count_na()", "\n", "\n")
# Example dataset
gene <-
c("ENSG0208234","ENSG0199674","ENSG0221622","ENSG0207604",
"ENSG0207431","ENSG0221312","ENSG00134940305","ENSG00394039490",
"ENSG09943004048")
hsap <- c(0,0,0, 0, 0, 0, 1,1, 1)
mmul <- c(NA,2 ,3, NA, 2, 1 , NA,2, NA)
mmus <- c(NA,2 ,NA, NA, NA, 2 , NA,3, 1)
rnor <- c(NA,2 ,NA, 1 , NA, 3 , NA,NA, 2)
cfam <- c(NA,2,NA, 2, 1, 2, 2,NA, NA)
ds_example <- data.frame(gene, hsap, mmul, mmus, rnor, cfam)
ds_example$gene <- as.character(ds_example$gene)
cat("\n", "\n", "Example dataset before function call", "\n", "\n")
print(ds_example)
cat("\n", "\n", "Function call", "\n", "\n")
ds_example <- t_count_na(dataset = ds_example,
variables = c("mmul", "mmus"))
cat("\n", "\n", "Example dataset after function call", "\n", "\n")
print(ds_example)
}
test(do_test = test)
# EOF .
-- cut --
Kind regards
Georg Maubach
Von:"Nordlund, Dan (DSHS/RDA)"
An: "r-help@r-project.org" ,
Datum: 24.05.2016 21:41
Betreff:Re: [R] Creating a data frame from scratch
Gesendet von: "R-help"
I would probably write the function something like this:
t_count_na <- function(dataset,
variables = "all") {
if (identical(variables, "all")) {
variable_list <- names(dataset)
} else {
variable_list <- variables
}
apply(dataset[,variable_list], 1, function(x) sum(is.na(x)))
}
Hope this is helpful,
Dan
Daniel Nordlund, PhD
Research and Data Analysis Division
Services & Enterprise Support Administration
Washington State Department of Social and Health Services
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> g.maub...@gmx.de
> Sent: Tuesday, May 24, 2016 11:55 AM
> To: r-help@r-project.org
> Subject: [R] Creating a data frame from scratch
>
> Hi All,
>
> I need to create a data frame from scratch and fill variables created on
the fly
> with values. What I have so far:
>
> -- schnipp --
>
> # Example dataset
> gene <-
> c("ENSG0208234","ENSG0199674","ENSG0221622","ENSG0
> 207604",
>
> "ENSG0207431","ENSG0221312","ENSG00134940305","ENSG0039403
> 9490",
> "ENSG09943004048")
> hsap <- c(0,0,0, 0, 0, 0, 1,1, 1)
> mmul <- c(NA,2 ,3, NA, 2, 1 , NA,2, NA)
> mmus <- c(NA,2 ,NA, NA, NA, 2 , NA,3, 1) rnor <- c(NA,2 ,NA, 1 , NA, 3 ,
> NA,NA, 2) cfam <- c(NA,2,NA, 2, 1, 2, 2,NA, NA)
>
> ds_example <- data.frame(gene, hsap, mmul, mmus, rnor, cfam)
> ds_example$gene <- as.character(ds_example$gene)
>
> t_count_na <- function(dataset,
>variables = "all")
> # credit: http://stackoverflow.com/questions/4862178/remove-rows-with-
> nas-in-data-frame
> {
> ds_na <- data.frame()
> # if variables = "all" create character vector of variable names
> if (variables == "all") {
> variable_list <- dimnames(dataset)[[ 2 ]]
> }
> # if a character vector with variable names is given
> # to run the function on a defined set of selected variables
> else {
> variable_list <- variables
> }
>
> for (var in variable_list) {
> new_name <- paste0("na_", var)
> ds_na[[ new_name ]] <- as.data.frame(is.na(datas
[R] Difference subsetting (dataset$variable vs. dataset["variable"]
Hi All,
I thought dataset$variable is the same as dataset["variable"]. I tried the
following:
> str(ZWW_Kunden$Branche)
chr [1:49673] "231" "151" "151" "231" "231" "111" "231" "111" "231" "231"
"151" "111" ...
> str(ZWW_Kunden["Branche"])
'data.frame':49673 obs. of 1 variable:
$ Branche: chr "231" "151" "151" "231" ...
and get different results: "chr {1:49673]" vs. "data.frame". First one is
a simple vector, second one is a data.frame.
This has consequences when subsetting a dataset and filter cases:
> ZWW_Kunden["Branche"] %in% c("315", "316", "317")
[1] FALSE
> head(ZWW_Kunden$Branche %in% c("315", "316", "317")) # head() only to
shorten output
[1] FALSE FALSE FALSE FALSE FALSE FALSE
I have thought dataset$variable is the same as dataset["variable"] but
actually it's not.
Can you explain what the difference is?
Kind regards
Georg
__
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and provide commented, minimal, self-contained, reproducible code.
[R] Variable labels and value labels
Hi All, I am using R for social sciences. In this field I am used to use short variable names like "q1" for question 1, "q2" for question 2 and so on and label the variables like q1 : "Please tell us your age" or q2 : "Could you state us your household income?" or something similar indicating which question is stored in the variable. Similar I am used to label values like 1: "Less than 18 years", 2 : "18 to 30 years", 3 : "31 to 60 years" and 4 : "61 years and more". I know that the packages Hmisc and memisc have a functionality for this but these labeling functions are limited to the packages they were defined for. Using the question tests as variable names is possible but very inconvenient. I there another way for labeling variables and values in R? Kind regards Georg Maubach __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Utility Functions
Hi All, I was new to R and this list a couple of mounths ago. When processing my data I got tremendous support from R-Help mailing list. The solutions I have worked out with your help might be also helpful for others. I have put the solutions in a couple of small functions with documentation and tests. You can find the software on Sourceforge.net at https://sourceforge.net/projects/r-project-utilities/files/?source=navbar You should download at least "r_toolbox.R" and store it in a directory like "r_toolbox" in your favourite project folder. Within "r_toolbox" folder put all the other files. You have to adjust the variable "t_toolbox_path" to your favourite project directory including the "r_toolbox" folder, e. g. "C:\My-Projects\t-toolbox\" on Windows or "/home/username/my-projects/r-toolbox" on Unix-like systems. You can use them for your projects. Although I developed them with great care these functions come with absolutely no warrenty. You need to use them at your own risk. As the functions are small and overseeable you will find out quickly by reading the source code that the functions are save to use. If you have any recommendations or improvement proposals please get back to me. Kind regards Georg Maubach __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Installing miniCRAN on Debian
Hi All, I am installng miniCRAN on Debian GNU Linux 8 Jessie (Linux analytics7 4.5.0-0.bpo.2-amd64 #1 SMP Debian 4.5.4-1~bpo8+1 (2016-05-13) x86_64 GNU/Linux) and R 3.3.0 -- cut -- > sessionInfo() R version 3.3.0 (2016-05-03) Platform: x86_64-pc-linux-gnu (64-bit) Running under: Debian GNU/Linux 8 (jessie) locale: [1] LC_CTYPE=de_DE.UTF-8 LC_NUMERIC=C LC_TIME=de_DE.UTF-8 [4] LC_COLLATE=de_DE.UTF-8 LC_MONETARY=de_DE.UTF-8 LC_MESSAGES=de_DE.UTF-8 [7] LC_PAPER=de_DE.UTF-8 LC_NAME=C LC_ADDRESS=C [10] LC_TELEPHONE=C LC_MEASUREMENT=de_DE.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_3.3.0 -- cut -- After running sudo apt-get install libssl-dev libcurl4-openssl-dev libxml2-dev libhunspell-dev and calling install.packages(pkgs = "miniCRAN", repos = "http://cran.csiro.au";, dependencies = TRUE) I get the message - ANTICONF ERROR --- Configuration failed because hunspell was not found. Try installing: * deb: libhunspell-dev (Debian, Ubuntu, etc) * rpm: hunspell-devel (Fedora, CentOS, RHEL) * brew: hunspell (Mac OSX) If hunspell is already installed, check that 'pkg-config' is in your PATH and PKG_CONFIG_PATH contains a hunspell.pc file. If pkg-config is unavailable you can set INCLUDE_DIR and LIB_DIR manually via: R CMD INSTALL --configure-vars='INCLUDE_DIR=... LIB_DIR=...' Running find / -name hunspell.pc gives /usr/lib/x86_64-linux-gnu/pkgconfig/hunspell.pc and running find / -name pkg-config gives /usr/share/bash-completion/completions/pkg-config How do I need to configure R correctly to get miniCRAN running? Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Unable to update R software to 3.3.0
Hi all, I did it today on Debian GNU Linux 8 Jessie this way: vim /etc/apt/sources.list deb http://cran.uni-muenster.de/bin/linux/debian jessie-cran3 ESC;:wq apt.get update apt-get install r-base r-base-dev This worked for me. When installing R packages from within R I found that R needed the following: apt-get install libssl-dev libcurl4-openssl-dev libhunspell-dev libxml2-dev You probably might to wish to install this also. HTH. Kind regards Georg Von:Marc Schwartz An: Sunish Kumar Bilandi , Kopie: R-help Datum: 01.06.2016 17:18 Betreff:Re: [R] Unable to update R software to 3.3.0 Gesendet von: "R-help" > On Jun 1, 2016, at 1:33 AM, Sunish Kumar Bilandi wrote: > > Hi Team, > > I am using RedHat 5 and installed R using YUM, (R version 3.2.3) Now I want to update R version tp 3.3.0, but I am unable to do that, Is there any alternate to do this? > > Hope to hear from your side. > > Regards, > > > Sunish Bilandi > Business Analyst, CIDA-01 > Evalueserve Hi, First, RHEL and related distributions (e.g. Fedora), have a dedicated R-SIG list: https://stat.ethz.ch/mailman/listinfo/r-sig-fedora Future queries in this domain should be submitted there, as many of the RH package maintainers (e.g. Tom Callaway, aka Spot) read that list. For R 3.3.0, it would appear that it is about a day away from being available for release: https://bodhi.fedoraproject.org/updates/FEDORA-EPEL-2016-6fc2c863b0 So for now, it would be available via the EPEL testing repos. Otherwise, you can wait until it is available via release in the next day or so, or download the RPMS directly here: http://koji.fedoraproject.org/koji/buildinfo?buildID=762521 Regards, Marc Schwartz __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: RE: Variable labels and value labels
Hi Petr,
I am looking for a general procedure that I can use with any package of R.
As to my current experience it probably will happen that I need a
procedure from another package than hmisc or memisc and the my solution
shall work even than so that I do need to find another way to do it.
Kind regards
Georg
Von:PIKAL Petr
An: "g.maub...@weinwolf.de" ,
"r-help@r-project.org" ,
Datum: 31.05.2016 14:56
Betreff:RE: [R] Variable labels and value labels
Hi
see in line
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> g.maub...@weinwolf.de
> Sent: Tuesday, May 31, 2016 2:01 PM
> To: r-help@r-project.org
> Subject: [R] Variable labels and value labels
>
> Hi All,
>
> I am using R for social sciences. In this field I am used to use short
variable
> names like "q1" for question 1, "q2" for question 2 and so on and label
the
> variables like q1 : "Please tell us your age" or q2 : "Could you state
us your
> household income?" or something similar indicating which question is
stored
> in the variable.
>
> Similar I am used to label values like 1: "Less than 18 years", 2 : "18
to
> 30 years", 3 : "31 to 60 years" and 4 : "61 years and more".
Seems to me that it is work for factors
nnn <- sample(1:4, 20, replace=TRUE)
q1 <-factor(nnn, labels=c("Less than 18 years", "18 to 30 years", "31 to
60 years","61 years and more"))
You can store such variables in data.frame with names "q1" to "qwhatever"
and possibly "Subject"
And you can store annotation of questions in another data frame with 2
columns e.g. "Question" and "Description"
Basically it is an approach similar to database and in R you can merge
those two data.frames by ?merge.
>
> I know that the packages Hmisc and memisc have a functionality for this
but
> these labeling functions are limited to the packages they were defined
for.
It seems to me strange. What prevents you to use functions from Hmisc?
Regards
Petr
> Using the question tests as variable names is possible but very
inconvenient.
>
> I there another way for labeling variables and values in R?
>
> Kind regards
>
> Georg Maubach
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
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[R] Antwort: Re: Variable labels and value labels
Hi Jim, many thanks for the hint. When looking at the documentation I did not get how I do control which value gets which label. Is it possible to define it? Kind regards Georg Von:Jim Lemon An: g.maub...@weinwolf.de, r-help mailing list , Datum: 01.06.2016 03:59 Betreff:Re: [R] Variable labels and value labels Hi Georg, You may find the "add.value.labels" function in the prettyR package useful. Jim On Tue, May 31, 2016 at 10:00 PM, wrote: > Hi All, > > I am using R for social sciences. In this field I am used to use short > variable names like "q1" for question 1, "q2" for question 2 and so on and > label the variables like q1 : "Please tell us your age" or q2 : "Could you > state us your household income?" or something similar indicating which > question is stored in the variable. > > Similar I am used to label values like 1: "Less than 18 years", 2 : "18 to > 30 years", 3 : "31 to 60 years" and 4 : "61 years and more". > > I know that the packages Hmisc and memisc have a functionality for this > but these labeling functions are limited to the packages they were defined > for. Using the question tests as variable names is possible but very > inconvenient. > > I there another way for labeling variables and values in R? > > Kind regards > > Georg Maubach > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Merging variables
Hi All,
I merged two datasets:
ds_merge1 <- merge(x = ds_bw_customer_4_match, y =
ds_zww_customer_4_match,
by.x = "customer", by.y = "customer",
all.x = TRUE, all.y = FALSE)
R created a new dataset with the variables customer.x and customer.y. I
would like to merge these two variable back together. I wrote a little
function (code can be run) for it:
-- cut --
customer.x <- c("Miller", "Smith", NA,"Bird", NA)
customer.y <- c("Miller", NA, "Doe", "Fish", NA)
ds_test <- data.frame(customer.x, customer.y, stringsAsFactors = FALSE)
t_merge_variables <-
function(dataset,
var1,
var2,
merged_var) {
# Initialize
dataset[[merged_var]] = rep(NA, nrow(dataset))
dataset[["mismatch"]] = rep(NA, nrow(dataset))
for (i in 1:nrow(dataset)) {
# Check 1: var1 missing, var2 missing
if (is.na(dataset[[i, var1]]) &
is.na(dataset[[i, var2]])) {
dataset[["mismatch"]] <- 1 # var1 & var2 are missing
# Check 2: var1 filled, var2 missing
} else if (!is.na(dataset[[i, var1]]) &
is.na(dataset[[i, var2]])) {
dataset[[i, merged_var]] <- dataset[[i, var1]]
dataset[["mismatch"]] <- 0
# Check 3: var1 missing, var2 filled
} else if (is.na(dataset[[i, var1]]) &
!is.na(dataset[i, var2])) {
dataset[[i, merged_var]] <- dataset[[i, var2]]
dataset[["mismatch"]] <- 0
# Check 4: var1 == var2
} else if (dataset[[i, var1]] == dataset[[i, var2]]) {
dataset[[i, merged_var]] <- dataset[[i, var1]]
dataset[["mismatch"]] <- 0
# Leftover: var1 != var2
} else {
dataset[[i, merged_var]] <- NA
dataset[["mismatch"]] <- 2 # var1 != var2
} # end if
} # end for
return(dataset)
}
ds_var_merge1 <- t_merge_variables(dataset = ds_test,
var1 = "customer.x",
var2 = "customer.y",
merged_var = "customer")
ds_var_merge1
-- cut --
It is executed without error but delivers the wrong values in the variable
"mismatch". This variable is always 1 although it should be NA, 1 or 2
respectively.
Can you tell me why the variable is not correctly set?
Kind regards
Georg
__
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[R] Antwort: RE: Merging variables
Hi David,
Hi Petr,
many thanks for your help. With your hints I got the idea how I could do
it and I came up with this solution:
-- cut --
#---
# Module: t_merge_variables.R
# Author: Georg Maubach
# Date : 2016-06-06
# Update: 2016-06-06
# Description : Merge two variables
# Source System : R 3.2.5 (64 Bit)
# Target System : R 3.2.5 (64 Bit)
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
#1-2-3-4-5-6-7-8
t_module_name = "t_merge_variables.R"
t_version = "2016-06-06"
cat(
paste0("\n",
t_module_name, " (Version: ", t_version, ")", "\n", "\n",
"This software comes with ABSOLUTELY NO WARRANTY.",
"\n", "\n"))
# If do_test is not defined globally define it here locally by
un-commenting it
# Switch t_do_test to TRUE to run test
t_do_test <- FALSE
# [ Function Defintion
]
t_merge_variables <-
function(dataset,
var1,
var2,
merged_var) {
# Merges two variables with identical, different or missing values
#
# Args:
# dataset (data frame, data table):
#Object with dimnames, e.g. data frame, data table.
# var1 (character):
#Variable 1 to be merged.
# var2 (character):
#Variable 2 to be merged.
# merged_var (class based on input variable, coercion done if
possible):
#Variable with the merged variables var1 and var2.
#
# Operation:
# Var1 and var2 are merged like follows:
# if var1 == var2: merged_var <- var1
# if var1 != var2: merged_var <- -900 (-900 = indicating mismatch)
# if var1 is filled & var2 is missing: merged_var <- var1
# if var1 is missing & var2 is filled: merged_var <- var2
# if var1 is missing & var2 is filled: merged_var <- -999
#(-999 = indicating NA)
#
# Returns:
# Original dataset and variable given in "merged_var" will be added.
#
# Error handling:
# None.
#
# Credits:
# https://www.mail-archive.com/r-help@r-project.org/msg236012.html
# Initialize
dataset[merged_var] = rep(NA, nrow(dataset))
dataset[merged_var] <-
# Check 1: var1 missing, var2 missing
ifelse(is.na(dataset[, var1]) & is.na(dataset[, var2]),
# then
dataset[[merged_var]] <- 0,
# Check 2: var1 filled, var2 missing
ifelse(!is.na(dataset[, var1]) & is.na(dataset[, var2]),
# then
dataset[[merged_var]] <- dataset[, var1],
# Check 3: var1 missing, var2 filled
ifelse(is.na(dataset[ , var1]) & !is.na(dataset[, var2]),
# then
dataset[[merged_var]] <- dataset[ , var2],
# Check 4: var1 == var2
ifelse(dataset[, var1] == dataset[, var2],
# then: use var1
dataset[[merged_var]] <- dataset[, var1],
#Leftover: var1 != var2
dataset[merged_var] <- 1
return(dataset)
}
# [ Test Defintion
]
t_test <- function(do_test = FALSE) {
if (do_test == TRUE) {
cat("\n", "\n", "Test function t_count_na()", "\n", "\n")
# Example dataset
customer.x <- c("Miller", "Smith", NA,"Bird", NA)
customer.y <- c("Miller", NA, "Doe", "Fish", NA)
ds_test <-
data.frame(customer.x, customer.y, stringsAsFactors = FALSE)
# Call function
ds_merge <- t_merge_variables(
dataset = ds_test,
var1 = "customer.x",
var2 = "customer.y",
merged_var = "customer"
)
# Dataset after function call
ds_merge
}
}
# [ Test Run
]--
t_test(do_test = t_do_test)
# [ Clean up
]--
rm("t_do_test", "t_module_name", "t_version", "t_test")
# EOF
-- cut --
It delivers the customer name if there is one or they match. If they don't
match it delivers 1. If both are missing it delivers 0.
This solution is for my applications sufficient.
Many thanks again for your help and giving me the ideas to solve my data
transformation task.
Kind regards
Georg
Von:PIKAL Petr
An: "g.maub...@weinwolf.de" ,
"r-help@r-project.org" ,
Datum: 06.06.2016 15:04
Betreff:RE: [R] Merging variables
Hi
Not sure if this is the most effective or general solution but
Here you get 2 if the value is same in both columns, 1 if it is only in
one column and the other is NA and 0 if there is mismatch of values.
temp <- (ds_test[,2] %in% ds_test[,1])+(ds_test[,1] %in% ds_test[,2])
here you get 0 if t
[R] Antwort: Re: Merging variables
Hi Michael,
yes, I was astonished about this behaviour either. I have worked with SPSS
a lot - and that works different.
I would like to share some of my data. Can you tell me how I can dump a
dataset in a way that I can post it here as text?
Kind regards
Georg
Von:Michael Dewey
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 06.06.2016 15:45
Betreff:Re: [R] Merging variables
X-Originating-<%= hostname %>-IP: [217.155.205.190]
Dear Georg
I find it a bit surprising that you end up with customer.x and
customer.y. Can you share with us a toy example of two data.frames which
exhibit this behaviour?
On 06/06/2016 13:29, g.maub...@weinwolf.de wrote:
> Hi All,
>
> I merged two datasets:
>
> ds_merge1 <- merge(x = ds_bw_customer_4_match, y =
> ds_zww_customer_4_match,
> by.x = "customer", by.y = "customer",
> all.x = TRUE, all.y = FALSE)
>
> R created a new dataset with the variables customer.x and customer.y. I
> would like to merge these two variable back together. I wrote a little
> function (code can be run) for it:
>
> -- cut --
>
> customer.x <- c("Miller", "Smith", NA,"Bird", NA)
> customer.y <- c("Miller", NA, "Doe", "Fish", NA)
> ds_test <- data.frame(customer.x, customer.y, stringsAsFactors = FALSE)
>
> t_merge_variables <-
> function(dataset,
>var1,
>var2,
>merged_var) {
>
> # Initialize
> dataset[[merged_var]] = rep(NA, nrow(dataset))
> dataset[["mismatch"]] = rep(NA, nrow(dataset))
>
> for (i in 1:nrow(dataset)) {
>
> # Check 1: var1 missing, var2 missing
> if (is.na(dataset[[i, var1]]) &
> is.na(dataset[[i, var2]])) {
> dataset[["mismatch"]] <- 1 # var1 & var2 are missing
>
> # Check 2: var1 filled, var2 missing
> } else if (!is.na(dataset[[i, var1]]) &
> is.na(dataset[[i, var2]])) {
> dataset[[i, merged_var]] <- dataset[[i, var1]]
> dataset[["mismatch"]] <- 0
>
> # Check 3: var1 missing, var2 filled
> } else if (is.na(dataset[[i, var1]]) &
> !is.na(dataset[i, var2])) {
> dataset[[i, merged_var]] <- dataset[[i, var2]]
> dataset[["mismatch"]] <- 0
>
> # Check 4: var1 == var2
> } else if (dataset[[i, var1]] == dataset[[i, var2]]) {
> dataset[[i, merged_var]] <- dataset[[i, var1]]
> dataset[["mismatch"]] <- 0
>
> # Leftover: var1 != var2
> } else {
> dataset[[i, merged_var]] <- NA
> dataset[["mismatch"]] <- 2 # var1 != var2
> } # end if
> } # end for
> return(dataset)
> }
>
> ds_var_merge1 <- t_merge_variables(dataset = ds_test,
> var1 = "customer.x",
> var2 = "customer.y",
> merged_var = "customer")
>
> ds_var_merge1
>
> -- cut --
>
> It is executed without error but delivers the wrong values in the
variable
> "mismatch". This variable is always 1 although it should be NA, 1 or 2
> respectively.
>
> Can you tell me why the variable is not correctly set?
>
> Kind regards
>
> Georg
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Michael
http://www.dewey.myzen.co.uk/home.html
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: RE: Merging variables
Hi Petr,
I would like to describe the data situation in brief:
I have an business warehouse dataset (referred to as BW data) containing
sales and an ERP customer master data dataset with additional information
(referred to as ERP data). Though customer IDs and customer names are
identical due to the fact that the business warehouse data is derived from
the ERP data. Due to selection criteria the BW data contains slightly
more customers than the ERP data. So customer names and all other
information is missing in the ERP data for some cases of the BW data. If
I merge those by customer ID variable customer names are duplicated using
customer.x and customer.y as variable names.
As both fields contains the same contents I would have expected R to merge
this into one variable, e. g. customer. But this is not the case.
Can I adjust the below given merge statement - which looks almost the same
in my script - that R does the merge of the variables if they are
identical automatically?
This is my code using left join:
-- cut --
ds_merge1 <- merge(x = ds_bw_customer_4_match, y =
ds_erp_customer_4_match,
by.x = "CustID", by.y = "CustID",
all.x = TRUE, all.y = FALSE)
-- cut --
Kind regards
Georg
Von:PIKAL Petr
An: Michael Dewey , "g.maub...@weinwolf.de"
, "r-help@r-project.org" ,
Datum: 06.06.2016 17:04
Betreff:RE: [R] Merging variables
Hi Michael
it is simple
set.seed(111)
let=sample(letters[1:10],6, replace=T)
dat1<-data.frame(let=let, customer=sample(1:10,6, replace=T))
let=sample(letters[1:10],6, replace=T)
dat2<-data.frame(let=let, customer=sample(1:10,6, replace=T))
merge(dat1, dat2, by.x="let", by.y="let", all=T)
Of course you could add customer variable to by parameter but sometimes it
is necessary to leave it out. When you have two sets of analytical results
and you have 2 variables operator but you want to merge those sets e.g. by
date/hour of analysis.
Regards
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael
> Dewey
> Sent: Monday, June 6, 2016 3:46 PM
> To: g.maub...@weinwolf.de; r-help@r-project.org
> Subject: Re: [R] Merging variables
>
> X-Originating-<%= hostname %>-IP: [217.155.205.190]
>
> Dear Georg
>
> I find it a bit surprising that you end up with customer.x and
customer.y. Can
> you share with us a toy example of two data.frames which exhibit this
> behaviour?
>
> On 06/06/2016 13:29, g.maub...@weinwolf.de wrote:
> > Hi All,
> >
> > I merged two datasets:
> >
> > ds_merge1 <- merge(x = ds_bw_customer_4_match, y =
> > ds_zww_customer_4_match,
> > by.x = "customer", by.y = "customer",
> > all.x = TRUE, all.y = FALSE)
> >
> > R created a new dataset with the variables customer.x and customer.y.
> > I would like to merge these two variable back together. I wrote a
> > little function (code can be run) for it:
> >
> > -- cut --
> >
> > customer.x <- c("Miller", "Smith", NA,"Bird", NA)
> > customer.y <- c("Miller", NA, "Doe", "Fish", NA)
> > ds_test <- data.frame(customer.x, customer.y, stringsAsFactors =
> > FALSE)
> >
> > t_merge_variables <-
> > function(dataset,
> >var1,
> >var2,
> >merged_var) {
> >
> > # Initialize
> > dataset[[merged_var]] = rep(NA, nrow(dataset))
> > dataset[["mismatch"]] = rep(NA, nrow(dataset))
> >
> > for (i in 1:nrow(dataset)) {
> >
> > # Check 1: var1 missing, var2 missing
> > if (is.na(dataset[[i, var1]]) &
> > is.na(dataset[[i, var2]])) {
> > dataset[["mismatch"]] <- 1 # var1 & var2 are missing
> >
> > # Check 2: var1 filled, var2 missing
> > } else if (!is.na(dataset[[i, var1]]) &
> > is.na(dataset[[i, var2]])) {
> > dataset[[i, merged_var]] <- dataset[[i, var1]]
> > dataset[["mismatch"]] <- 0
> >
> > # Check 3: var1 missing, var2 filled
> > } else if (is.na(dataset[[i, var1]]) &
> > !is.na(dataset[i, var2])) {
> > dataset[[i, merged_var]] <- dataset[[i, var2]]
> > dataset[["mismatch"]] <- 0
> >
> > # Check 4: var1 == var2
> > } else if (dataset[[i, var1]] == dataset[[i, var2]]) {
> > dataset[[i, merged_var]] <- dataset[[i, var1]]
> > dataset[["mismatch"]] <- 0
> >
> > # Leftover: var1 != var2
> > } else {
> > dataset[[i, merged_var]] <- NA
> > dataset[["mismatch"]] <- 2 # var1 != var2
> > } # end if
> > } # end for
> > return(dataset)
> > }
> >
> > ds_var_merge1 <- t_merge_variables(dataset = ds_test,
> > var1 = "customer.x",
> > var2 = "customer.y",
> > merged_var = "customer")
> >
> > ds_var_merge1
> >
> > -- cut --
> >
> > It is executed without error but delivers the wrong values in the
> > variable "mismatch". This variable is always 1 although it should be
> > NA, 1 or 2 respectively.
> >
> > Can you tell me why the variable is not correctly set?
> >
> > Kind regards
> >
> > Georg
>
[R] Antwort: RE: Antwort: Re: Merging variables
Hi Petr,
thanks for your reply.
I prepared little example for you:
-- cut --
ds_temp_1 <-
structure(list(
CustId = c(1001, 1002, 1003, 1004, 1005, 1006),
CustName = c("Miller", "Smith", "Doe", "White", "Black",
"Nobody"),
sales = c(100, 500, 300, 50, 700, 10)
),
.Names = c("CustId",
"CustName", "sales"), row.names = c(NA, 6L), class =
"data.frame")
ds_temp_2 <-
structure(
list(
CustId = c(1001, 1002, 1003),
CustName = c("Miller",
"Smith", "Doe"),
CustGroup = c(1, 2, 3)
),
.Names = c("CustId",
"CustName", "CustGroup"),
row.names = c(NA, 3L),
class = "data.frame"
)
ds_merge <- merge(ds_temp_1, ds_temp_2,
by.x = "CustId", all.x = TRUE,
by.y = "CustId", all.y = FALSE)
ds_merge
-- cut --
which gives
ds_merge
CustId CustName.x sales CustName.y CustGroup
1 1001 Miller 100 Miller 1
2 1002 Smith 500 Smith 2
3 1003Doe 300Doe 3
4 1004 White50 NA
5 1005 Black 700 NA
6 1006 Nobody10 NA
where CustName is split into CustName.x and CustName.y.
What I would like to have is:
ds_merge
CustId CustName sales CustGroup
1 1001 Miller 100 1
2 1002 Smith 500 2
3 1003Doe 300 3
4 1004 White50 NA
5 1005 Black 700 NA
6 1006 Nobody10 NA
That is CustName in a single variable cause the values within that
variable are identical. I guess because of NA for some cases in ds_temp_2
R generates CustName.x and CustName.y.
Is there a simple way of merging a dataset and having R return a single
variable is the values are identical or missing in either one of the
datasets?
Kind regards
Georg
Von:PIKAL Petr
An: "g.maub...@weinwolf.de" ,
Kopie: "r-help@r-project.org"
Datum: 07.06.2016 13:11
Betreff:RE: [R] Antwort: Re: Merging variables
Hi
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> g.maub...@weinwolf.de
> Sent: Tuesday, June 7, 2016 8:19 AM
> To: Michael Dewey
> Cc: r-help@r-project.org
> Subject: [R] Antwort: Re: Merging variables
>
> Hi Michael,
>
> yes, I was astonished about this behaviour either. I have worked with
SPSS a
> lot - and that works different.
If you want to join two data frames by common names you can use use
merge(dat1, dat2, )
without specifing by. From help page:
By default the data frames are merged on the columns with names they both
have, but separate specifications of the columns can be given by by.x and
by.y. The rows in the two data frames that match on the specified columns
are extracted, and joined together.
>
> I would like to share some of my data. Can you tell me how I can dump a
> dataset in a way that I can post it here as text?
copy result of dput directly to your mail
dput(dat)
structure(list(hz = c(0, 25, 50), vykon = c(0, 11.6, 22.6)), .Names =
c("hz",
"vykon"), row.names = c(NA, -3L), class = "data.frame")
We can use
dat <- structure(list(hz = c(0, 25, 50), vykon = c(0, 11.6, 22.6)), .Names
= c("hz",
"vykon"), row.names = c(NA, -3L), class = "data.frame")
to reconstruct the object.
Regards
Petr
>
> Kind regards
>
> Georg
>
>
>
>
> Von:Michael Dewey
> An: g.maub...@weinwolf.de, r-help@r-project.org,
> Datum: 06.06.2016 15:45
> Betreff:Re: [R] Merging variables
>
>
>
> X-Originating-<%= hostname %>-IP: [217.155.205.190]
>
> Dear Georg
>
> I find it a bit surprising that you end up with customer.x and
customer.y. Can
> you share with us a toy example of two data.frames which exhibit this
> behaviour?
>
> On 06/06/2016 13:29, g.maub...@weinwolf.de wrote:
> > Hi All,
> >
> > I merged two datasets:
> >
> > ds_merge1 <- merge(x = ds_bw_customer_4_match, y =
> > ds_zww_customer_4_match,
> > by.x = "customer", by.y = "customer",
> > all.x = TRUE, all.y = FALSE)
> >
> > R created a new dataset with the variables customer.x and customer.y.
> > I would like to merge these two variable back together. I wrote a
> > little function (code can be run) for it:
> >
> > -- cut --
> >
> > customer.x <- c("Miller", "Smith", NA,"Bird", NA)
> > customer.y <- c("Miller", NA, "Doe", "Fish", NA)
> > ds_test <- data.frame(customer.x, customer.y, stringsAsFactors =
> > FALSE)
> >
> > t_merge_variables <-
> > function(dataset,
> >var1,
> >var2,
> >merged_var) {
> >
> > # Initialize
> > dataset[[merged_var]] = rep(NA, nrow(dataset))
> > dataset[["mismatch"]] = rep(NA, nrow(dataset))
> >
> > for (i in 1:nrow(dataset)) {
> >
> > # Check 1: var1 missing, var2 missing
> > if (is.na(dataset[[i, var1]]) &
> > is.na(dataset[[i, var2]])) {
> > dataset[["mismatch"]] <- 1 # var1 & var2 are missing
[R] Warning message in openxlsx
Hi All,
I get the warning message
Warning message:
In styles$font : partial match of 'font' to 'fonts'
when executing
> xls_workbook <- t_create_workbook()
> xls_sheetname <- "Kunden"
> xls_ds_to_save <- ds_merge1
> xls_filename <- paste0(data_created,
"_Merge1_BW-SAP-Kunden_cleaned.xlsx")
> t_add_sheet(workbook = xls_workbook,
+ sheetname = xls_sheetname,
+ dataset = xls_ds_to_save)
> t_write_xlsx(workbook = xls_workbook,
+ path = path_output,
+ filename = xls_filename,
+ overwrite = TRUE)
where t_create_workbook() is
return(createWorkbook())
and t_add_sheet() is
addWorksheet(workbook,
sheetName = sheetname)
writeDataTable(workbook,
sheet = sheetname,
x = dataset)
### writeDataTable writes data to a sheet an adds
### autofilter to the first line
if (freeze_row <= 1 | freeze_col <= 1) {
NULL # do nothing
}
else {
freezePane(workbook,
sheet = sheetname,
firstActiveRow = freeze_row,
firstActiveCol = freeze_col)
}
setColWidths(workbook,
sheet = sheetname,
cols = 1:ncol(dataset),
widths = "auto")
and t_write_xlsx is
saveWorkbook(workbook,
file = file.path(path, filename),
overwrite = overwrite)
I am woundring what "partial match of 'font' to 'fonts'" means cause I do
not call it in the functions calls. I use these calls a lot in my programs
but never got this message before.
What does this message mean? How can I avoid this message?
Kind regards
Georg Maubach
PS: You can find more information about the used functions by going to
https://sourceforge.net/projects/r-project-utilities/files/?source=navbar
.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Installation of package "rio" broken
Hi all, today I wanted to install package "rio". As it depends on package "feather" which is only available as source I have chosen to install "rio" from source. The installations fails with the following messages: -- cut -- * installing *source* package 'feather' ... ** Paket 'feather' erfolgreich entpackt und MD5 Summen überprüft ** libs *** arch - i386 g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c RcppExports.cpp -o RcppExports.o g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather-read.cpp -o feather-read.o g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather-types.cpp -o feather-types.o g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather-write.cpp -o feather-write.o g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather/buffer.cc -o feather/buffer.o g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather/feather-c.cc -o feather/feather-c.o g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather/io.cc -o feather/io.o feather/io.cc:18:0: warning: "NOMINMAX" redefined [enabled by default] c:\program files\rtools\gcc-4.6.3\bin\../lib/gcc/i686-w64-mingw32/4.6.3/../../../../include/c++/4.6.3/i686-w64-mingw32/bits/os_defines.h:46:0: note: this is the location of the previous definition g++ -m32 -std=c++0x -I"C:/PROGRA~1/R/R-32~1.2/include" -DNDEBUG -I. -I"C:/Users/admin/Documents/R/win-library/3.2/Rcpp/include" -I"d:/RCompile/r-compiling/local/local320/include" -O2 -Wall -mtune=core2 -c feather/metadata.cc -o feather/metadata.o feather/metadata.cc:29:7: error: expected nested-name-specifier before 'FBString' feather/metadata.cc:29:7: error: 'FBString' has not been declared feather/metadata.cc:29:16: error: expected ';' before '=' token feather/metadata.cc:29:16: error: expected unqualified-id before '=' token feather/metadata.cc:32:7: error: expected nested-name-specifier before 'ColumnVector' feather/metadata.cc:32:7: error: 'ColumnVector' has not been declared feather/metadata.cc:32:20: error: expected ';' before '=' token feather/metadata.cc:32:20: error: expected unqualified-id before '=' token feather/metadata.cc:178:3: error: 'ColumnVector' does not name a type feather/metadata.cc: In member function 'feather::Status feather::metadata::TableBuilder::Impl::Finish()': feather/metadata.cc:146:5: error: 'FBString' was not declared in this scope feather/metadata.cc:146:14: error: expected ';' before 'desc' feather/metadata.cc:148:7: error: 'desc' was not declared in this scope feather/metadata.cc:154:9: error: 'desc' was not declared in this scope feather/metadata.cc:156:27: error: 'columns_' was not declared in this scope feather/metadata.cc:157:34: error: unable to deduce 'auto' from '' feather/metadata.cc: In member function 'void feather::metadata::TableBuilder::Impl::add_column(const flatbuffers::Offset&)': feather/metadata.cc:173:5: error: 'columns_' was not declared in this scope feather/metadata.cc: In constructor 'feather::metadata::TableBuilder::TableBuilder()': feather/metadata.cc:190:5: error: type 'feather::metadata::TableBuilder' is not a direct base of 'feather::metadata::TableBuilder' make: *** [feather/metadata.o] Error 1 Warnung: Ausführung von Kommando 'make -f "Makevars" -f "C:/PROGRA~1/R/R-32~1.2/etc/i386/Makeconf" -f "C:/PROGRA~1/R/R-32~1.2/share/make/winshlib.mk" CXX='$(CXX1X) $(CXX1XSTD)' CXXFLAGS='$(CXX1XFLAGS)' CXXPICFLAGS='$(CXX1XPICFLAGS)' SHLIB_LDFLAGS='$(SHLIB_CXX1XLDFLAGS)' SHLIB_LD='$(SHLIB_CXX1XLD)' SHLIB="feather.dll" OBJECTS="RcppExports.o feather-read.o feather-types.o feather-write.o"' ergab Status 2 ERROR: compilation failed for package 'feather' * removing 'C:/Users/admin/Documents/R/win-library/3.2/feather' Warning in install.packages : running command '"C:/PROGRA~1/R/R-32~1.2/bin/x64/R" CMD INSTALL -l "C:\Users\admin\Documents\R\win-library\3.2" C:\Users\admin\AppData\Local\
[R] Building a binary vector out of dichotomous variables
Hi All,
I need to build a binary vector made of a set of dichotomous variables.
What I have so far is:
-- cut --
ds_example <-
structure(
list(
year2013 = c(0, 0, 0, 1, 1, 1, 1, 0),
year2014 = c(0,
0, 1, 1, 0, 0, 1, 1),
year2015 = c(0, 1, 1, 1, 0, 1, 0, 0)
),
.Names = c("year2013",
"year2014", "year2015"),
row.names = c(NA, 8L),
class = "data.frame"
)
attach(ds_example)
base <- 1000
binary_vector <- base + year2013 * 100 + year2014 * 10 + year2015
detach(ds_example)
binary_vector
ds_example <- cbind(ds_example, binary_vector)
varlist <- c("year2013", "year2014", "year2015")
base <- 10^length(varlist)
binary_vector <- NULL
for (i in 1:3) {
binary_vector <-
base +
ds_example [[varlist[i]]] * base / (10 ^ i)
}
ds_example <- cbind(ds_example, binary_vector)
message("Wrong result!")
ds_example
-- cut --
How do I get vectors like 1000 1001 1011 1100 1101 1110 1010 for
each case?
Is there a better approach than mine?
Kind regards
Georg
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[R] Fw: Aw: Re: Building a binary vector out of dichotomous variables
> Hi Tom,
>
> thanks for your reply.
>
> Yes, that's exactly what I am looking for. I did not know about the automatic
> type conversion in R.
>
> #-- cut --
> ds_example <-
> structure(
> list(
> year2013 = c(0, 0, 0, 1, 1, 1, 1, 0),
> year2014 = c(0,
>0, 1, 1, 0, 0, 1, 1),
> year2015 = c(0, 1, 1, 1, 0, 1, 0, 0)
> ),
> .Names = c("year2013",
>"year2014", "year2015"),
> row.names = c(NA, 8L),
> class = "data.frame"
> )
>
> #-- Proposal: works!
> as.numeric(with(ds_example,paste(1,year2013,year2014,year2015,sep='')))
>
> # I store my know-how about R in functions for later use.
>
> #--´ Putting it in a function - does not work!
> t_make_binary_vector <- function(dataset,
> input_variables,
> output_variable = "binary_vector") {
> dataset[output_variable] <- "1"
> print(dataset[output_variable])
>
> for (variable in input_variables) {
> print(variable)
> dataset[output_variable] <- paste(dataset[output_variable],
> dataset[variable],
> sep='')
> }
>
> # print(dataset[output_variable])
>
> dataset[output_variable] <- as.integer(dataset[output_variable])
>
> return(dataset)
> }
>
> t_make_binary_vector(dataset = ds_example,
> input_variables = c("year2013", "year2014", "year2015"),
> output_variable = "binary_vector")
>
>
> #-- Doesn't work either.
> t_make_binary_vector <- function(dataset,
> input_variables,
> output_variable = "binary_vector") {
> dataset[output_variable] <- as.integer(paste(1, dataset[ ,
> input_variables], sep = ''))
>
> return(dataset)
> }
>
> t_make_binary_vector(dataset = ds_example,
> input_variables = c("year2013", "year2014", "year2015"),
> output_variable = "binary_vector")
>
> #-- cut --
>
> Why is R taking the parameter value itself to paste it together instead of
> referencing the variable within the dataset?
>
> What did I get wrong about R? How can I fix it?
>
> Kind regards
>
> Georg
>
>
> > Gesendet: Donnerstag, 16. Juni 2016 um 16:13 Uhr
> > Von: "Tom Wright"
> > An: g.maub...@weinwolf.de
> > Cc: "R. Help"
> > Betreff: Re: [R] Building a binary vector out of dichotomous variables
> >
> > Does this do what you want?
> >
> > as.numeric(with(ds_example,paste(1,year2013,year2014,year2015,sep='')))
> >
> > On Thu, Jun 16, 2016 at 8:57 AM, wrote:
> > > Hi All,
> > >
> > > I need to build a binary vector made of a set of dichotomous variables.
> > >
> > > What I have so far is:
> > >
> > > -- cut --
> > >
> > > ds_example <-
> > > structure(
> > > list(
> > > year2013 = c(0, 0, 0, 1, 1, 1, 1, 0),
> > > year2014 = c(0,
> > >0, 1, 1, 0, 0, 1, 1),
> > > year2015 = c(0, 1, 1, 1, 0, 1, 0, 0)
> > > ),
> > > .Names = c("year2013",
> > >"year2014", "year2015"),
> > > row.names = c(NA, 8L),
> > > class = "data.frame"
> > > )
> > >
> > > attach(ds_example)
> > > base <- 1000
> > > binary_vector <- base + year2013 * 100 + year2014 * 10 + year2015
> > > detach(ds_example)
> > >
> > > binary_vector
> > >
> > > ds_example <- cbind(ds_example, binary_vector)
> > >
> > > varlist <- c("year2013", "year2014", "year2015")
> > >
> > > base <- 10^length(varlist)
> > >
> > > binary_vector <- NULL
> > >
> > > for (i in 1:3) {
> > > binary_vector <-
> > >base +
> > >ds_example [[varlist[i]]] * base / (10 ^ i)
> > > }
> > >
> > > ds_example <- cbind(ds_example, binary_vector)
> > >
> > > message("Wrong result!")
> > > ds_example
> > >
> > > -- cut --
> > >
> > > How do I get vectors like 1000 1001 1011 1100 1101 1110 1010 for
> > > each case?
> > >
> > > Is there a better approach than mine?
> > >
> > > Kind regards
> > >
> > > Georg
> > >
> > > __
> > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
__
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and provide commented, minimal, self-contained, reproducible
[R] (Off-Topic] Introducing a new R Blog
Hi All, today I would like to announce a now R blog. I contains a few entries about the findings during my course of studies and my daily work: https://github.com/gmaubach/R-Know-How/wiki/R-Blog I hope you'll find my hints usefull. In addition you could have a look at a small R collection of functions I found usefull when working with my data: https://github.com/gmaubach/R-Project-Utilities Kind regards Georg Maubach __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Subscripting problem with is.na()
Hi All,
I would like to recode my NAs to 0. Using a single vector everything is
fine.
But if I use a data.frame things go wrong:
-- cut --
var1 <- c(1:3, NA, 5:7, NA, 9:10)
var2 <- c(1:3, NA, 5:7, NA, 9:10)
ds_test <-
data.frame(var1, var2)
test <- var1
test[is.na(test)] <- 0
test # NA recoded OK
# First try
ds_test[is.na(ds_test$var1)] <- 0 # duplicate subscripts WRONG
# Second try
ds_test[is.na("var1")] <- 0
ds_test$var1 # not recoded WRONG
# Third try: to me the most intuitive approach
is.na(ds_test["var1"]) <- 0 # attempt to select less than one element in
integerOneIndex WRONG
# Fourth try
ds_test[is.na(var1)] <- 0 # duplicate subscripts for columns WRONG
-- cut --
How can I do it correctly?
Where could I have found something about it?
Kind regards
Georg
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[R] r_toolbox: Update
Hi folks, I have updated the functions of the r_toolbox.R set of utilities: https://sourceforge.net/projects/r-project-utilities/files/?source=navbar Naming was changed with some functions to reflect similar functions in SAS or SPSS, e. g. t_n_miss, t_n_valid. In addition I added functions for reporting memory usage, selecting variables by type and getting an overview over the levels of factors. I hope you find these functions useful. Please get back to me if you have suggestions or encounter any difficulties. Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subscripting problem with is.na()
Hi Bert,
many thanks for all your help and your comments. I learn at lot this way.
My question was about is.na() at the first sight but the actual task looks like
this:
I have two variables in my customer data that signal if the customer accout was
closed by master data management or by sales. Say these variables are
closed_mdm and closed_sls. They contain NA if the customer account is still
open or a closing code from "01" to "08" if the customer account was closed and
why.
For my analysis I need a variable that combines the two variables closed_mdm
and closed_sls to set a filter easily on those who are closed not matter what
the reason was nor who closed the account.
As I always encounter problems when dealing with ifelse statements and NA I
decided to merge these two variables to one variable containing 0 = not closed
and 1 = closed. In my context this seems to be - at least to me - a reasonable
approach.
Replacement of missing values and merging the variables is the easiest way for
me.
-- cut --
cust_id <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20)
closed_mdm <- c("01", NA, NA, NA, "08", "07", NA, NA, "05", NA, NA, NA, "04",
NA, NA, NA, NA, NA, NA, NA)
closed_sls <- c(NA, "08", NA, NA, "08", "07", NA, NA, NA, NA, "03", NA, NA, NA,
"05", NA, NA, NA, NA, NA)
# 1st try
ds_temp1 <- data.frame(cust_id, closed_mdm, closed_sls)
ds_temp1
ds_temp1$closed <- closed_mdm | closed_sls # WRONG
# 2nd try
closed_mdm_fac1 <- as.factor(closed_mdm)
closed_sls_fac1 <- as.factor(closed_sls)
ds_temp2 <- data.frame(cust_id, closed_mdm_fac1, closed_sls_fac1)
ds_temp2
ds_temp2$closed <- ds_temp$closed_mdm_fac1 | ds_temp$closed_sls_fac1 # WRONG
# 3rd try
closed_mdm_num1 <- as.numeric(closed_mdm) # OK
closed_sls_num1 <- as.numeric(closed_sls) # OK
ds_temp3 <- data.frame(cust_id, closed_mdm_num1, closed_sls_num1)
ds_temp3
ds_temp3$closed <- ds_temp$closed_mdm_num1 | ds_temp$closed_sls_num1 # WRONG
# 4th try
ds_temp4 <- ds_temp3
ds_temp4
# Does not run due to not allowed NA in subscripts
ds_temp4[is.na(ds_temp4$closed_mdm_num1), ds_temp4$closed_mdm_num1] <- 0
ds_temp4[is.na(ds_temp4$closed_sls_num1), ds_temp4$closed_sls_num1] <- 0
# 5th try
ds_temp4$closed_mdm_num1 <- ifelse(is.na(ds_temp4$closed_mdm_num1), 1, 0)
ds_temp4$closed_sls_num1 <- ifelse(is.na(ds_temp4$closed_sls_num1), 1, 0)
ds_temp4
ds_temp4$closed <- ifelse(ds_temp4$closed_mdm_num1 == 1 |
ds_temp4$closed_sls_num1 == 1, 1, 0)
ds_temp4
-- cut --
Is there a better way to do it?
Kind regards
Georg
> Gesendet: Donnerstag, 23. Juni 2016 um 23:55 Uhr
> Von: "Bert Gunter"
> An: "David L Carlson"
> Cc: "R Help"
> Betreff: Re: [R] Subscripting problem with is.na()
>
> ... actually, FWIW, I would say that this little discussion mostly
> demonstrates why the OP's request is probably not a good idea in the
> first place. Usually, NA's should be left as NA's to be dealt with
> properly by R and packages. In biological measurements, for example,
> NA's often mean "below the ability to reliably measure." Biologists
> with whom I've worked over many years often want to convert these to 0
> or omit the cases, both of which lead to biased estimates and/or
> underestimates of variability and excess claims of "statistical
> significance" (for those who belong to this religious persuasion). One
> should never say never, but I suspect that there are relatively few
> circumstances where the conversion the OP requested is actually wise.
>
> Feel free to ignore/reject such extraneous comments of course.
>
> Cheers,
> Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Thu, Jun 23, 2016 at 12:14 PM, David L Carlson wrote:
> > Good point. I did not think about factors. Also your example raises another
> > issue since column c is logical, but gets silently converted to numeric.
> > This would seem to get the job done assuming the conversion is intended for
> > numeric columns only:
> >
> >> test <- data.frame(a=c(1,NA,2), b = c("A","b",NA), c= rep(NA,3))
> >> sapply(test, class)
> > a b c
> > "numeric" "factor" "logical"
> >> num <- sapply(test, is.numeric)
> >> test[, num][is.na(test[, num])] <- 0
> >> test
> > ab c
> > 1 1A NA
> > 2 0b NA
> > 3 2 NA
> >
> > David C
> >
> > -Original Message-
> > From: Bert Gunter [mailto:bgunter.4...@gmail.com]
> > Sent: Thursday, June 23, 2016 1:48 PM
> > To: David L Carlson
> > Cc: Ivan Calandra; R Help
> > Subject: Re: [R] Subscripting problem with is.na()
> >
> > Not in general, David:
> >
> > e.g.
> >
> >> test <- data.frame(a=c(1,NA,2), b = c("A","b",NA), c= rep(NA,3))
> >
> >> is.na(test)
> > a bc
> > [1,] FALSE FALSE TRUE
> > [2,] TRUE FALSE TRUE
> > [3,] FALSE TRUE TRUE
> >
> >> test[is.na(test)]
> > [1] NA NA NA NA NA
> >
> >> test[is.na(test)
[R] Antwort: Fw: Re: Subscripting problem with is.na()
Hi David,
Hi Bert,
many thanks for the valuable discussion on NA in R (please see extract
below). I follow your arguments leaving NA as they are for most of the
time. In special occasions however I want to replace the NA with another
value. To preserve the newly acquired knowledge for me I wrote this
function:
-- cut --
t_replace_na <- function(dataset, variable, value) {
if(inherits(dataset[[variable]], "factor") == TRUE) {
dataset[variable] <- as.character(dataset[variable])
print(class(dataset[variable]))
dataset[, variable][is.na(dataset[, variable])] <- value
dataset[variable] <- as.factor(dataset[variable])
print(class(dataset[variable]))
} else {
dataset[, variable][is.na(dataset[, variable])] <- value
}
return(dataset)
}
ds_test <- data.frame(a=c(1,NA,2), b = rep(NA,3), c = c("A","b",NA))
print(sapply(ds_test, class))
t_replace_na(ds_test, "a", value = -1)
t_replace_na(ds_test, "b", value = -2)
t_replace_na(ds_test, "c", value = -3)
-- cut --
Unfortunately the if-statement does not work due to a wrong class
definition within the function. When finding out what is going on I did
this:
-- cut --
test_class <- function(dataset, variable) {
if(inherits(dataset[, variable], "factor") == TRUE) {
return(c(class(dataset[variable]), TRUE))
} else {
return(c(class(dataset[variable]), FALSE))
}
}
ds_test <- data.frame(a=c(1,NA,2), b = rep(NA,3), c = c("A","b",NA))
print(sapply(ds_test, class))
# -- Test a --
class(ds_test[, "a"])
if(inherits(ds_test[, "a"], "factor")) {
print(c(class(ds_test[, "a"]), "TRUE"))
} else {
print(c(class(ds_test[, "a"]), "FALSE"))
}
test_class(ds_test, "a")
warning("'a' should be numeric NOT data.frame!")
# -- Test b --
if(inherits(ds_test[, "b"], "factor")) {
print(c(class(ds_test[, "b"]), "TRUE"))
} else {
print(c(class(ds_test[, "b"]), "FALSE"))
}
class(ds_test[, "b"])
test_class(ds_test, "b")
warning("'b' should be logical NOT data.frame!")
# -- Test c --
if(inherits(ds_test[, "c"], "factor")) {
print(c(class(ds_test[, "c"]), "TRUE"))
} else {
print(c(class(ds_test[, "c"]), "FALSE"))
}
class(ds_test[, "c"])
test_class(ds_test, "c")
warning("'c' should be factor NOT data.frame.
In addition data.frame != factor")
-- cut --
Why do I get different results for the same function if it is inside or
outside my own function definition?
Kind regards
Georg
> Gesendet: Donnerstag, 23. Juni 2016 um 21:14 Uhr
> Von: "David L Carlson"
> An: "Bert Gunter"
> Cc: "R Help"
> Betreff: Re: [R] Subscripting problem with is.na()
>
> Good point. I did not think about factors. Also your example raises
another issue since column c is logical, but gets silently converted to
numeric. This would seem to get the job done assuming the conversion is
intended for numeric columns only:
>
> > test <- data.frame(a=c(1,NA,2), b = c("A","b",NA), c= rep(NA,3))
> > sapply(test, class)
> a b c
> "numeric" "factor" "logical"
> > num <- sapply(test, is.numeric)
> > test[, num][is.na(test[, num])] <- 0
> > test
> ab c
> 1 1A NA
> 2 0b NA
> 3 2 NA
>
> David C
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: RE: Antwort: Fw: Re: Subscripting problem with is.na()
Hi Petr,
many thanks for your reply and the examples.
My subscripting problems drive me nuts.
I have understood that dataset[variable] is semantically identical to
dataset[, variable] cause dataset[variable] takes all cases because no
other subscripts are given.
Where can I lookup the rules when to use the comma and when not?
Kind regards
Georg
Von:PIKAL Petr
An: "g.maub...@weinwolf.de" ,
Kopie: "r-help@r-project.org"
Datum: 27.06.2016 11:03
Betreff:RE: [R] Antwort: Fw: Re: Subscripting problem with
is.na()
Hi
see in line
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> g.maub...@weinwolf.de
> Sent: Monday, June 27, 2016 10:45 AM
> To: David L Carlson ; Bert Gunter
>
> Cc: r-help@r-project.org
> Subject: [R] Antwort: Fw: Re: Subscripting problem with is.na()
>
> Hi David,
> Hi Bert,
>
> many thanks for the valuable discussion on NA in R (please see extract
> below). I follow your arguments leaving NA as they are for most of the
> time. In special occasions however I want to replace the NA with another
> value. To preserve the newly acquired knowledge for me I wrote this
> function:
>
> -- cut --
> t_replace_na <- function(dataset, variable, value) {
> if(inherits(dataset[[variable]], "factor") == TRUE) {
>dataset[variable] <- as.character(dataset[variable])
>print(class(dataset[variable]))
>dataset[, variable][is.na(dataset[, variable])] <- value
>dataset[variable] <- as.factor(dataset[variable])
>print(class(dataset[variable]))
> } else {
>dataset[, variable][is.na(dataset[, variable])] <- value
> }
> return(dataset)
> }
>
> class(ds_test[, "c"])
> test_class(ds_test, "c")
> warning("'c' should be factor NOT data.frame.
> In addition data.frame != factor")
> -- cut --
>
> Why do I get different results for the same function if it is inside or
> outside my own function definition?
Because you still are missing the way how to subscript data frames.
test_class <- function(dataset, variable) {
if(inherits(dataset[, variable], "factor") == TRUE) {
return(c(class(dataset[,variable]), TRUE))
} else {
return(c(class(dataset[,variable]), FALSE))
##
}
}
> test_class(ds_test, "a")
[1] "numeric" "FALSE"
> test_class(ds_test, "c")
[1] "factor" "TRUE"
>
If you properly arrange commas in your function you get desired result
p_replace_na <- function(dataset, variable, value) {
if(inherits(dataset[,variable], "factor") == TRUE) {
dataset[,variable] <- as.character(dataset[,variable])
print(class(dataset[,variable]))
dataset[, variable][is.na(dataset[, variable])] <- value
dataset[, variable] <- as.factor(dataset[, variable])
print(class(dataset[, variable]))
} else {
dataset[, variable][is.na(dataset[, variable])] <- value
}
return(dataset)
}
> p_replace_na(ds_test, "c", value = -3)
[1] "character"
[1] "factor"
a b c
1 1 NA A
2 NA NA b
3 2 NA -3
> t_replace_na(ds_test, "c", value = -3)
[1] "data.frame"
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?
>
Cheers
Petr
>
> Kind regards
>
> Georg
>
>
>
> > Gesendet: Donnerstag, 23. Juni 2016 um 21:14 Uhr
> > Von: "David L Carlson"
> > An: "Bert Gunter"
> > Cc: "R Help"
> > Betreff: Re: [R] Subscripting problem with is.na()
> >
> > Good point. I did not think about factors. Also your example raises
> another issue since column c is logical, but gets silently converted to
> numeric. This would seem to get the job done assuming the conversion is
> intended for numeric columns only:
> >
> > > test <- data.frame(a=c(1,NA,2), b = c("A","b",NA), c= rep(NA,3))
> > > sapply(test, class)
> > a b c
> > "numeric" "factor" "logical"
> > > num <- sapply(test, is.numeric)
> > > test[, num][is.na(test[, num])] <- 0
> > > test
> > ab c
> > 1 1A NA
> > 2 0b NA
> > 3 2 NA
> >
> > David C
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
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[R] Antwort: RE: Antwort: Fw: Re: Subscripting problem with is.na()
Hi All,
Petr, Bert, David, Ivan, Duncan and Rui helped me to develop a function
able to replace NA's in variables IF NEEDED:
#---
# Module: t_replace_na.R
# Author: Georg Maubach
# Date : 2016-06-27
# Update: 2016-06-27
# Description : Replace NA with another value
# Source System : R 3.3.0 (64 Bit)
# Target System : R 3.3.0 (64 Bit)
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
#1-2-3-4-5-6-7-8
t_version = "2016-06-27"
t_module_name = "t_replace_na.R"
cat(
paste0("\n",
t_module_name, " (Version: ", t_version, ")", "\n", "\n",
"This software comes with ABSOLUTELY NO WARRANTY.",
"\n", "\n"))
# If do_test is not defined globally define it here locally by
un-commenting it
t_do_test <- FALSE
# [ Function Defintion
]
t_replace_na <- function(dataset, variables, value) {
# Replace NA with another given value
#
# Args:
# dataset (data frame, data table):
# Object with dimnames, e.g. data frame, data table.
# variables (character vector):
# List of variable names.
#
# Operation:
# NA is replaced by the value given with the parameter "value".
#
# A factor is converted explicitly with as.character(), the missing
value
# replacement is done and then the character vector is converted back
with
# as.factor(). Thus NA becomes a category of the new factor variable.
#
# Caution:
# Please check your data in case you replace NA within factors due to
# explicit type conversion. Tests were done only for the below given
# dataset.
#
# Returns:
# Original dataset.
#
# Error handling:
# None.
#
# Credits:
https://www.mail-archive.com/r-help@r-project.org/msg236537.html
for (variable in variables) {
if (inherits(dataset[, variable], "factor") == TRUE) {
dataset[, variable] <- as.character(dataset[, variable])
print(class(dataset[, variable]))
dataset[, variable][is.na(dataset[, variable])] <- value
dataset[, variable] <- as.factor(dataset[, variable])
print(class(dataset[, variable]))
} else {
dataset[, variable][is.na(dataset[, variable])] <- value
}
}
return(dataset)
}
# [ Test Defintion
]
t_test <- function(do_test = FALSE) {
if (do_test == TRUE) {
cat("\n", "\n", "Test function t_count_na()", "\n", "\n")
# Example dataset
ds_example <- data.frame(a=c(1,NA,2), b = rep(NA,3), c =
c("A","b",NA))
cat("\n", "\n", "Example dataset before function call", "\n", "\n")
cat("Variables and their classes:\n")
print(sapply(ds_example, class))
cat("Dataset:\n")
print(ds_example)
cat("\n", "\n", "Function call", "\n", "\n")
ds_result <- t_replace_na(ds_example, "a", value = -1)
cat("\n", "\n", "Dataset after function call", "\n", "\n")
print(ds_result)
cat("\n", "\n", "Function call", "\n", "\n")
ds_result <- t_replace_na(ds_example, "b", value = -2)
cat("\n", "\n", "Example dataset after function call", "\n", "\n")
print(ds_result)
cat("\n", "\n", "Function call", "\n", "\n")
ds_result <- t_replace_na(ds_example, "c", value = -3)
cat("\n", "\n", "Example dataset after function call", "\n", "\n")
print(ds_result)
}
}
# [ Test Run
]--
t_test(do_test = t_do_test)
# [ Clean up
]--
rm("t_module_name", "t_version", "t_do_test", "t_test")
# EOF .
Please note: R has capabilities to handle NA correctly. There is often no
need to recode NA. Also NA might or might not have meaning. You have to
decide with regard to the meaning of the original data and the business
problem.
Kind regards
Georg
Von:PIKAL Petr
An: "g.maub...@weinwolf.de" ,
Kopie: "r-help@r-project.org"
Datum: 27.06.2016 11:03
Betreff:RE: [R] Antwort: Fw: Re: Subscripting problem with
is.na()
Hi
see in line
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> g.maub...@weinwolf.de
> Sent: Monday, June 27, 2016 10:45 AM
> To: David L Carlson ; Bert Gunter
>
> Cc: r-help@r-project.org
> Subject: [R] Antwort: Fw: Re: Subscripting problem with is.na()
>
> Hi David,
> Hi Bert,
>
> many thanks for the valuable discussion on NA in R (please see extract
> below). I follow your arguments leaving NA as they are for most of the
> time. In special occasions however I want to replace the NA with another
> value. To preserve the newly acquired knowledge for me I wrote this
> function:
>
> --
[R] Installing from source on Windows 7: tibble
Hi All,
I would like to install R packages from source on Windows 7 64-Bit.
Currently my settings are:
-- cut --
> sessionInfo()
R version 3.3.0 (2016-05-03)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
locale:
[1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252
[3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C
[5] LC_TIME=German_Germany.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] tools_3.3.0
-- cut --
The environment variable PATH on Windows 7 is set to:
C:\R-Project\Rtools\mingw_32\bin;C:\R-Project\Rtools\mingw_64\bin;C:\R-Project\Rtools\bin;C:\R-Project\Rtools\gcc-4.6.3\bin;C:\Program
Files\Python 3.5\Scripts\;C:\Program Files\Python
3.5\;C:\Python27\;C:\Python27\Scripts; etc. etc.
RTools is installed in C:\R-Project\RTools
The call of
C:\R-Project\Rtools\mingw_64\bin\g++.exe --version
results in
g++ (x86_64-posix-seh, Built by MinGW-W64 project) 4.9.3
If I do
> install.packages("tibble", type = "source")
I get
-- cut --
trying URL 'https://cran.uni-muenster.de/src/contrib/tibble_1.0.tar.gz'
Content type 'application/x-gzip' length 38038 bytes (37 KB)
downloaded 37 KB
* installing *source* package 'tibble' ...
** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
** libs
*** arch - i386
c:/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include" -DNDEBUG
-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
-I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2 -c
RcppExports.cpp -o RcppExports.o
c:/Rtools/mingw_32/bin/g++: not found
make: *** [RcppExports.o] Error 127
Warnung: Ausführung von Kommando 'make -f
"C:/R-PROJ~1/R-33~1.0/etc/i386/Makeconf" -f
"C:/R-PROJ~1/R-33~1.0/share/make/winshlib.mk"
SHLIB_LDFLAGS='$(SHLIB_CXXLDFLAGS)' SHLIB_LD='$(SHLIB_CXXLD)'
SHLIB="tibble.dll" OBJECTS="RcppExports.o matrixToDataFrame.o"' ergab
Status 2
ERROR: compilation failed for package 'tibble'
* removing 'C:/R-Project/R-3.3.0/library/tibble'
* restoring previous 'C:/R-Project/R-3.3.0/library/tibble'
Warning in install.packages :
running command '"C:/R-PROJ~1/R-33~1.0/bin/x64/R" CMD INSTALL -l
"C:\R-Project\R-3.3.0\library"
C:\Users\MAUBAC~1.WEI\AppData\Local\Temp\Rtmp23SQxM/downloaded_packages/tibble_1.0.tar.gz'
had status 1
Warning in install.packages :
installation of package ‘tibble’ had non-zero exit status
-- cut --
There is no make.conf in "C:\R-Project\Rtools\mingw_64\etc". I found "
Makeconf" in "C:\R-Project\R-3.3.0\etc\x64". Do I need it? How do I need
to configure the settings in this file?
I searched old aunt Google but did not understand what to do and how to
configure R environment variables correctly.
What do I need to do to install packages from source?
Kind regards
Georg
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Installing from source on Windows 7: tibble
Hi Duncan,
many thanks for your reply.
I did insert die paths to the g++ compiler because I got the message about
the not existent compiler.
I took the directories for the compiler out again:
C:\R-Project\Rtools\bin;C:\ProgramData\Oracle\Java\javapath;C:\Program
Files\Python 3.5\Scripts\;C:\Program Files\Python
3.5\;C:\Python27\;C:\Python27\Scripts, etc. etc.
Calling
install.packages("tibble", type = "source")
gives this message:
-- cut --
* installing *source* package 'tibble' ...
** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
** libs
*** arch - i386
c:/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include" -DNDEBUG
-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
-I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2 -c
RcppExports.cpp -o RcppExports.o
c:/Rtools/mingw_32/bin/g++: not found
make: *** [RcppExports.o] Error 127
Warnung: Ausführung von Kommando 'make -f
"C:/R-PROJ~1/R-33~1.0/etc/i386/Makeconf" -f
"C:/R-PROJ~1/R-33~1.0/share/make/winshlib.mk"
SHLIB_LDFLAGS='$(SHLIB_CXXLDFLAGS)' SHLIB_LD='$(SHLIB_CXXLD)'
SHLIB="tibble.dll" OBJECTS="RcppExports.o matrixToDataFrame.o"' ergab
Status 2
ERROR: compilation failed for package 'tibble'
* removing 'C:/R-Project/R-3.3.0/library/tibble'
* restoring previous 'C:/R-Project/R-3.3.0/library/tibble'
Warning in install.packages :
running command '"C:/R-PROJ~1/R-33~1.0/bin/x64/R" CMD INSTALL -l
"C:\R-Project\R-3.3.0\library"
C:\Users\MAUBAC~1.WEI\AppData\Local\Temp\RtmpGqOlOW/downloaded_packages/tibble_1.0.tar.gz'
had status 1
Warning in install.packages :
installation of package ‘tibble’ had non-zero exit status
-- cut --
What else could I do?
Kind regards
Georg
Von:Duncan Murdoch
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 29.06.2016 13:07
Betreff:Re: [R] Installing from source on Windows 7: tibble
On 29/06/2016 5:49 AM, g.maub...@weinwolf.de wrote:
> Hi All,
>
> I would like to install R packages from source on Windows 7 64-Bit.
> Currently my settings are:
>
> -- cut --
>> sessionInfo()
> R version 3.3.0 (2016-05-03)
> Platform: x86_64-w64-mingw32/x64 (64-bit)
> Running under: Windows 7 x64 (build 7601) Service Pack 1
>
> locale:
> [1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252
> [3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C
> [5] LC_TIME=German_Germany.1252
>
> attached base packages:
> [1] stats graphics grDevices utils datasets methods base
>
> loaded via a namespace (and not attached):
> [1] tools_3.3.0
> -- cut --
>
> The environment variable PATH on Windows 7 is set to:
>
>
C:\R-Project\Rtools\mingw_32\bin;C:\R-Project\Rtools\mingw_64\bin;C:\R-Project\Rtools\bin;C:\R-Project\Rtools\gcc-4.6.3\bin;C:\Program
> Files\Python 3.5\Scripts\;C:\Program Files\Python
> 3.5\;C:\Python27\;C:\Python27\Scripts; etc. etc.
Take the mingw_32, mingw_64 and gcc-4.6.3 directories off your path.
They aren't needed; the first two could conceivably be harmful.
>
> RTools is installed in C:\R-Project\RTools
>
> The call of
>
> C:\R-Project\Rtools\mingw_64\bin\g++.exe --version
>
> results in
>
> g++ (x86_64-posix-seh, Built by MinGW-W64 project) 4.9.3
>
> If I do
>
>
>> install.packages("tibble", type = "source")
>
> I get
>
> -- cut --
> trying URL 'https://cran.uni-muenster.de/src/contrib/tibble_1.0.tar.gz'
> Content type 'application/x-gzip' length 38038 bytes (37 KB)
> downloaded 37 KB
>
> * installing *source* package 'tibble' ...
> ** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
> ** libs
>
> *** arch - i386
> c:/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include" -DNDEBUG
> -I"C:/R-Project/R-3.3.0/library/Rcpp/include"
> -I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2
-c
> RcppExports.cpp -o RcppExports.o
> c:/Rtools/mingw_32/bin/g++: not found
> make: *** [RcppExports.o] Error 127
> Warnung: Ausführung von Kommando 'make -f
> "C:/R-PROJ~1/R-33~1.0/etc/i386/Makeconf" -f
> "C:/R-PROJ~1/R-33~1.0/share/make/winshlib.mk"
> SHLIB_LDFLAGS='$(SHLIB_CXXLDFLAGS)' SHLIB_LD='$(SHLIB_CXXLD)'
> SHLIB="tibble.dll" OBJECTS="RcppExports.o matrixToDataFrame.o"' ergab
> Status 2
> ERROR: compilation failed for package 'tibble'
> * removing 'C:/R-Project/R-3.3.0/library/tibble'
> * restoring previous 'C:/R-Project/R-3.3.0/library/tibble'
> Warning in install.packages :
> running command '"C:/R-PROJ~1/R-33~1.0/bin/x64/R" CMD INSTALL -l
> "C:\R-Project\R-3.3.0\library"
>
C:\Users\MAUBAC~1.WEI\AppData\Local\Temp\Rtmp23SQxM/downloaded_packages/tibble_1.0.tar.gz'
> had status 1
> Warning in install.packages :
> installation of package ‘tibble’ had non-zero exit status
> -- cut --
>
> There is no make.conf in "C:\R-Project\Rtools\mingw_64\etc". I found "
> Makeconf" in "C:\R-Project\R-3.3.0\etc\x64". Do I need it? How do I need
> to configure the settings in this file?
Yes, since you haven't installed Rtools in the default location, you
should edit two Makeconf files. In
C:\R-Project\R-3.3.0\etc\x64
[R] Antwort: Re: Antwort: Re: Installing from source on Windows 7: tibble [SOLVED]
Hi Duncan,
indeed, I did not see the other part of your message.
I did
BINPREF ?= C:/R-Project/Rtools/mingw_32/bin/
COMPILED_BY = g++ # instead of gcc-4.9.3
in "C:\R-Project\R-3.3.0\etc\i386\Makeconf"
and
BINPREF ?= C:/R-Project/Rtools/mingw_64/bin/
COMPILED_BY = g++ # instead of gcc-4.9.3
in "C:\R-Project\R-3.3.0\etc\x64\Makeconf"
Now I could compile the package with no futher errors.
Messages are
-- cut --
* installing *source* package 'tibble' ...
** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
** libs
*** arch - i386
C:/R-Project/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
-DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
-I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2 -c
RcppExports.cpp -o RcppExports.o
C:/R-Project/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
-DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
-I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2 -c
matrixToDataFrame.cpp -o matrixToDataFrame.o
C:/R-Project/Rtools/mingw_32/bin/g++ -shared -s -static-libgcc -o
tibble.dll tmp.def RcppExports.o matrixToDataFrame.o
-Ld:/Compiler/gcc-4.9.3/local330/lib/i386
-Ld:/Compiler/gcc-4.9.3/local330/lib -LC:/R-PROJ~1/R-33~1.0/bin/i386 -lR
installing to C:/R-Project/R-3.3.0/library/tibble/libs/i386
*** arch - x64
C:/R-Project/Rtools/mingw_64/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
-DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
-I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2 -c
RcppExports.cpp -o RcppExports.o
C:/R-Project/Rtools/mingw_64/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
-DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
-I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2 -c
matrixToDataFrame.cpp -o matrixToDataFrame.o
C:/R-Project/Rtools/mingw_64/bin/g++ -shared -s -static-libgcc -o
tibble.dll tmp.def RcppExports.o matrixToDataFrame.o
-Ld:/Compiler/gcc-4.9.3/local330/lib/x64
-Ld:/Compiler/gcc-4.9.3/local330/lib -LC:/R-PROJ~1/R-33~1.0/bin/x64 -lR
installing to C:/R-Project/R-3.3.0/library/tibble/libs/x64
** R
** inst
** preparing package for lazy loading
** help
*** installing help indices
** building package indices
** installing vignettes
** testing if installed package can be loaded
*** arch - i386
*** arch - x64
* DONE (tibble)
-- cut --
So - complete success.
Many thanks for your help.
One last questions: Why did Rtools.exe not create a directory named
"gcc-4.9.3" in "C:\R-Project\Rtools" and putting "
C:\R-Project\Rtools\mingw_32" and "C:\R-Project\Rtools\mingw_64" directly
in "C:\R-Project\Rtools\"? gcc-4.6.3 was installed that way.
Kind regards
Georg
Von:Duncan Murdoch
An: g.maub...@weinwolf.de,
Kopie: r-help@r-project.org
Datum: 29.06.2016 16:21
Betreff:Re: Antwort: Re: [R] Installing from source on Windows 7:
tibble
On 29/06/2016 10:17 AM, g.maub...@weinwolf.de wrote:
> Hi Duncan,
>
> many thanks for your reply.
>
> I did insert die paths to the g++ compiler because I got the message
about
> the not existent compiler.
>
> I took the directories for the compiler out again:
>
> C:\R-Project\Rtools\bin;C:\ProgramData\Oracle\Java\javapath;C:\Program
> Files\Python 3.5\Scripts\;C:\Program Files\Python
> 3.5\;C:\Python27\;C:\Python27\Scripts, etc. etc.
>
> Calling
>
> install.packages("tibble", type = "source")
>
>
> gives this message:
>
> -- cut --
> * installing *source* package 'tibble' ...
> ** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
> ** libs
>
> *** arch - i386
> c:/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include" -DNDEBUG
> -I"C:/R-Project/R-3.3.0/library/Rcpp/include"
> -I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2
-c
> RcppExports.cpp -o RcppExports.o
> c:/Rtools/mingw_32/bin/g++: not found
> make: *** [RcppExports.o] Error 127
> Warnung: Ausführung von Kommando 'make -f
> "C:/R-PROJ~1/R-33~1.0/etc/i386/Makeconf" -f
> "C:/R-PROJ~1/R-33~1.0/share/make/winshlib.mk"
> SHLIB_LDFLAGS='$(SHLIB_CXXLDFLAGS)' SHLIB_LD='$(SHLIB_CXXLD)'
> SHLIB="tibble.dll" OBJECTS="RcppExports.o matrixToDataFrame.o"' ergab
> Status 2
> ERROR: compilation failed for package 'tibble'
> * removing 'C:/R-Project/R-3.3.0/library/tibble'
> * restoring previous 'C:/R-Project/R-3.3.0/library/tibble'
> Warning in install.packages :
>running command '"C:/R-PROJ~1/R-33~1.0/bin/x64/R" CMD INSTALL -l
> "C:\R-Project\R-3.3.0\library"
>
C:\Users\MAUBAC~1.WEI\AppData\Local\Temp\RtmpGqOlOW/downloaded_packages/tibble_1.0.tar.gz'
> had status 1
> Warning in install.packages :
>installation of package ‘tibble’ had non-zero exit status
> -- cut --
>
> What else could I do?
You seem to have missed the second part of my advice, describing what to
do with the two Makeconf files.
Duncan Murdoch
>
> Kind regards
>
> Georg
>
>
>
>
>
> Von:Duncan Murdoch
> An: g.maub...@weinwolf.de, r-help@r-project.org,
> Datum: 29.06.2016 13:07
>
[R] Antwort: Re: Antwort: Re: Antwort: Re: Installing from source on Windows 7: tibble [RE OPENED]
Hi Duncan,
I would not have changed the COMPILED_BY option unless I thought I have
to.
In my "C:\R-Project\Rtools\mingw_32\bin" I have
c++.exe
g++.exe
gcc.exe
i686-w64-mingw32-c++.exe
i686-w64-mingw32-g++.exe
i686-w64-mingw32-gcc-4.9.3.exe
i686-w64-mingw32-gcc.exe
In my "C:\R-Project\Rtools\mingw_64\bin" I have
c++.exe
cpp.exe
g++.exe
gcc.exe
x86_64-w64-mingw32-c++.exe
x86_64-w64-mingw32-g++.exe
x86_64-w64-mingw32-gcc-4.9.3.exe
x86_64-w64-mingw32-gcc.exe
Which one should I configure and use?
Kind regards
Georg
Von:Duncan Murdoch
An: g.maub...@weinwolf.de,
Kopie: r-help@r-project.org
Datum: 29.06.2016 17:34
Betreff:Re: Antwort: Re: Antwort: Re: [R] Installing from source
on Windows 7: tibble [SOLVED]
On 29/06/2016 10:48 AM, g.maub...@weinwolf.de wrote:
> Hi Duncan,
>
> indeed, I did not see the other part of your message.
>
> I did
>
> BINPREF ?= C:/R-Project/Rtools/mingw_32/bin/
> COMPILED_BY = g++ # instead of gcc-4.9.3
I wouldn't change the COMPILED_BY; some packages use it to configure
themselves for gcc-4.9.3, as opposed to the previous version gcc-4.6.3.
>
> in "C:\R-Project\R-3.3.0\etc\i386\Makeconf"
>
> and
>
> BINPREF ?= C:/R-Project/Rtools/mingw_64/bin/
> COMPILED_BY = g++ # instead of gcc-4.9.3
>
> in "C:\R-Project\R-3.3.0\etc\x64\Makeconf"
>
> Now I could compile the package with no futher errors.
>
> Messages are
>
> -- cut --
> * installing *source* package 'tibble' ...
> ** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
> ** libs
>
> *** arch - i386
> C:/R-Project/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
> -DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
> -I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2
-c
> RcppExports.cpp -o RcppExports.o
> C:/R-Project/Rtools/mingw_32/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
> -DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
> -I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2
-c
> matrixToDataFrame.cpp -o matrixToDataFrame.o
> C:/R-Project/Rtools/mingw_32/bin/g++ -shared -s -static-libgcc -o
> tibble.dll tmp.def RcppExports.o matrixToDataFrame.o
> -Ld:/Compiler/gcc-4.9.3/local330/lib/i386
> -Ld:/Compiler/gcc-4.9.3/local330/lib -LC:/R-PROJ~1/R-33~1.0/bin/i386 -lR
> installing to C:/R-Project/R-3.3.0/library/tibble/libs/i386
>
> *** arch - x64
> C:/R-Project/Rtools/mingw_64/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
> -DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
> -I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2
-c
> RcppExports.cpp -o RcppExports.o
> C:/R-Project/Rtools/mingw_64/bin/g++ -I"C:/R-PROJ~1/R-33~1.0/include"
> -DNDEBUG-I"C:/R-Project/R-3.3.0/library/Rcpp/include"
> -I"d:/Compiler/gcc-4.9.3/local330/include" -O2 -Wall -mtune=core2
-c
> matrixToDataFrame.cpp -o matrixToDataFrame.o
> C:/R-Project/Rtools/mingw_64/bin/g++ -shared -s -static-libgcc -o
> tibble.dll tmp.def RcppExports.o matrixToDataFrame.o
> -Ld:/Compiler/gcc-4.9.3/local330/lib/x64
> -Ld:/Compiler/gcc-4.9.3/local330/lib -LC:/R-PROJ~1/R-33~1.0/bin/x64 -lR
> installing to C:/R-Project/R-3.3.0/library/tibble/libs/x64
> ** R
> ** inst
> ** preparing package for lazy loading
> ** help
> *** installing help indices
> ** building package indices
> ** installing vignettes
> ** testing if installed package can be loaded
> *** arch - i386
> *** arch - x64
> * DONE (tibble)
> -- cut --
>
> So - complete success.
>
> Many thanks for your help.
>
> One last questions: Why did Rtools.exe not create a directory named
> "gcc-4.9.3" in "C:\R-Project\Rtools" and putting"
> C:\R-Project\Rtools\mingw_32" and "C:\R-Project\Rtools\mingw_64"
directly
> in "C:\R-Project\Rtools\"? gcc-4.6.3 was installed that way.
The 4.6.3 compiler was compiled for "multilib" operation: the same
compiler took command line options to distinguish between 32 bit and 64
bit compiles. The newer version doesn't support that, so we need two
separate installs.
Duncan Murdoch
> Kind regards
>
> Georg
>
>
>
>
>
> Von:Duncan Murdoch
> An: g.maub...@weinwolf.de,
> Kopie: r-help@r-project.org
> Datum: 29.06.2016 16:21
> Betreff:Re: Antwort: Re: [R] Installing from source on Windows
7:
> tibble
>
>
>
> On 29/06/2016 10:17 AM, g.maub...@weinwolf.de wrote:
> > Hi Duncan,
> >
> > many thanks for your reply.
> >
> > I did insert die paths to the g++ compiler because I got the message
> about
> > the not existent compiler.
> >
> > I took the directories for the compiler out again:
> >
> > C:\R-Project\Rtools\bin;C:\ProgramData\Oracle\Java\javapath;C:\Program
> > Files\Python 3.5\Scripts\;C:\Program Files\Python
> > 3.5\;C:\Python27\;C:\Python27\Scripts, etc. etc.
> >
> > Calling
> >
> > install.packages("tibble", type = "source")
> >
> >
> > gives this message:
> >
> > -- cut --
> > * installing *source* package 'tibble' ...
> > ** Paket 'tibble' erfolgreich entpackt und MD5 Summen überprüft
> > ** libs
> >
> > *** arch - i386
> > c:/Rt
[R] Writing a formula to Excel
Hi All,
I am using excel.link to work seemslessly with Excel.
In addition to values, like numbers and strings, I would like to insert a
full operational formula into a cell.
xlc["G14"] <- print(paste("=G9*100/G6"), quote = FALSE)
The strings is put into the cell, but the cell is not evaluated. Thus the
string is show as result of the computation.
If I open that cell b pressing "F2" or by double-clicking the cell and
pressing RETURN will start the evaluation of the expession.
xlc["G14"] <- parse("=G9*100/G6") # does not run
How can I put a formula into Excel that is evaluated right away?
Kind regards
Georg
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Documenting data
Hi Pito,
Dear Readers,
as other have already mentioned, there are good practices for documenting code
and data. I would like to summarize them and add a few not mentioned earlier:
1. You should have always two things: your raw data and your R script/s. The
raw data is immutable whereas the R script/s produce the results.
2. You might want to distinguish between documentating your CODE and
documenting your DATA. Documenting code is similar to what you already know
from your programmng experiences. Documenting data is somewhat different cause
you store information about the meaning of you data directly in your data.
Example
You have a variable with codes ranging from 1 to 5. But what do they mean?
Perhaps it could be
1 = Strongly agree
2 = Agree
3 = Neither agree/nor disagree
4 = Disagree
5 = Strongly Disagree
But it could also be the other way round:
1 = Strongly Disagree
2 = Disagree
3 = Nether agree/nor disagree
4 = Agree
5 = Strongly Agree
What the codes in your variable means depends on the systems oder processes you
derived your data from.
Within R there are some limitations for storing the informtation about what a
variable or a value within a variable means. Possibilities to store this
information is in other software packages like SAS or SPSS much broader
implemented. In R you can work with meaningful variable names and the data
type/class factor which can store mappings between values and value
descriptions.
Example
-- cut --
var1 <- c(rep(1:5, 3))
ds_example <- data.frame(var1)
var1_labels <- c("1 = Strongly Agree",
"2 = Agree",
"3 = Neither agree/nor disagree",
"4 = Disagree",
"5 = Strongly disagree")
ds_example[["var1"]] <- factor(ds_example[["var1"]],
levels = c(1, 2, 3, 4, 5),
labels = var1_labels)
summary(ds_example["var1"])
-- cut --
In addition you find methods to work with variable labels and value labels in
the pacakges Hmisc and memisc. They can also produce a thing called codebook
which contains all variable names, variable labels, values, value labels and
summaries of the distribution of values within the variables.
3. In addition to this you could structure your script in a modular way
according to the analysis process, e. g.
importing, cleaning, preparation for analysis, analysis, reporting. Other
structure may be more sufficient in your case. These modules could have a
number in the file name indicating in which sequence the scripts should be run.
4. I find it valuable to use a software repository like Github, Sourceforge or
others to keep the revisions save and seucre in case you would like to go back
to a version with code you deleted before and figure out that you need it now
again. The R Studio IDE has an interface to git if you like to go with that.
Good commit message can help you track what has changed. Commits also help you
to prepare precise steps when developing your scripts.
5. I have no experience with Sweave or knitr but you could also compile a
simple documentation through copying comments to an Excel sheet using R-2-Excel
libraries like excel.link or others.
Example
install.packages("excel.link")
library(excel.link)
xlc["A1"] <- "Project Documentation"
xlc["A2"] <- "Step XY"
xlc["A3"] <- "Some explanation about step xy"
This way you have the documentation in your code and in an external source.
Which approach you chose depends on your experience with R and its libraries as
well as the size of your project and the need for documentation.
6. It can be helpful to store interim results in a format that can be read by
non-R-users, e. g. Excel.
7. Documenting code can be done using roxygen2.
If there are different opinions to my suggestions please say so.
Kind regards
Georg
> Gesendet: Donnerstag, 30. Juni 2016 um 16:51 Uhr
> Von: "Pito Salas"
> An: r-help@r-project.org
> Betreff: [R] Documenting data
>
> I am studying statistics and using R in doing it. I come from software
> development where we document everything we do.
>
> As I “massage” my data, adding columns to a frame, computing on other data,
> perhaps cleaning, I feel the need to document in detail what the meaning, or
> background, or calculations, or whatever of the data is. After all it is now
> derived from my raw data (which may have been well documented) but it is
> “new.”
>
> Is this a real problem? Is there a “best practice” to address this?
>
> Thanks!
>
> Pito Salas
> Brandeis Computer Science
> Feldberg 131
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE
Re: [R] Documenting data
Hi Bert,
Hi Readers,
I did not know much about attributes in R and how to use them. If it is that
flexible you are right and I have learnt something.
Kind regards
Georg
> Gesendet: Donnerstag, 30. Juni 2016 um 20:06 Uhr
> Von: "Bert Gunter"
> An: g.maub...@gmx.de
> Cc: "Pito Salas" , "R Help"
> Betreff: Re: [R] Documenting data
>
> I believe Georg's pronouncements are wrong. See inline below.
>
> -- Bert
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> "...
>
> > Within R there are some limitations for storing the informtation about what
> > a variable or a value within a variable means.
>
> That is FALSE. There are no limitations. For example, just attach a
> "doc" attribute to your data that says whatever you wish to about
> them. e.g.
>
> > somedata <- runif(10)
> > attr(somedata,"doc") <- "Anything you want to say about the data"
>
> > attr(somedata,"doc")
> [1] "Anything you want to say about the data"
>
>
> You can go as crazy as you want to with this, e.g. creating a (S3 or
> S4 )class "documented" with appropriate methods for printing it from
> classes that inherit from data frames, lists, etc. See also the
> roxygen2 package for data documentation and R's ?promptData function
> for data documentation file in Rd format.
>
> R is Turing complete -- so it can do anything any other programming
> language can do. You could program SAS in R if you wanted. The
> difference is that SAS has pre-programmed some capabilities that R
> leaves for users, including contributed packages -- like Sweave,
> knitr, etc. You may or may not like this extra flexibility (and extra
> work, depending on whether someone else has already done the work for
> you), and efficiency may or may not be an issue; but to say that R has
> "limitations" is a gross misrepresentation, imho.
>
>
>
> Possibilities to store this information is in other software packages
> like SAS or SPSS much broader implemented. In R you can work with
> meaningful variable names and the data type/class factor which can
> store mappings between values and value descriptions.
> >
> > Example
> > -- cut --
> > var1 <- c(rep(1:5, 3))
> > ds_example <- data.frame(var1)
> >
> > var1_labels <- c("1 = Strongly Agree",
> > "2 = Agree",
> > "3 = Neither agree/nor disagree",
> > "4 = Disagree",
> > "5 = Strongly disagree")
> >
> > ds_example[["var1"]] <- factor(ds_example[["var1"]],
> >levels = c(1, 2, 3, 4, 5),
> >labels = var1_labels)
> >
> > summary(ds_example["var1"])
> > -- cut --
> >
> > In addition you find methods to work with variable labels and value labels
> > in the pacakges Hmisc and memisc. They can also produce a thing called
> > codebook which contains all variable names, variable labels, values, value
> > labels and summaries of the distribution of values within the variables.
> >
> > 3. In addition to this you could structure your script in a modular way
> > according to the analysis process, e. g.
> > importing, cleaning, preparation for analysis, analysis, reporting. Other
> > structure may be more sufficient in your case. These modules could have a
> > number in the file name indicating in which sequence the scripts should be
> > run.
> >
> > 4. I find it valuable to use a software repository like Github, Sourceforge
> > or others to keep the revisions save and seucre in case you would like to
> > go back to a version with code you deleted before and figure out that you
> > need it now again. The R Studio IDE has an interface to git if you like to
> > go with that. Good commit message can help you track what has changed.
> > Commits also help you to prepare precise steps when developing your scripts.
> >
> > 5. I have no experience with Sweave or knitr but you could also compile a
> > simple documentation through copying comments to an Excel sheet using
> > R-2-Excel libraries like excel.link or others.
> >
> > Example
> > install.packages("excel.link")
> > library(excel.link)
> > xlc["A1"] <- "Project Documentation"
> > xlc["A2"] <- "Step XY"
> > xlc["A3"] <- "Some explanation about step xy"
> >
> > This way you have the documentation in your code and in an external source.
> >
> > Which approach you chose depends on your experience with R and its
> > libraries as well as the size of your project and the need for
> > documentation.
> >
> > 6. It can be helpful to store interim results in a format that can be read
> > by non-R-users, e. g. Excel.
> >
> > 7. Documenting code can be done using roxygen2.
> >
> > If there are different opinions to my suggestions please say so.
> >
> > Kind regards
> >
> > Georg
> >
> >
> >> Gesendet: Donnerstag, 30. Juni 2016 um 16:51 Uhr
> >> Von: "Pito Salas"
> >> An: r-help@r-project.org
> >> Betreff
[R] Dump of new Methods
Dear Readers,
Hi All,
to drive my R knowlegde a bit further I followed the advice of some of you
by reading Chambers: Programming with data.
I tried some examples from the book:
-- cut --
setClass("track", representation (x = "numeric",
y = "numeric"))
track <- function(x, y) {
# an object representing measurements 'y', tracked at positions 'x'
x <- as(x, "numeric")
y <- as(y, "numeric")
if(length(x) != length(y)) {
stop("x, y should have equal length!")
}
new("track", x = x, y = y)
}
dumpMethod("track", "track")
setMethod("show", "track",
function(object) {
xy = rbind(object@x, object@y)
dimanmes(xy) = list(c("x", "y"),
1:ncol(y))
show(xy)
})
setMethod("plot",
signature(x = "track", y = "missing"),
function(x, y, ...)
plot(unclass(x), xlab = "Position", ylab = "Value", ...)
)
dumpMethod("plot", "track")
-- cut --
Where do I find the dumped data? Is it in a single file or is every dump
stored in a separate file? Where is it stored on my drive?
Kind regards
Georg
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Dump of new Methods (SOLVED)
Hi Bert,
many thanks.
Found them.
Kind regards
Georg
Von:Bert Gunter
An: g.maub...@weinwolf.de,
Datum: 04.07.2016 16:43
Betreff:Re: [R] Dump of new Methods
?getwd
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Mon, Jul 4, 2016 at 1:34 AM, wrote:
> Dear Readers,
> Hi All,
>
> to drive my R knowlegde a bit further I followed the advice of some of
you
> by reading Chambers: Programming with data.
>
> I tried some examples from the book:
>
> -- cut --
>
> setClass("track", representation (x = "numeric",
> y = "numeric"))
>
> track <- function(x, y) {
> # an object representing measurements 'y', tracked at positions 'x'
> x <- as(x, "numeric")
> y <- as(y, "numeric")
> if(length(x) != length(y)) {
> stop("x, y should have equal length!")
> }
> new("track", x = x, y = y)
> }
>
> dumpMethod("track", "track")
>
> setMethod("show", "track",
> function(object) {
> xy = rbind(object@x, object@y)
> dimanmes(xy) = list(c("x", "y"),
> 1:ncol(y))
> show(xy)
> })
>
> setMethod("plot",
> signature(x = "track", y = "missing"),
> function(x, y, ...)
> plot(unclass(x), xlab = "Position", ylab = "Value", ...)
> )
>
> dumpMethod("plot", "track")
>
> -- cut --
>
> Where do I find the dumped data? Is it in a single file or is every dump
> stored in a separate file? Where is it stored on my drive?
>
> Kind regards
>
> Georg
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: dplyr : row total for all groups in dplyr summarise
Hi guys,
I checked out your example but I can't follow the results.:
> mtcars %>%
+ group_by (am, gear) %>%
+ summarise (n=n()) %>%
+ mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%")) %>%
+ ungroup() %>%
+ mutate(row.tot = sum(n))
Source: local data frame [4 x 5]
am gear n rel.freq row.tot
(dbl) (dbl) (int)(chr) (int)
1 0 315 79% 32
2 0 4 4 21% 32
3 1 4 8 62% 32
4 1 5 5 38% 32
We have a total of 32 cases and 15 * 100 / 32 = 48,9 % instead of 79 %.
The same with the other columns. How is 79 % calculated?
When searching the web I saw this example:
-- cut --
#-- not run --
url <- "http://www.lock5stat.com/datasets/HollywoodMovies2011.csv";
response <- GET(url)
Hollywoodmovies2011 <- content(x = GET(url), as = data.frame)
#-- end not run
Hollywoodmovies2011 %>%
group_by(genre) %>%
summarize(count = n()) %>%
mutate(rf = count / sum(count))
-- cut --
which gives
Source: local data frame [9 x 3]
Genre count %
(fctr) (int) (dbl)
1Action32 0.235294118
2 Adventure 1 0.007352941
3 Animation12 0.088235294
4Comedy27 0.198529412
5 Drama21 0.154411765
6 Fantasy 2 0.014705882
7Horror17 0.12500
8 Romance11 0.080882353
9 Thriller13 0.095588235
Here the % correspond to the count and the sum of count, e. g. sum = 136
and 32 / 136 = 0,2352941.
What is the difference when counting? What do the relative counts in the
first example mean?
Kind regards
Georg
Von:Ulrik Stervbo
An: David Winsemius ,
Kopie: r-help@r-project.org, mai...@infomed.sld.cu
Datum: 05.07.2016 06:06
Betreff:Re: [R] dplyr : row total for all groups in dplyr
summarise
Gesendet von: "R-help"
That will give you the wrong result when used on summarised data
David Winsemius schrieb am Di., 5. Juli 2016
02:10:
> I thought there was an nrow() function?
>
> Sent from my iPhone
>
> On Jul 4, 2016, at 9:59 AM, Ulrik Stervbo
wrote:
>
> If you want the total number of rows in the original data.frame after
> counting the rows in each group, you can ungroup and sum the row counts,
> like:
>
> library("dplyr")
>
>
> mtcars %>%
>group_by (am, gear) %>%
>summarise (n=n()) %>%
>mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%")) %>%
>ungroup() %>%
>mutate(row.tot = sum(n))
>
> HTH
> Ulrik
>
> On Mon, 4 Jul 2016 at 18:23 David Winsemius
> wrote:
>
>>
>> > On Jul 4, 2016, at 6:56 AM, mai...@infomed.sld.cu wrote:
>> >
>> > Hello,
>> > How can I aggregate row total for all groups in dplyr summarise ?
>>
>> Row total … of what? Aggregate … how? What is the desired answer?
>>
>>
>>
>> > library(dplyr)
>> > mtcars %>%
>> > group_by (am, gear) %>%
>> > summarise (n=n()) %>%
>> > mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%"))
>> >
>> > best regard
>> > Maicel Monzon
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > --
>> > Este mensaje le ha llegado mediante el servicio de correo electronico
>> que ofrece Infomed para respaldar el cumplimiento de las misiones del
>> Sistema Nacional de Salud. La persona que envia este correo asume el
>> compromiso de usar el servicio a tales fines y cumplir con las
regulaciones
>> establecidas
>> >
>> > Infomed: http://www.sld.cu/
>> >
>> > __
>> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
[[alternative HTML version deleted]]
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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[R] WG: Fw: Re: dplyr : row total for all groups in dplyr summarise
Hi All,
if I run the suggested code
mtcars %>%
group_by (am, gear) %>%
summarise (n = n()) %>%
mutate(rel.freq = paste0(round(100 * n / sum(n), 0), "%")) %>%
ungroup() %>%
plyr::rbind.fill(data.frame(n = nrow(mtcars), rel.freq =
"100%”))
I get
> mtcars %>%
+ group_by (am, gear) %>%
+ summarise (n = n()) %>%
+ mutate(rel.freq = paste0(round(100 * n / sum(n), 0), "%")) %>%
+ ungroup() %>%
+ plyr::rbind.fill(data.frame(n = nrow(mtcars), rel.freq =
+ "100%”))
+
R stops execution cause something within the prgram syntax is missing.
What has to be changed to be able to run the code?
Kind regards
Georg Maubach
> Gesendet: Dienstag, 05. Juli 2016 um 18:30 Uhr
> Von: "David Winsemius"
> An: mai...@infomed.sld.cu
> Cc: r-help@r-project.org
> Betreff: Re: [R] dplyr : row total for all groups in dplyr summarise
>
>
>
> mtcars %>%
>group_by (am, gear) %>%
>summarise (n=n()) %>%
>mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%")) %>%
>ungroup() %>% plyr::rbind.fill(data.frame(
n=nrow(mtcars),rel.freq="100%”))
>
>
> > On Jul 5, 2016, at 4:47 AM, mai...@infomed.sld.cu wrote:
> >
> > Sorry, what I wanted to do was to add a total row at the end of the
summary. The marginal totals by columns correspond to 100% and the sum of
levels.
> > best reagard
> > Maicel Monzon
> >
> >
> > Ulrik Stervbo escribió:
> >
> >> Yes. But in the sample code the data is summarised. In which case you
get 4
> >> rows and not the correct 32.
> >>
> >> On Tue, 5 Jul 2016, 07:48 David Winsemius,
wrote:
> >>
> >>> nrow(mtcars)
> >>>
> >>>
> >>> Sent from my iPhone
> >>>
> >>> On Jul 4, 2016, at 9:03 PM, Ulrik Stervbo
wrote:
> >>>
> >>> That will give you the wrong result when used on summarised data
> >>>
> >>> David Winsemius schrieb am Di., 5. Juli
2016
> >>> 02:10:
> >>>
> I thought there was an nrow() function?
>
> Sent from my iPhone
>
> On Jul 4, 2016, at 9:59 AM, Ulrik Stervbo
> wrote:
>
> If you want the total number of rows in the original data.frame
after
> counting the rows in each group, you can ungroup and sum the row
counts,
> like:
>
> library("dplyr")
>
>
> mtcars %>%
> group_by (am, gear) %>%
> summarise (n=n()) %>%
> mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%")) %>%
> ungroup() %>%
> mutate(row.tot = sum(n))
>
> HTH
> Ulrik
>
> On Mon, 4 Jul 2016 at 18:23 David Winsemius
> wrote:
>
> >
> > > On Jul 4, 2016, at 6:56 AM, mai...@infomed.sld.cu wrote:
> > >
> > > Hello,
> > > How can I aggregate row total for all groups in dplyr summarise
?
> >
> > Row total ? of what? Aggregate ? how? What is the desired answer?
> >
> >
> >
> > > library(dplyr)
> > > mtcars %>%
> > > group_by (am, gear) %>%
> > > summarise (n=n()) %>%
> > > mutate(rel.freq = paste0(round(100 * n/sum(n), 0), "%"))
> > >
> > > best regard
> > > Maicel Monzon
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > --
> > > Este mensaje le ha llegado mediante el servicio de correo
electronico
> > que ofrece Infomed para respaldar el cumplimiento de las misiones
del
> > Sistema Nacional de Salud. La persona que envia este correo asume
el
> > compromiso de usar el servicio a tales fines y cumplir con las
regulaciones
> > establecidas
> > >
> > > Infomed: http://www.sld.cu/
> > >
> > > __
> > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more,
see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible
code.
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
> >>
> >
> >
> >
> >
> > This message was sent using IMP, the Internet Messaging Program.
> >
> >
> >
> > --
> > Este mensaje le ha llegado mediante el servicio de correo electronico
que ofrece Infomed para respaldar el cumplimiento de las misiones del
Sistema Nacional de Salud. La persona que envia este correo asume el
compromiso de usar el servicio a tales fines y cumplir con las
regulaciones establecidas
> >
> > Infomed: http://www.sld.cu/
> >
>
[R] Formatting ggplot2 graph
Hi All,
my current code looks lke this:
freq_ls <- structure(list(Var1 = c("zldkkd", "aakdkdk",
"aaakdkd", "aaieiwo", "vöalsl",
"ssddkdk",
"glowowp", "laoiw", "ruklow",
"rolsl",
"delk", "inslvnz"), Anzahl = c(1772L,
761L,
536L, 317L, 197L, 160L, 30L, 20L, 10L, 6L, 6L, 1L), Prozent = c(46.4,
19.9, 14,
8.3, 5.2, 4.2, 0.8, 0.5, 0.3, 0.2, 0.2, 0)), .Names = c("Var1",
"Anzahl",
"Prozent"), class = c("tbl_df", "data.frame"), row.names = c(NA,
-12L))
ggplot(freq_ls) +
geom_bar(aes(x = Var1,
y = Anzahl),
stat = "identity",
fill = "gray") +
theme(axis.text.x = element_text(angle = 45, hjust = 1)) +
ggtitle("Title of the Plot")
I would like to add the abolute and relative frequencies on top of the
bars. In addition I want the values printed in descending ording according
to the data.
I searched the web and found:
geom_text(stat='bin',aes(label=..count..),vjust=-1)
(Source:
http://stackoverflow.com/questions/26553526/how-to-add-frequency-count-labels-to-the-bars-in-a-bar-graph-using-ggplot2
)
but this does not work in my case. Inserting the code
ggplot(freq_ls) +
geom_bar(aes(x = Var1,
y = Anzahl),
stat = "identity",
fill = "gray") +
geom_text(stat='bin',aes(label=..count..),vjust=-1) +
theme(axis.text.x = element_text(angle = 45, hjust = 1)) +
ggtitle("Title of the Plot")
results in
`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Warning messages:
1: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
2: Removed 1 rows containing missing values (geom_text).
I looked in the book Wickhan: ggplot2 but could find an answer to the
question:
- How to show number if tey are pre-calculated?
- How to sort the bars according to the sequence of values in descending
order or if - pre-ordered - in the given order?
What do I have to change in my code to do it?
Kind regards
Georg
__
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and provide commented, minimal, self-contained, reproducible code.
[R] Choropleth: Turnover by ZipCode
Hi All,
Dear Readers,
I need to create a choropleth graph with turnover by zipcode. This is what
I have so far:
# Not run (Begin)
# Install packages if needed
# install.packages(pkgs = c("maptools", "rgdal", "RColorBrewer",
"grDevices"))
# Not run (End)
# Load libraries
library(maptools); library(rgdal); library(RColorBrewer);
library(grDevices)
# Configuration
# Adjust if needed!
file_path <- file.path("C:", "temp")
# Read data
# Source: http://arnulf.us/PLZ
url <- "http://www.metaspatial.net/download/plz.tar.gz";
file_name_gzip <- basename(url)
file_name_extract <- "post_pl.shp"
download.file(url, file.path(file_path, file_name_gzip))
untar(tarfile = file.path(file_path, file_name_gzip),
compressed = "gzip",
exdir = file_path)
# Dataset
# I have the data for all zipcodes available in my region
ds_temp <-
structure(
list(
ZipCode = c(1099, 10178, 13125, 21406, 32429, 41569),
Sales = c(4, 2, 9, 5, 7, 3),
Revenue = c(12, 9, 100, 80, 90,
25)
),
.Names = c("ZipCode", "Sales", "Revenue"),
row.names = c(NA,
6L),
class = "data.frame"
)
print(ds_temp)
# Prepare graphic
file_name_pdf <- file.path(file_path, "sales-and-revenue-by-zipcodes.pdf")
cairo_pdf(bg = "grey98", file_name_pdf, width = 16, height = 9)
y <- readShapeSpatial(file.path(file_path, file_name_extract),
proj4string = CRS("+proj=longlat"))
x <- spTransform(y,CRS=CRS("+proj=merc"))
# How do I need to change this line?
# Needs to be replaced by turnover from ds_temp
color <- sample(1:7, length(x), replace=T)
# Create graphic
plot(x,
col = brewer.pal(7, "Oranges")[color],
border = F) # How to I tell R to plot turnover from ds_temp?
# Title
mtext(
"Turnover by Zipcodes",
side = 3,
line = -4,
adj = 0,
cex = 1.7
)
# Write to disc
dev.off()
# Cleanup
rm("ds_temp", "color", "file_name_extract",
"file_name_gzip", "file_name_pdf", "file_path",
"url", "x", "y")
unlink(file.path(file_path, "plz.tar.gz"))
unlink(file.path(file_path, "post_pl.dbf"))
unlink(file.path(file_path, "post_pl.shp"))
unlink(file.path(file_path, "post_pl.shx"))
# unlink(file.path(file_path, "sales-and-revenue-by-zipcodes.pdf"))
What do I need to do to color the amount of turnover or the frequencies of
sales from the ds_temp dataset in the graph?
Kind regards
Georg Maubach
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and provide commented, minimal, self-contained, reproducible code.
[R] R Toolbox (Release 2 of 2016-07-21)
Hi All,
I have uploaded a new release of the R Toolbox.
R Toolbox is a collection of simple but useful functions which I developed
for myself to shorten the develoment process. Currently all functions use
base R. No other packages are needed. One exception is "t_openxlsx" cause
this module deals explicitly with the openxlsx package.
It is simple to install the functions. Just copy them to an appropriety
place on your hard disk and adjust the variable "t_toolbox_location" to
the place you stored the toolbox in. Running "r_toolbox.R" from that
location will load all modules.
In addition to new functions (see Release Comparison below) some functions
were improved. The are called with their package names, e. g.
openxlsx::read.xlsx() instead of "read.xlsx()". This way confusion with
functions having the same name but comming from other packages is avoided.
Pleae be aware that I have include some not tested function in this
release. All modules have a variable "t_status" now, stating the
development status, e. g. "development", "testing", "release".
Here is a Releae Comparison:
-- cut --
release_comparison <-
structure(list(Module = c("r_toolbox.R", "t_adjust_packages.R",
"t_conventions.r",
"t_create_variable.R", "t_definitions.R",
"t_find_originals_and_duplicates.R",
"t_get_factor_levels.R",
"t_merge_variables.R", "t_n_miss.R",
"t_n_valid.R", "t_openxlsx_shortcuts.r",
"t_rename_variables.R",
"t_replace_na.R", "t_report_memory.R",
"t_select_vars_by_type.R"), Release1 =
c(TRUE, FALSE, FALSE,
FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, FALSE,
FALSE, FALSE), Release2 = c(TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE)), .Names = c("Module",
"Release1", "Release2"), row.names = c(NA, 15L),
class = "data.frame")
edit(release_comparison)
-- cut ---
Release 1 is of 2016-05-31, Releae 2 of 2016-06-21.
You can download the toolbox from
https://sourceforge.net/projects/r-project-utilities/
Kind regards
Georg Maubach
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and provide commented, minimal, self-contained, reproducible code.
[R] Error when installing packages
Hi All, I try to install packages on Debian GNU Linux 8 (Kernel 3.16.0-4-amd64). My sessionInfo() is R version 3.3.1 (2016-06-21) Platform: x86_64-pc-linux-gnu (64-bit) Running under: Debian GNU/Linux 8 (jessie) locale: [1] LC_CTYPE=de_DE.UTF-8 LC_NUMERIC=C [3] LC_TIME=de_DE.UTF-8LC_COLLATE=de_DE.UTF-8 [5] LC_MONETARY=de_DE.UTF-8LC_MESSAGES=de_DE.UTF-8 [7] LC_PAPER=de_DE.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=de_DE.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_3.3.1 Installing the following packages Warning in install.packages : packages ‘excel.link’, ‘installr’ are not available (for R version 3.3.1) Warning in install.packages : dependencies ‘latticist’, ‘graph’, ‘RBGL’, ‘pkgDepTools’, ‘Rgraphviz’ are not available also installing the dependencies ‘RCurl’, ‘RWekajars’ results in the following messages: (1) * installing *source* package ‘RCurl’ ... checking for curl-config... no Cannot find curl-config (2) * installing *source* package ‘RWekajars’ ... ./configure: 1: ./configure: /usr/lib/jvm/default-java/jre/bin/java: not found ./configure: 50: test: -ge: unexpected operator ./configure: 51: test: -eq: unexpected operator Need at least Java version 1.6/6.0. ERROR: configuration failed for package ‘RWekajars’ Annotation: I have openjdk-8-jre installed. (3) * installing *source* package ‘cairoDevice’ ... ERROR: gtk+2. not found by pkg-config. ERROR: configuration failed for package ‘cairoDevice’ (4) * installing *source* package ‘rgdal’ ... configure: CC: gcc -std=gnu99 configure: CXX: g++ configure: rgdal: 1.1-10 checking for /usr/bin/svnversion... no configure: svn revision: 622 checking for gdal-config... no no configure: error: gdal-config not found or not executable. ERROR: configuration failed for package ‘rgdal’ (5) * installing *source* package ‘rgeos’ ... configure: CC: gcc -std=gnu99 configure: CXX: g++ configure: rgeos: 0.3-19 checking for /usr/bin/svnversion... no configure: svn revision: 524 checking for geos-config... no no configure: error: geos-config not found or not executable. ERROR: configuration failed for package ‘rgeos’ ... and much more. Do all these error messages have something in common? How could I fix the installation? Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Spread data.frame on 2 variables
Hi All,
I need to spread a data.frame on 2 variables, e. g. "channel" and "unit".
If I do it in two steps spreads keeps all cases that does not look like
the one before although it contains the same values for a specific case.
Here is what I have right now:
-- cut --
test1$dummy <- 1
test2 <- spread(data = test1, key = 'channel', value = "dummy")
test2
cat("First spread is OK!")
test2$dummy <- 1
test3 <- spread(data = test2, key = 'unit', value = 'dummy')
test1
# test2
test3
warning(paste0("Second spread is not OK cause spread does not merge
cases\n",
"with CustID 700 and 800 into one case,\n",
"cause they have values on different variables,\n",
"although the corresponding values of the cases with",
"custID 700 and 800 are missing."))
cat("What I would like to have is:\n")
target4 <- structure(list(custID = c(100, 200, 300, 500, 600, 700, 800,
900),
`10` = c(1, NA, NA, NA, NA, NA, NA, NA),
`20` = c(1, NA, NA, NA, NA, NA, NA, NA),
`30` = c(NA, NA, NA, NA, NA, NA, 1, 1),
`40` = c(NA, NA, NA, NA, 1, NA, 1, 1),
`50` = c(NA, NA, 1, NA, NA, NA, 1, 1),
`60` = c(NA, NA, NA, NA, NA, 1, NA, NA),
`70` = c(NA, NA, NA, NA, NA, 1, NA, NA),
`99` = c(NA, 1, NA, 1, NA, NA, NA, NA),
`1000` = c(1, NA, NA, NA, NA, NA, 1, 1),
`2000` = c(NA, NA, NA, NA, 1, 1, 1, NA),
`3000` = c(NA, NA, 1, NA, NA, 1, NA, NA),
`4000` = c(NA, NA, 1, NA, NA, NA, NA, NA),
`6000` = c(NA, NA, NA, NA, 1, NA, NA, NA),
`` = c(NA, 1, NA, 1, NA, NA, NA, NA)),
.Names = c("custID",
"10", "20", "30", "40", "50", "60", "70", "99",
"1000", "2000", "3000", "4000", "6000", ""),
row.names = c(NA, 8L), class = "data.frame")
target4
cat("What would be a proper way to create target4 from test1?")
-- cut --
What would be the proper way to create target4 from test1?
Kind regards
Georg
__
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[R] Antwort: Re: Re: Spread data.frame on 2 variables (SOLVED)
Hi Ulrik,
many thanks for your help.
The problem was that R regards a dataset with a combination like
caseID custID channel unit
1 100010 10
2 100020 10
3 100020 30
as two diffrenet sets of cases: 1 set = case 1, 2 set = case 2 and 3 due
to the different values of unit in case 3 value 30, althought all cases
should be restructured based just on custID.
To get a dataset like
caseID custID channel -10 channel-20 unit-10
unit-30
1 10001 1 1 1
instead of
caseID custID channel -10 channel-20 unit-10
unit-30
1 10001 1 1 NA
2 1000NA 1 NA 1
I used the approach you suggested:
1. I created a subset of my data with the first variable to be
restructured:
d_temp1 <- dataset[ , c("custID", "channel"))
2. I deleted all the cases the were dupliates
d_temp1 <- duplicated(d_temp1, c("custID", "channel")
3. I introduced a dummy variable delivering the values for the new
variables created by dplyr:spread()
d_temp1$dummy <- 1
4. Then I restructured the subset
d_temp1 <- dplyr::spread(d_temp1, key_variable = "channel", value =
d_temp1$dummy)
5. I repeaed steps 1 to 4 with the other variable "unit" (instead of
"channel") creating a new dataset named d_temp2.
6. I deleted the variables used for restructuring in steps 1 to 5
"channel" and "unit" from the original dataset "dataset".
dataset$channel <- NULL
dataset$unit <- NULL
7. I checked if I still had duplicates
duplicates <- duplicated(dataset, key_variable = c("Debitor"))
sum(duplicates) # was 0 it this time
8. I merged the datasets back together
dataset_2 <- merge(x = dataset, y = d_temp1, by.x = "Debitor", by.y =
"Debitor", all.x = TRUE, all.y = TRUE) # leaving out all.y would be fine
dataset_2 <- merge(x = dataset2, y = d_temp2, by.x = "Debitor", by.y =
"Debitor", all.x = TRUE, all.y = TRUE) # leaving out all.y would be fine
There might be a combination of commands and functions doing the same
thing in one step but I find that this is clear, comprehensible and
reproducable even at a later date or by other readers willing to use base
R for their work.
Many thanks again for your help.
Kind regards
Georg
Von:Ulrik Stervbo
An: g.maub...@weinwolf.de, R-help ,
Datum: 28.07.2016 14:20
Betreff:Re: Re: [R] Spread data.frame on 2 variables
Hi Georg,
it is difficult to figure out what happens between your expectation and
the outcome if we cannot see a minimal dataset.
Based on your description I did this
library(tidyr)
library(dplyr)
test_df <- data_frame(channel = LETTERS[1:5], unit = letters[1:5], custID
= c(1:5), dummy = 1)
test_df %>% spread(channel, dummy) %>% mutate(dummy = 1) %>% spread(unit,
dummy)
which seems to be working fine as I get wide data. If a combination is
missing in the long form it will also be missing in the wide form. Maybe
you are looking for something like this:
channel_wide <- test_df %>% select(channel, custID) %>% spread(channel,
custID)
unit_wide <- test_df %>% select(unit, custID) %>% spread(unit, custID)
bind_cols(channel_wide, unit_wide)
Apologies for the HTML - it's gmail
Best wishes,
Ulrik
On Thu, 28 Jul 2016 at 13:54 wrote:
Hi Ulrik,
I have included a reproducable example. I ran the code and it did exactly
what I wanted to show you.
You are right: the solution shall merge cases in the end cause the values
on the variables are either missing or the same.
Example 1: Values are the same
If you look at 6 and 7 and variable 70 the value is 1 in both cases. This
is in this context the same information and cases 6 and 7 with custID can
be merged to 1 for variable 70.
Example 2: Values are missing and not missing
If you look at cases 8 and 9 the value for case 8 at variable 40, 50 and
2000 is missing whereas the variables 40, 50 and 2000 have all 1 for case
9. Case 8 and 9 could be merged together cause the missing values are
overwritten what is correct in this case.
The solution I am looking for is to transform the data from long into wide
form and keep all but missing value information.
Did I explain my problem in a comprehensible way? Are there any further
questions?
Kind regards
Georg
Von:Ulrik Stervbo
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 28.07.2016 12:59
Betreff:Re: [R] Spread data.frame on 2 variables
Hi Georg,
it's hard to tell without a reproducible example.
Should spread really merge elements? Does spread know anything about
CustID? Maybe you need to make a useful key of the CustIDs first and
spread on that?
Maybe I'm all off, because I'm really just guessing.
Best,
Ulrik
On Thu, 28 Jul 2016 at 12:36 wrote:
Hi All,
I need to spread a data.frame o
[R] Accessing an object using a string
Hi All,
I would like to access an object using a sting.
# Create example dataset
var1 <- c(1, 2, 3)
var2 <- c(4, 5, 6)
data1 <- data.frame(var1, var2)
var3 <- c(7, 8, 9)
var4 <- c(10, 11, 12)
data2 <- data.frame(var3, var4)
save(file = "c:/temp/test.RData", list = c("data1", "data2"))
# Define function
t_load_dataset <- function(file_path,
file_name) {
file_location <- file.path(file_path, file_name)
print(paste0('Loading ', file_location, " ..."))
cat("\n")
object_list <- load(file = file_location,
envir = .GlobalEnv)
print(paste(length(object_list), "dataset(s) loaded from",
file_location))
cat("\n")
print("The following objects were loaded:")
print(object_list)
cat("\n")
for (i in object_list) {
print(paste0("Object '", i, "' in '", file_name, "' contains:"))
str(i)
names(i) # does not work
}
}
I have only the character vector object_list containing the names of the
objects as strings. I would like to access the objects in object_list to
be able to print the names of the variables within the object (usuallly a
data frame).
Is it possible to do this? How is it done?
Kind regards
Georg
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[R] Antwort: Accessing an object using a string (SOLVED)
Hi All,
I found the function get() which returns an object.
My whole function looks like this:
-- cut --
#---
# Module: t_load_dataset.R
# Author: Georg Maubach
# Date : 2016-08-15
# Update: 2016-08-15
# Description : Load dataset and print information on contents
# Source System : R 3.3.0 (64 Bit)
# Target System : R 3.3.0 (64 Bit)
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
#1-2-3-4-5-6-7-8
t_module_name = "t_load_dataset"
t_version = "2016-08-15"
t_status = "released"
cat(
paste0("\n",
t_module_name, " (Version: ", t_version, ", Status: ", t_status,
")", "\n", "\n",
"Copyright (C) Georg Maubach 2016
This software comes with ABSOLUTELY NO WARRANTY.", "\n", "\n"))
# If do_test is not defined globally define it here locally by
un-commenting it
# Switch t_do_test to TRUE to run test
t_do_test <- FALSE
# [ Function Defintion
]
t_load_dataset <- function(file_path,
file_name) {
# Loads and RData file with all objects in it and prints information on
its
# contents
#
# Args:
# file_path (string):
#String with path name.
# file_name (string):
#String with file name.
#
# Operation:
# Loads the RData file with all its objects, stores the objects in the
# global environment .GlobalEnv and prints information about the
objects.
#
# Usage:
# The function is designed to work only on data frames.
#
# Returns:
# Nothing, but stores loaded objects directly into the global
environment.
#
# Error handling:
# None.
#-
cat("--- [ t_load_dataset() ]
--\n\n")
file_location <- file.path(file_path, file_name)
cat(paste0('Loading ', file_location, " ...\n\n"))
dataset_list <- load(file = file_location,
envir = .GlobalEnv)
cat(paste0(
length(dataset_list),
" dataset(s) loaded:\n"))
cat(dataset_list)
cat("\n\n")
for (dataset in dataset_list) {
cat(paste0("Dataset '", dataset, "' contains ",
nrow(get(dataset, envir = .GlobalEnv)),
" cases in ",
ncol(get(dataset, envir = .GlobalEnv)),
" variables:\n"))
cat(names(get(dataset, envir = .GlobalEnv)))
cat("\n\n")
}
cat("-- [ Done ]
---\n\n")
}
# [ Test Defintion
]
t_test <- function(do_test = FALSE) {
if (do_test == TRUE) {
# Example dataset
var1 <- c(1, 2, 3)
var2 <- c(4, 5, 6)
d_data1 <- data.frame(var1, var2)
var3 <- c(7, 8, 9)
var4 <- c(10, 11, 12)
d_data2 <- data.frame(var3, var4)
# Save datasets
v_file_name <- "test_t_load_dataset.RData"
save(file = file.path(getwd(),
v_file_name),
list = c("d_data1", "d_data2"))
# Call function
t_load_dataset(file_path = getwd(), file_name = v_file_name)
# Cleanup
unlink(file.path(getwd(), v_file_name))
}
}
# [ Test Run
]--
t_test(do_test = t_do_test)
# [ Clean up
]--
rm("t_module_name", "t_version", "t_status", "t_do_test", "t_test")
# EOF
-- cut --
I will include it later the toolbox of R function on Sourceforge.net.
Kind regards
Georg
Von:g.maub...@weinwolf.de
An: r-help@r-project.org,
Datum: 15.08.2016 10:51
Betreff:[R] Accessing an object using a string
Gesendet von: "R-help"
Hi All,
I would like to access an object using a sting.
# Create example dataset
var1 <- c(1, 2, 3)
var2 <- c(4, 5, 6)
data1 <- data.frame(var1, var2)
var3 <- c(7, 8, 9)
var4 <- c(10, 11, 12)
data2 <- data.frame(var3, var4)
save(file = "c:/temp/test.RData", list = c("data1", "data2"))
# Define function
t_load_dataset <- function(file_path,
file_name) {
file_location <- file.path(file_path, file_name)
print(paste0('Loading ', file_location, " ..."))
cat("\n")
object_list <- load(file = file_location,
envir = .GlobalEnv)
print(paste(length(object_list), "dataset(s) loaded from",
file_location))
cat("\n")
print("The following objects were loaded:")
print(object_list)
cat("\n")
for (i in object_list) {
print(paste0("Object '", i, "' in '", file_name, "' contains:"))
str(i)
names(i) # does not work
}
}
I have only the character vector object_list containin
[R] Antwort: Re: Accessing an object using a string
Hi Greg
and all others who replied to my question,
many thanks for all your answers and help. Currently I store all my
objects in .GlobalEnv = Workspace. I am not yet familiar working with
different environments nor did I see that this would be necessary for my
analysis.
Could you explain why working with different environments would be
helpful?
You suggested to read variables into lists rather than storing them in
global variables. This sounds interesting. Could you provide an example of
how to define and use this?
Kind regards
Georg
Von:Greg Snow <538...@gmail.com>
An: g.maub...@weinwolf.de,
Kopie: r-help
Datum: 15.08.2016 20:33
Betreff:Re: [R] Accessing an object using a string
The names function is a primitive, which means that if it does not
already do what you want, it is generally not going to be easy to
coerce it to do it.
However, the names of an object are generally stored as an attribute
of that object, which can be accessed using the attr or attributes
functions. If you change your code to not use the names function and
instead use attr or attributes to access the names then it should work
for you.
You may also want to consider changing your workflow to have your data
objects read into a list rather than global variables, then process
using lapply/sapply (this would require a change in how your data is
saved from your example, but if you can change that then everything
after can be cleaner/simpler/easier/more fool proof/etc.)
On Mon, Aug 15, 2016 at 2:49 AM, wrote:
> Hi All,
>
> I would like to access an object using a sting.
>
> # Create example dataset
> var1 <- c(1, 2, 3)
> var2 <- c(4, 5, 6)
> data1 <- data.frame(var1, var2)
>
> var3 <- c(7, 8, 9)
> var4 <- c(10, 11, 12)
> data2 <- data.frame(var3, var4)
>
> save(file = "c:/temp/test.RData", list = c("data1", "data2"))
>
> # Define function
> t_load_dataset <- function(file_path,
>file_name) {
> file_location <- file.path(file_path, file_name)
>
> print(paste0('Loading ', file_location, " ..."))
> cat("\n")
>
> object_list <- load(file = file_location,
> envir = .GlobalEnv)
>
> print(paste(length(object_list), "dataset(s) loaded from",
> file_location))
> cat("\n")
>
> print("The following objects were loaded:")
> print(object_list)
> cat("\n")
>
> for (i in object_list) {
> print(paste0("Object '", i, "' in '", file_name, "' contains:"))
> str(i)
> names(i) # does not work
> }
> }
>
> I have only the character vector object_list containing the names of the
> objects as strings. I would like to access the objects in object_list to
> be able to print the names of the variables within the object (usuallly
a
> data frame).
>
> Is it possible to do this? How is it done?
>
> Kind regards
>
> Georg
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
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and provide commented, minimal, self-contained, reproducible code.
[R] Installation of rJava fails
Hi All,
I try to install RWeka on Debian GNU Linux 8 Jessie (uname -a: 3.16.0-4-amd64
#1 SMP Debian 3.16.7-ckt25-2+deb8u3 (2016-07-02) x86_64) which has a dependency
to "rJava".
I did
apt-get install openjdk-8-jre
which went OK.
Java is installed in:
/var/lib/dpkg/alternatives/java
/usr/lib/jvm/java-8-openjdk-amd64/jre/bin/java
/usr/lib/jvm/java-8-openjdk-amd64/bin/java
/etc/alternatives/java
When doing this
install.packages("rJava")
I get
* installing *source* package ‘rJava’ ...
** Paket ‘rJava’ erfolgreich entpackt und MD5 Summen überprüft
interpreter : '/usr/lib/jvm/default-java/jre/bin/java'
archiver: '/usr/lib/jvm/default-java/bin/jar'
compiler: '/usr/lib/jvm/default-java/bin/javac'
header prep.: '/usr/lib/jvm/default-java/bin/javah'
cpp flags : '-I/usr/lib/jvm/default-java/include'
java libs : '-L/usr/lib/jvm/default-java/jre/lib/amd64/server -ljvm'
checking whether Java run-time works...
./configure: line 3736: /usr/lib/jvm/default-java/jre/bin/java: No such file or
directory
no
configure: error: Java interpreter '/usr/lib/jvm/default-java/jre/bin/java'
does not work
ERROR: configuration failed for package ‘rJava’
* removing ‘/usr/local/lib/R/site-library/rJava’
Warning in install.packages :
installation of package ‘rJava’ had non-zero exit status
Do I need to use another Java version or installation? How do I tell
install.packages() where my Java installation resides?
Kind regards
Georg
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and provide commented, minimal, self-contained, reproducible code.
[R] Iteration over variables
Hi All,
I would like to write a program that iterates over a set of dynamically
generated variables and produces some stats or prints parts of the data.
# --- data
v_turnover_2011 <- c(10, 20, 30, 40 , 50)
v_customer_2011 <- c(0, 1, NA, 0, 1)
v_turnover_2012 <- c(10, 20, 30, 40 , 50)
v_customer_2012 <- c(0, 1, NA, 0, 1)
d_dataset <- data.frame(v_turnover_2011, v_turnover_2012,
v_customer_2011, v_customer_2012)
# -- Aim is to iterate over dynamically generated variables and compute
# -- statistics or print parts of the data
# -- Does not produce any output
for (year in 2011:2012) {
head(d_dataset[, c(paste0("v_turnover_", year),
paste0("v_customer_", year))])
}
# -- Does not produce any output
aux_func <- function(year) {
head(d_dataset[, c(paste0("v_turnover_", year),
paste0("v_customer_", year))])
}
for (year in 2011:2012) {
aux_func(year = year)
}
d_results <- data.frame()
for (year in 2011:2012) {
d_results <- rbind(d_results,
paste0("mean", year) = mean(d_dataset[,
c(paste0("v_turnover_", year))]))
}
Is there a way to iterate over variables and compute statistics and print
parts of the dataset?
Kind regards
Georg
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and provide commented, minimal, self-contained, reproducible code.
[R] Putting a bunch of Excel files as data.frames into a list fails
Hi All,
I need to read a bunch of Excel files and store them in R.
I decided to store the different Excel files in data.frames in a named
list where the names are the file names of each file (and that is
different from the sources as far as I can see):
-- cut --
# Sources:
# -
http://stackoverflow.com/questions/11433432/importing-multiple-csv-files-into-r
# -
http://stackoverflow.com/questions/9564489/opening-all-files-in-a-folder-and-applying-a-function
# -
http://stackoverflow.com/questions/12945687/how-to-read-all-worksheets-in-an-excel-workbook-into-an-r-list-with-data-frame-e
v_file_path <- "H:/2016/Analysen/Neukunden/Input"
v_file_pattern <- "*.xlsx"
v_files <- list.files(path = v_file_path,
pattern = v_file_pattern,
ignore.case = TRUE)
print(v_files)
v_list_of_files <- list()
for (v_file in v_files) {
v_list_of_files[v_file] <- openxlsx::read.xlsx(
file.path(v_file_path,
v_file))
}
This code does not work cause it stores only the first variable of each
Excel file in a named list.
What do I need to change to get it running?
Kind regards
Georg
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and provide commented, minimal, self-contained, reproducible code.
[R] Filtering Cases with != NA
Dear All, I am new to "R" and search for a solution to exclude cases if a certain variable contains NA for a case. Example No Name Turnover 1 Smith 1500 2 Mayor 200 3 Miller 4 Batic 750 I would like to create a subset excluding case 3 Miller NA. I tried to following: new_dataset <- subset(dataset, subset = Turnover != NA) This does not work. The new_dataset contains all variables but not cases are left. R responds "Variables with all observations missing". How could I do it right? Kind regards Georg [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Merging Data Sets with Full Outer Join
Hi All, I would like to match some datasets. Both deliver variables AND cases which might or might not be present in all datasets: This sequence Kunden <- Kunden_2011 Kunden <- merge(Kunden, Kunden_2012, by.x = "Debitor", by.y = "Debitor") Kunden <- merge(Kunden, Kunden_2013, by.x = "Debitor", by.y = "Debitor") Kunden <- merge(Kunden, Kunden_2014, by.x = "Debitor", by.y = "Debitor") Kunden <- merge(Kunden, Kunden_2015, by.x = "Debitor", by.y = "Debitor") delivers too few cases. So I guess it does an equi-join. How can I join the datasets and keep the variables as well as the cases? I am looking forward to your reply. Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Creating variables on the fly
Hi all,
I would like to use a loop for tasks that occurs repeatedly:
# Groups
# Umsatz <= 0: 1 (NICHT kaufend)
# Umsatz > 0: 2 (kaufend)
for (year in c("2011", "2012", "2013", "2014", "2015")) {
paste0("Kunden$Kunde_real_", year) <- (paste0("Kunden$Umsatz_", year) <= 0) *
1 +
(paste0("Kunden$Umsatz_", year) > 0) *
2
paste0("Kunden$Kunde_real_", year) <- factor(paste0("Kunden$Umsatz_", year),
levels = c(1, 2),
labels = c("NICHT kaufend",
"kaufend"))
}
This actually does not work due to the fact that the expression
"paste0("Kunden$Kunde_real_", year)" ist not interpreted as a variable name by
the R script language interpreter.
Is there a way to assembly variable names on the fly in R?
Regards
Georg
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[R] Antwort: Fw: Re: Creating variables on the fly (SOLVED)
Hi Don,
Hi to all readers,
many thanks for all your answers and all your help.
I adapted Don's code to my data and Don's code does the trick:
str(Kunden01)
for (year in 2011:2015) {
Reeller_Kunde <- paste0("Reeller_Kunde_", year)
Umsatz <- paste0("Umsatz_", year)
cat('Creating', Reeller_Kunde,'from', Umsatz,'\n')
Kunden01[[ Reeller_Kunde ]] <- ifelse( Kunden01[[ Umsatz ]] >= 0, 1, 2)
Kunden01[[ Reeller_Kunde ]] <- factor( Kunden01[[ Reeller_Kunde ]],
levels=c(1,2),
labels= c("NICHT kaufend",
"kaufend")
)
}
str(Kunden01)
This way a new variable is created by building it from a string
concatenation.
I also like the cat() function to document the process within the loop
while running the program.
Many thanks for your help.
Kind regards
Georg
Von:g.maub...@gmx.de
An: g.maub...@weinwolf.de,
Datum: 25.04.2016 21:37
Betreff:Fw: Re: [R] Creating variables on the fly
> Gesendet: Montag, 25. April 2016 um 19:35 Uhr
> Von: "MacQueen, Don"
> An: "g.maub...@gmx.de" , "r-help@r-project.org"
> Betreff: Re: [R] Creating variables on the fly
>
> I'm going to assume that Kunden is a data frame, and it has columns
> (variables) with names like
> Umstatz_2011
> and that you want to create new columns with names like
> Kunde_real_2011
>
> If that is so, then try this (not tested):
>
> for (year in 2011:2015) {
> nmK <- paste0("Kunde_real_", year)
> nmU <- paste0("Umsatz_", year)
> cat('Creating',nmK,'from',nmU,'\n')
> Kunden[[ nmK ]] <- ifelse( Kunden[[ nmU ]] <= 0, 1, 2)
> Kunden[[ nmK ]] <- factor( Kunden[[ nmK ]],
>levels=c(1,2),
>labels= c("NICHT kaufend", "kaufend")
>)
>
> }
>
> This little example should illustrate the method:
>
>
> > foo <- data.frame(a=1:4)
> > foo
> a
> 1 1
> 2 2
> 3 3
> 4 4
> > foo[['b']] <- foo[['a']]*3
> > foo
> a b
> 1 1 3
> 2 2 6
> 3 3 9
> 4 4 12
>
>
>
> --
> Don MacQueen
>
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
>
>
>
>
>
> On 4/22/16, 8:52 AM, "R-help on behalf of g.maub...@gmx.de"
> wrote:
>
> >Hi all,
> >
> >I would like to use a loop for tasks that occurs repeatedly:
> >
> ># Groups
> ># Umsatz <= 0: 1 (NICHT kaufend)
> ># Umsatz > 0: 2 (kaufend)
> >for (year in c("2011", "2012", "2013", "2014", "2015")) {
> > paste0("Kunden$Kunde_real_", year) <- (paste0("Kunden$Umsatz_", year)
> ><= 0) * 1 +
> >(paste0("Kunden$Umsatz_", year)
>
> > 0) * 2
> > paste0("Kunden$Kunde_real_", year) <- factor(paste0("Kunden$Umsatz_",
> >year),
> > levels = c(1, 2),
> > labels = c("NICHT
> >kaufend", "kaufend"))
> > }
> >
> >This actually does not work due to the fact that the expression
> >"paste0("Kunden$Kunde_real_", year)" ist not interpreted as a variable
> >name by the R script language interpreter.
> >
> >Is there a way to assembly variable names on the fly in R?
> >
> >Regards
> >
> >Georg
> >
> >__
> >R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >https://stat.ethz.ch/mailman/listinfo/r-help
> >PLEASE do read the posting guide
> >http://www.R-project.org/posting-guide.html
> >and provide commented, minimal, self-contained, reproducible code.
>
>
__
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[R] Missing Values in Logical Expressions
Hi All,
I need to evaluate missing values in my data. I am able to filter these
values and do simple statistics on it. But I do need new variables based
on variables with missing values in my dataset:
Check_Kunde_2011 <- ifelse(is.na(Umsatz_2011) == TRUE & Kunde_2011 == 1,
1, 0)
Check_Kunde_2011 <- factor(Check_Kunde_2011, levels = c(1,0), labels =
c("Check", "OK"))
The new variable is not correctly created. It contains no values:
table(Check_Kunde_2011)
< table of extent 0 >
I searched the web but could not find a solution.
How can I work with variables and missing values in logical expressions?
Where could I find something about this?
Kind regards
Georg
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[R] Antwort: RE: Missing Values in Logical Expressions
Hi Petr,
Hi Jim,
many thanks for your help. Today I constructed a sample dataset and tested
your suggestions. Everything worked OK.
Then I took the code and testet on the original data. And - it worked OK
this morning also.
I went back to my script of Thuesday and ran it again. OK. Then I used my
script of Monday and ran it.. OK.
I have no idea what was wrong yesterday. To see that there is a problem
and not being able to replicate it a day later even if it did not work all
day before is very strange.
If the problem arises again, I will raise my hand.
Many thanks again for your help.
Kind regards
Georg
Von:PIKAL Petr
An: "g.maub...@weinwolf.de" ,
"r-help@r-project.org" ,
Datum: 26.04.2016 11:11
Betreff:RE: [R] Missing Values in Logical Expressions
Hm
Based on Jim's data your construction gives me correct result.
> Umsatz_2011<-c(1,2,3,4,5,NA,7,8,NA,10)
> Kunde_2011<-rep(0:1,5)
> Check_Kunde_2011 <- ifelse(is.na(Umsatz_2011) == TRUE & Kunde_2011 == 1,
1, 0)
> Check_Kunde_2011 <- factor(Check_Kunde_2011, levels = c(1,0), labels =
c("Check", "OK"))
> table(Check_Kunde_2011)
Check_Kunde_2011
CheckOK
1 9
So I presume that the problem lies in your data.
You should provide some sample of your data either by posting result of
str(yourdata)
or
dput(head(yourdata))
if you want some advice why with correct code you did not get appropriate
result.
Instead ifelse you can also use
Check_Kunde_2011 <- (is.na(Umsatz_2011)&(Kunde_2011==1))*1
to get desired 0/1 vector.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> g.maub...@weinwolf.de
> Sent: Tuesday, April 26, 2016 10:10 AM
> To: r-help@r-project.org
> Subject: [R] Missing Values in Logical Expressions
>
> Hi All,
>
> I need to evaluate missing values in my data. I am able to filter these
values
> and do simple statistics on it. But I do need new variables based on
variables
> with missing values in my dataset:
>
> Check_Kunde_2011 <- ifelse(is.na(Umsatz_2011) == TRUE & Kunde_2011 ==
> 1, 1, 0)
> Check_Kunde_2011 <- factor(Check_Kunde_2011, levels = c(1,0), labels =
> c("Check", "OK"))
>
> The new variable is not correctly created. It contains no values:
>
> table(Check_Kunde_2011)
> < table of extent 0 >
>
> I searched the web but could not find a solution.
>
> How can I work with variables and missing values in logical expressions?
>
> Where could I find something about this?
>
> Kind regards
>
> Georg
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
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[R] R Script Template
Hi All, I am addressing this post to all who are new to R. When learing R in the last weeks I took some notes for myself to have code snippets ready for the data analysis process. I put these snippets together as a script template for future use. Almost all of the given command prototypes are tested. The template script contains snippets for best practices and leaves out the commands that should not be used. Relying on the given snippets shall lead to high quality code. The code is based on examples from the ressources given in the template. I highly recommend to read the books or take the online courses to see how everything works and fits together. Despite putting everything together with care, the script is provided as-is with no warrenty or liability whatsoever. Please address any remarks or suggestions for improvement to the R-Help mailing list. Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Interdependencies of variable types, logical expressions and NA
Hi All,
my script tries to do the following on factors:
> ## Check for case 3: Umsatz = 0 & Kunde = 1
> for (year in 2011:2015) {
+ Umsatz <- paste0("Umsatz_", year)
+ Kunde <- paste0("Kunde01_", year)
+ Check <- paste0("Check_U_0__Kd_1_", year)
+
+ cat('Creating', Check, 'from', Umsatz, "and", Kunde, '\n')
+
+ Kunden01[[ Check ]] <- ifelse(Kunden01[[ Umsatz ]] == 0 &
+ Kunden01[[ Kunde ]] == 1,
+ 1, 0
+ )
+ Kunden01[[ Check ]] <- factor(Kunden01[[ Check ]],
+ levels=c(1, 0),
+ labels= c("Check 0", "OK")
+ )
+
+ }
Creating Check_U_0__Kd_1_2011 from Umsatz_2011 and Kunde01_2011
Creating Check_U_0__Kd_1_2012 from Umsatz_2012 and Kunde01_2012
Creating Check_U_0__Kd_1_2013 from Umsatz_2013 and Kunde01_2013
Creating Check_U_0__Kd_1_2014 from Umsatz_2014 and Kunde01_2014
Creating Check_U_0__Kd_1_2015 from Umsatz_2015 and Kunde01_2015
>
> table(Kunden01$Check_U_0__Kd_1_2011, useNA = "ifany")
Check 0 OK
1 16 13
> table(Kunden01$Check_U_0__Kd_1_2012, useNA = "ifany")
Check 0 OK
1 17 12
> table(Kunden01$Check_U_0__Kd_1_2013, useNA = "ifany")
Check 0 OK
2 17 13
> table(Kunden01$Check_U_0__Kd_1_2014, useNA = "ifany")
Check 0 OK
1 15 14
> table(Kunden01$Check_U_0__Kd_1_2015, useNA = "ifany")
Check 0 OK
2 15 13
>
> Kunden01$Check_U_0__Kd_1_all <- ifelse(Kunden01$Check_U_0__Kd_1_2011 ==
1 |
+Kunden01$Check_U_0__Kd_1_2012 ==
1 |
+Kunden01$Check_U_0__Kd_1_2013 ==
1 |
+Kunden01$Check_U_0__Kd_1_2014 ==
1 |
+Kunden01$Check_U_0__Kd_1_2015 ==
1,
+1, 0)
>
> table(Kunden01$Check_U_0__Kd_1_all, useNA = "ifany")
0
723
(Ann.: I made the values up. But the relations equal real world data.)
I had expected to get back a factor or at least a numeric variable
containing 0, 1 and NA, instead 1 is not included.
I searched the web for information on the treatment of logical expressions
when the data contains NA. I found:
1.
https://stat.ethz.ch/R-manual/R-devel/library/base/html/NA.html
Examples
# Some logical operations do not return NA
c(TRUE, FALSE) & NA
c(TRUE, FALSE) | NA
2.
https://stat.ethz.ch/R-manual/R-devel/library/base/html/Logic.html
NA is a valid logical object. Where a component of x or y is NA, the
result will be NA if the outcome is ambiguous. In other words NA & TRUE
evaluates to NA, but NA & FALSE evaluates to FALSE. See the examples
below.
## construct truth tables :
x <- c(NA, FALSE, TRUE)
names(x) <- as.character(x)
outer(x, x, "&") ## AND table
outer(x, x, "|") ## OR table
Ann. Not very useful. How should it be read?
3.
http://www.ats.ucla.edu/stat/r/faq/missing.htm
Good explanation for NA in general and in analysis, but no information
about NA in logical expressions.
Then I made some tests with different data types and variables with NA:
-- cut --
# 2016-04-27-001_truth_table_for_logicals_and_NA.R
# Test 1
var2 <- c(TRUE, FALSE)
var3 <- c(NA, NA)
var1 <- c(1, 1)
ds <- data.frame(var1, var2, var3)
ds
ds$value_and_logical <- ifelse(ds$var1 | ds$var2, TRUE, FALSE)
ds$logical_and_na <- ifelse(ds$var2 | ds$var3, TRUE, FALSE)
ds$value_and_na <- ifelse(ds$var1 | ds$var3, TRUE, FALSE)
print(ds)
# Output
# var1 var2 var3 value_and_logical logical_and_na value_and_na
# 11 TRUE NA TRUE TRUE TRUE
# 21 FALSE NA TRUE NA TRUE
# Test 2
ds$var1 <- factor(ds$var1, levels = c(0, 1), labels = c("NOT ok", "OK"))
ds$var2 <- factor(ds$var2, levels = c(0, 1), labels = c("NOT ok", "OK"))
ds$var3 <- factor(ds$var3, levels = c(0, 1), labels = c("NOT ok", "OK"))
ds$value_and_logical <- ifelse(ds$var1 | ds$var2, TRUE, FALSE)
ds$logical_and_na <- ifelse(ds$var2 | ds$var3, TRUE, FALSE)
ds$value_and_na <- ifelse(ds$var1 | ds$var3, TRUE, FALSE)
# Output (abbrev.)
# Warning message:
# In Ops.factor(ds$var1, ds$var3) : ?|? ist nicht sinnvoll für Faktoren
print(ds)
# Output
# var1 var2 var3 value_and_logical logical_and_na value_and_na
# 1 OK NA NA NA
# 2 OK NA NA NA
-- cut --
I had expected to get the same result in Test 2 as in Test 1.
Where can I find information and documentation about NA handling in
logical expressions on different variable types?
Kind regards
Georg
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[R] Antwort: Re: R Script Template
Hi All,
please find enclosed the missing attachment.
Kind regards
Georg
-- cut --
#-[ Header ]
--
# Program : Framework for R scripts
# Author: Georg Maubach
# Date : 2016-03-03
# Update: 2016-04-27
# Description : Foundation for the analysis process
# Source System : R 3.2.5 (64 Bit)
# Target System : R 3.2.5 (64 Bit)
# Release : 1
# License : CC-BY-NC-SA
# File Name : 2016-04-27_Template_Scipt.R
#---
#- [ Purpose of the document ]
# This document provides a framework for a script able to handle real
world
# data throughout the complete analysis process. In each step examples or
# prototypes of needed or helpful commands are given. Chapters and
sections in
# this document can be regarded as a toolbox. The needed tools shall be
adapted
# to the processed data. Commands are ordered an a consistent way to
support the
# user to produce high quality output.
#---
# - [ At hand ]
# help("function")# Extract or Replace Parts of an Object
# example("function") # Examples on "Extract"
# demo(package = .packages(all.available = TRUE)) # Show demos of packages
#---
# - [ Editing Marks ]
--
# %ROTA% : Result of the analysis in text form if needed to explain
further
# steps
# %ToDo% : ToDo's
#---
# - [ Warrenty Disclaimer ]
# The software is provided "as-is". The author disclaims to the fullest
extent
# authorized by law any and all warranties, whether express or implied,
# including, without limitation, any implied warranties of merchantability
or
# fitness for a particular purpose. Without limitation of the foregoing,
the
# author expressly does not warrant that:
#
# (a) the software will meet your requirements or expectations;
# (b) the software or the software content will be free of bugs, errors,
# viruses or other defects;
# (c) any results, output, or data provided through or generated by the
software
# will be accurate, up-to-date, complete or reliable;
# (d) the software will be compatible with third party software;
# (e) any errors in the software will be corrected.
#---
# - [ Limitation of Liability ]
# In no event will the author be liable for any direct, indirect,
consequential,
# incidental, special, exemplary, or punitive damages or liabilities
whatsoever
# arising from or relating to the software, the software content or this
# agreement, whether based on contract, tort (including negligence),
strict
# liability or other theory, even if the author has been advised of the
# possibility of such damages.
#
# The use of the software goes to the whole risk of the user.
#---
#1-2-3-4-5-6-7-8
#---#
# Setup #
#---#
# Environment
# Please make sure that RTools is installed
Sys.getenv("R_ZIPCMD", "zip")
# needed for openxlsx::write.xlsx()
Sys.setenv(R_ZIPCMD= "C:/R-Project/Rtools/bin/zip")
.libPaths() # Install directory for libraries
# .libPaths("new path if needed")
# Workplace
sessionInfo()# Environment
list.files(R.home()) # Show R home directory
getwd() # Get working directory
list.dirs() # List directories in working directory
list.files() # List files in working directory
library()# List all installed packages
search() # List all loaded packages
ls() # List objects in environment
#---#
# Configure #
#---#
path <- file.path("path", "to","directory")
setwd(path) # Set working directory
options(width = 65) # Set output width
#-#
# Install #
#-#
available.packages()
# Desired packages
my_packages <- c(
"ctv"# Package to install packages based on themes
"data.table",# Fast manipulation of large datasets
"dplyr", # Data manipulation for data frames
"geoR",
"haven", # import data from stastical packages
"Hmisc",
"httr", # package to deal with HTTP requests
"installr", # Dependency of openxlsx::write.xlsx()
"lubridate",
"mapdata", # data for high-quality maps
"maps", # draw maps
"maptools", # import ESRI data
"memisc",# package data import
[R] Antwort: Re: selecting columns from a data frame or data table by type, ie, numeric, integer
Hi All,
Hi Carl,
I am not sure if this is useful to you, but I followed your conversation
and thought of you when I read this:
for (i in 1:ncol(dataset)) {
if(class(dataset) == "character|numeric|factor|or whatsoever") {
dataset[, i] <- as.factor(dataset[, i])
}
}
Source: Zumel, Nina / Mount, John: Practical Data Science with R, Manning
Publications: Shelter Island, 2014, Chapter 2: Loading data into R, p. 25
This way you can select variables of a certain class only and do
transformations. I found that this approach is not applicable if used with
statistical functions like head(). Transformations worked fine for me.
I found reading the above given source worthwile.
Kind regards
Georg
PS: I am not related to the above given authors. I am just a reader
reporting on - at least to me - a valuable ressource.
Von:Carl Sutton via R-help
An: William Dunlap ,
Kopie: "r-help@r-project.org"
Datum: 29.04.2016 22:08
Betreff:Re: [R] selecting columns from a data frame or data table
by type, ie, numeric, integer
Gesendet von: "R-help"
Thank you Bill Dunlap. So simple I never tried that approach. Tried
dozens of others though, read manuals till I was getting headaches, and of
course the answer was simple when one is competent. Learning, its a
struggle, but slowly getting there.
Thanks again
Carl Sutton CPA
On Friday, April 29, 2016 10:50 AM, William Dunlap
wrote:
> dt1[ vapply(dt1, FUN=is.numeric, FUN.VALUE=NA) ]a c1 1 1.12 2
1.0...10 10 0.2
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Apr 29, 2016 at 9:19 AM, Carl Sutton via R-help
wrote:
Good morning RGuru's
I have a data frame of 575 columns. I want to extract only those columns
that are numeric(double) or integer to do some machine learning with. I
have searched the web for a couple of days (off and on) and have not found
anything that shows how to do this. Lots of ways to extract rows, but
not columns. I have attempted to use "(x == y)" indices extraction method
but that threw error that == was for atomic vectors and lists, and I was
doing this on a data frame.
My test code is below
# a technique to get column classes
library(data.table)
a <- 1:10
b <- c("a","b","c","d","e","f","g","h","i","j")
c <- seq(1.1, .2, length = 10)
dt1 <- data.table(a,b,c)
str(dt1)
col.classes <- sapply(dt1, class)
head(col.classes)
dt2 <- subset(dt1, typeof = "double" | "numeric")
str(dt2)
dt2 # not subset
dt2 <- dt1[, list(typeof = "double")]
str(dt2)
class_data <- dt1[,sapply(dt1,is.integer) | sapply(dt1, is.numeric)]
class_data
sum(class_data)
typeof(class_data)
names(class_data)
str(class_data)
Any help is appreciated
Carl Sutton CPA
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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__
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[R] Antwort: Antwort: Re: selecting columns from a data frame or data table by type, ie, numeric, integer
Hi Martin,
many thanks for your answer and your broad explanation.
I am a newbie to "R" and got help on this list and thought I could give
something back what looked OK to me.
regarding 0)
You're right, it's pseudo code. I assumed that anybody on the list would
be able to adapt the code to their needs so that it worked. Next time I
will post runnable code.
regarding 1)
Your right: "[, i]" is missing. My fault. Sorry.
regarding 3)
I got your point and will do better in the future.
One question: What books do you recommend to read to get to know "R"
better?
Kind regards
Georg
Von:Martin Maechler
An: ,
Kopie: Carl Sutton , "r-help@r-project.org"
Datum: 04.05.2016 09:05
Betreff:[R] Antwort: Re: selecting columns from a data frame or
data table by type, ie, numeric, integer
>
> on Wed, 4 May 2016 08:30:50 +0200 writes:
> Hi All,
> Hi Carl,
>
> I am not sure if this is useful to you, but I followed your conversation
> and thought of you when I read this:
>
> for (i in 1:ncol(dataset)) {
> if(class(dataset) == "character|numeric|factor|or whatsoever") {
> dataset[, i] <- as.factor(dataset[, i])
> }
> }
Ouch -- so many problems in such a short piece of R code !!!
> Source: Zumel, Nina / Mount, John: Practical Data Science with R,
Manning
> Publications: Shelter Island, 2014, Chapter 2: Loading data into R, p.
25
Sorry, but after reading the above, I'd strongly recommend getting
better books about R...
{{maybe do not take those containing "data science" ;-)}}
Compared to the nice and efficient solution of Bill Dunlap,
the above is really bad-bad-bad in at least four ways :
0) They way you write it above, you cannot use it,
== "variant1|variant2|..."
is pseudocode and does not really work
1) Note the missing "[, i]" in the 2nd line: It should be
if(class(dataset[, i]) ...
2) A for loop changing each column at a time is really slow for
largish data sets
3) [last but not at all least!]
Please ... many of you readers, do learn:
Using checks such as
if ( class(x) == "numeric" )
are (almost) always wrong by design !!!
Instead you really should (almost) always use
if(inherits(x, "numeric"))
Why? Because classes in R (S3 or S4) can *extend* other classes.
Example: Many of you know that after fm <- glm(...)
class(fm) is c("glm", "lm") and so
> if(class(fm) == "lm")
+ "yes"
Warning message:
In if (class(fm) == "lm") "yes" :
the condition has length > 1 and only the first element will be used
Similarly, in your case
y <- 1:10
class(y) <- c("myNumber", "numeric")
when that 'y' is a column in your data frame,
the test for if(class(dataset[,i]) == "numeric") will *not*
work but actually produce the above warning.
However, one could als have had
Num <- setClass("Num", contains="numeric")
N <- Num(1:10)
> Num <- setClass("Num", contains="numeric")
> N <- Num(1:10)
> N
An object of class "Num"
[1] 1 2 3 4 5 6 7 8 9 10
> if(class(N) == "numeric") "yes" else "no"
[1] "no"
>
I hope that many of the readers --- including *MANY* authors of
R packages !! --- have understood the above and will fix their R
code -- and even more their books where applicable !!
Martin Maechler,
ETH Zurich & R Core Team
>
> This way you can select variables of a certain class only and do
> transformations. I found that this approach is not applicable if used
with
> statistical functions like head(). Transformations worked fine for me.
>
> I found reading the above given source worthwile.
>
> Kind regards
>
> Georg
>
> PS: I am not related to the above given authors. I am just a reader
> reporting on - at least to me - a valuable ressource.
>
>
>
> Von:Carl Sutton via R-help
> An: William Dunlap ,
> Kopie: "r-help@r-project.org"
> Datum: 29.04.2016 22:08
> Betreff:Re: [R] selecting columns from a data frame or data
table
> by type, ie, numeric, integer
> Gesendet von: "R-help"
>
>
>
> Thank you Bill Dunlap. So simple I never tried that approach. Tried
> dozens of others though, read manuals till I was getting headaches, and
of
> course the answer was simple when one is competent. Learning, its a
> struggle, but slowly getting there.
> Thanks again
> Carl Sutton CPA
>
>
> On Friday, April 29, 2016 10:50 AM, William Dunlap
> wrote:
>
>
>
> > dt1[ vapply(dt1, FUN=is.numeric, FUN.VALUE=NA) ]a c1 1 1.12 2
> 1.0...10 10 0.2
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
> On Fri, Apr 29, 2016 at 9:19 AM, Carl Sutton via R-help
> wrote:
>
> Good morning RGuru's
> I have a data frame of 575 columns. I want to extract only those
columns
> that are numeric(double) or integer to do some machine learning with. I
> have searched the web for a couple of days (off and on) and have not
found
> anything that shows how to do this. Lots of ways
[R] sink(): Cannot open file
Hi All,
I would like to route the output to a file using sink(). When using the
example from the ?sink documentation:
sink("sink-examp.txt")
i <- 1:10
outer(i, i, "*")
sink()
unlink("sink-examp.txt")
## capture all the output to a file.
zz <- file("all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message")
sink()
file.show("all.Rout")
I can not open the file in Windows Explorer. The error message is:
"Cannot open file. File is in use be another proces."
How can I close the file in a manner that I can open it right after it was
created?
Kind regards
Georg
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: sink(): Cannot open file
Hi Jim,
thanks for your reply.
ad 1)
"all.Rout" was created in the correct directory. It exists properly with
correct file properties on Windows, e.g. creation date and time and file
size information.
ad 2)
I can not access the file with Notepad.exe directly after it was created
by R. The error message is (translated):
"Cannot access file "all.Rout". The file is opened by another process."
ad 3)
If I close R completely the file access is released. Then I can read the
file using Notepad.exe. The contents is:
Error in log("a") : non-numeric argument to mathematical function
I tried
close(zz)
but the error persists.
To me it looks like R is still accessing the file and not releasing the
connection for other programs. close(zz) should have solved the problem
but unfortantely it doesn't.
What else could I try?
Kind regards
Georg
Von:Jim Lemon
An: g.maub...@weinwolf.de,
Kopie: r-help mailing list
Datum: 10.05.2016 12:50
Betreff:Re: [R] sink(): Cannot open file
Hi Georg,
I don't suppose that you have:
1) checked that the file "all.Rout" exists somewhere?
2) if so, looked at the file with Notepad, perhaps?
3) let us in on the secret by pasting the contents of "all.Rout" into
your message if it is not too big?
At a guess, trying:
close(zz)
might get you there.
Jim
On Tue, May 10, 2016 at 5:25 PM, wrote:
> Hi All,
>
> I would like to route the output to a file using sink(). When using the
> example from the ?sink documentation:
>
> sink("sink-examp.txt")
> i <- 1:10
> outer(i, i, "*")
> sink()
> unlink("sink-examp.txt")
>
> ## capture all the output to a file.
> zz <- file("all.Rout", open = "wt")
> sink(zz)
> sink(zz, type = "message")
> try(log("a"))
> ## back to the console
> sink(type = "message")
> sink()
> file.show("all.Rout")
>
> I can not open the file in Windows Explorer. The error message is:
>
> "Cannot open file. File is in use be another proces."
>
> How can I close the file in a manner that I can open it right after it
was
> created?
>
> Kind regards
>
> Georg
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Re: sink(): Cannot open file
Hi Jim,
I tried:
sink("all.Rout")
try(log("a"))
sink()
The program executes without warning or error. The file "all.Rout" is
begin created. Nothing will be written to it. The file is accessable
rights after the execution of the program by notepad.exe.
The program
zz <- file("all.Rout", open = "wt")
sink(zz, type = "message")
try(log("a"))
sink()
close(zz)
unlink(zz)
creates the file, does not write anything to it and is not accessable
after program execution in R with notepad.exe.
Any ideas what happens behind the szenes?
Kind regards
Georg
Von:Jim Lemon
An: g.maub...@weinwolf.de,
Kopie: r-help mailing list
Datum: 10.05.2016 13:16
Betreff:Re: Re: [R] sink(): Cannot open file
Have you tried:
sink("all.Rout")
try(log("a"))
sink()
Jim
On Tue, May 10, 2016 at 9:05 PM, wrote:
> Hi Jim,
>
> thanks for your reply.
>
> ad 1)
> "all.Rout" was created in the correct directory. It exists properly with
> correct file properties on Windows, e.g. creation date and time and file
> size information.
>
> ad 2)
> I can not access the file with Notepad.exe directly after it was created
> by R. The error message is (translated):
>
> "Cannot access file "all.Rout". The file is opened by another process."
>
> ad 3)
> If I close R completely the file access is released. Then I can read the
> file using Notepad.exe. The contents is:
>
> Error in log("a") : non-numeric argument to mathematical function
>
> I tried
>
> close(zz)
>
> but the error persists.
>
> To me it looks like R is still accessing the file and not releasing the
> connection for other programs. close(zz) should have solved the problem
> but unfortantely it doesn't.
>
> What else could I try?
>
> Kind regards
>
> Georg
>
>
>
>
> Von:Jim Lemon
> An: g.maub...@weinwolf.de,
> Kopie: r-help mailing list
> Datum: 10.05.2016 12:50
> Betreff:Re: [R] sink(): Cannot open file
>
>
>
> Hi Georg,
> I don't suppose that you have:
>
> 1) checked that the file "all.Rout" exists somewhere?
>
> 2) if so, looked at the file with Notepad, perhaps?
>
> 3) let us in on the secret by pasting the contents of "all.Rout" into
> your message if it is not too big?
>
> At a guess, trying:
>
> close(zz)
>
> might get you there.
>
> Jim
>
> On Tue, May 10, 2016 at 5:25 PM, wrote:
>> Hi All,
>>
>> I would like to route the output to a file using sink(). When using the
>> example from the ?sink documentation:
>>
>> sink("sink-examp.txt")
>> i <- 1:10
>> outer(i, i, "*")
>> sink()
>> unlink("sink-examp.txt")
>>
>> ## capture all the output to a file.
>> zz <- file("all.Rout", open = "wt")
>> sink(zz)
>> sink(zz, type = "message")
>> try(log("a"))
>> ## back to the console
>> sink(type = "message")
>> sink()
>> file.show("all.Rout")
>>
>> I can not open the file in Windows Explorer. The error message is:
>>
>> "Cannot open file. File is in use be another proces."
>>
>> How can I close the file in a manner that I can open it right after it
> was
>> created?
>>
>> Kind regards
>>
>> Georg
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Antwort: Re: Re: sink(): Cannot open file
Hi Sarah, John, Jim,
Hi All,
I have set my envrionment variable
path <- file.path("H:", "2016", "Analysis")
setwd(dir = path)
This works well cause the file is created in that directory.
I have tried
close(zz)
unlink(zz)
and neither worked nor did it work out using them together.
I had this before when working with IBM SPSS Statistics. There was a
workaround for the problem in SPSS.
Is there one for R?
Kind regards
Georg
Von:Sarah Goslee
An: g.maub...@weinwolf.de,
Kopie: r-help mailing list
Datum: 10.05.2016 17:17
Betreff:Re: [R] Antwort: Re: Re: sink(): Cannot open file
Try closing the type of sink you're actually opening:
zz <- file("all.Rout", open = "wt")
sink(zz, type = "message")
try(log("a"))
sink(type = "message")
close(zz)
unlink(zz)
If you look carefully at the example in?sink, there are two close
statements, one for each stream being sent to that file.
Sarah
- Weitergeleitet von Georg Maubach/WWBO/WW/HAW am 10.05.2016 18:29
-
Von:"John Sorkin"
An: , ,
Kopie:
Datum: 10.05.2016 17:20
Betreff:Re: [R] Antwort: Re: Re: sink(): Cannot open file
George,
I do not know what operating system you are working with, but when I use
sink() under windows, I need to specify a valid path which I don't see in
your code. I might, for example specify:
sink("c:\myfile.txt")
R code goes here
sink()
with the expectation that I would create a file myfile.txt that would
contain the output of my R program.
John
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
On Tue, May 10, 2016 at 11:05 AM, wrote:
> Hi Jim,
>
> I tried:
>
> sink("all.Rout")
> try(log("a"))
> sink()
>
> The program executes without warning or error. The file "all.Rout" is
> begin created. Nothing will be written to it. The file is accessable
> rights after the execution of the program by notepad.exe.
>
> The program
>
> zz <- file("all.Rout", open = "wt")
> sink(zz, type = "message")
> try(log("a"))
> sink()
> close(zz)
> unlink(zz)
>
> creates the file, does not write anything to it and is not accessable
> after program execution in R with notepad.exe.
>
> Any ideas what happens behind the szenes?
>
> Kind regards
>
> Georg
>
>
>
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Re: Antwort: Re: Re: sink(): Cannot open file
Hi Sarah,
yes, I followed your suggestion.
If I do exactly what is in the example of the documentation:
sink("C:/Temp/sink-examp.txt")
i <- 1:10
outer(i, i, "*")
sink()
unlink("C:/Temp/sink-examp.txt")
it does not write anything, i. e. no file is created in "C:/Temp/". The
script is executed without an error or warning message.
If I run
## capture all the output to a file.
zz <- file("C:/Temp/all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message") # I think ,this was your suggestion
sink()
unlink("C:/Temp/all.Rout")
the script is executed without error or warning message, the file is
created in "C:/Temp/" but if I try to open it right away after the script
is done the message
DE: "Auf das Dokument "C:\Temp\all.Rout" kann nicht zugegriffen werden, da
es von einer anderen Anwendung verwendet wird."
EN: "Cannot access the document "C:\Temp\all.Rout" cause it is used by
another application."
What do I do wrong?
Kind regards
Georg
Von:Sarah Goslee
An: g.maub...@weinwolf.de,
Datum: 10.05.2016 18:46
Betreff:Re: Re: [R] Antwort: Re: Re: sink(): Cannot open file
On Tue, May 10, 2016 at 12:34 PM, wrote:
> sink(type = "message")
But did you do that ^^ as I suggested?
If you start a message sink with
sink(zz, type="message")
as you did, you need to explicitly close that stream. Just using
sink()
doesn't do it.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Antwort: Re: Re: sink(): Cannot open file
Duncan,
thanks for the hint.
I have done it correctly in R fashion
## capture all the output to a file.
zz <- file("C:/Temp/all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message")
sink()
unlink("C:/Temp/all.Rout")
But the error persits.
Kind regards
Georg
Von:Duncan Murdoch
An: John Sorkin , drjimle...@gmail.com,
g.maub...@weinwolf.de,
Kopie: r-help@r-project.org
Datum: 10.05.2016 19:03
Betreff:Re: [R] Antwort: Re: Re: sink(): Cannot open file
On 10/05/2016 11:15 AM, John Sorkin wrote:
> George,
> I do not know what operating system you are working with, but when I use
sink() under windows, I need to specify a valid path which I don't see in
your code. I might, for example specify:
>
> sink("c:\myfile.txt")
Note that the backslash should be doubled (so it isn't interpreted as an
escape for the "m" that follows it), or replaced with a forward slash.
Duncan Murdoch
> R code goes here
> sink()
>
> with the expectation that I would create a file myfile.txt that would
contain the output of my R program.
>
> John
>
>
> John David Sorkin M.D., Ph.D.
> Professor of Medicine
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
> Baltimore VA Medical Center
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> (Phone) 410-605-7119
> (Fax) 410-605-7913 (Please call phone number above prior to faxing)
> >>> 05/10/16 11:10 AM >>>
> Hi Jim,
>
> I tried:
>
> sink("all.Rout")
> try(log("a"))
> sink()
>
> The program executes without warning or error. The file "all.Rout" is
> begin created. Nothing will be written to it. The file is accessable
> rights after the execution of the program by notepad.exe.
>
> The program
>
> zz <- file("all.Rout", open = "wt")
> sink(zz, type = "message")
> try(log("a"))
> sink()
> close(zz)
> unlink(zz)
>
> creates the file, does not write anything to it and is not accessable
> after program execution in R with notepad.exe.
>
> Any ideas what happens behind the szenes?
>
> Kind regards
>
> Georg
>
>
>
>
> Von: Jim Lemon
> An: g.maub...@weinwolf.de,
> Kopie: r-help mailing list
> Datum: 10.05.2016 13:16
> Betreff: Re: Re: [R] sink(): Cannot open file
>
>
>
> Have you tried:
>
> sink("all.Rout")
> try(log("a"))
> sink()
>
> Jim
>
> On Tue, May 10, 2016 at 9:05 PM, wrote:
> > Hi Jim,
> >
> > thanks for your reply.
> >
> > ad 1)
> > "all.Rout" was created in the correct directory. It exists properly
with
> > correct file properties on Windows, e.g. creation date and time and
file
> > size information.
> >
> > ad 2)
> > I can not access the file with Notepad.exe directly after it was
created
> > by R. The error message is (translated):
> >
> > "Cannot access file "all.Rout". The file is opened by another
process."
> >
> > ad 3)
> > If I close R completely the file access is released. Then I can read
the
> > file using Notepad.exe. The contents is:
> >
> > Error in log("a") : non-numeric argument to mathematical function
> >
> > I tried
> >
> > close(zz)
> >
> > but the error persists.
> >
> > To me it looks like R is still accessing the file and not releasing
the
> > connection for other programs. close(zz) should have solved the
problem
> > but unfortantely it doesn't.
> >
> > What else could I try?
> >
> > Kind regards
> >
> > Georg
> >
> >
> >
> >
> > Von: Jim Lemon
> > An: g.maub...@weinwolf.de,
> > Kopie: r-help mailing list
> > Datum: 10.05.2016 12:50
> > Betreff: Re: [R] sink(): Cannot open file
> >
> >
> >
> > Hi Georg,
> > I don't suppose that you have:
> >
> > 1) checked that the file "all.Rout" exists somewhere?
> >
> > 2) if so, looked at the file with Notepad, perhaps?
> >
> > 3) let us in on the secret by pasting the contents of "all.Rout" into
> > your message if it is not too big?
> >
> > At a guess, trying:
> >
> > close(zz)
> >
> > might get you there.
> >
> > Jim
> >
> > On Tue, May 10, 2016 at 5:25 PM, wrote:
> >> Hi All,
> >>
> >> I would like to route the output to a file using sink(). When using
the
> >> example from the ?sink documentation:
> >>
> >> sink("sink-examp.txt")
> >> i <- 1:10
> >> outer(i, i, "*")
> >> sink()
> >> unlink("sink-examp.txt")
> >>
> >> ## capture all the output to a file.
> >> zz <- file("all.Rout", open = "wt")
> >> sink(zz)
> >> sink(zz, type = "message")
> >> try(log("a"))
> >> ## back to the console
> >> sink(type = "message")
> >> sink()
> >> file.show("all.Rout")
> >>
> >> I can not open the file in Windows Explorer. The error message is:
> >>
> >> "Cannot open file. File is in use be another proces."
> >>
> >> How can I close the file in a manner that I can open it right after
it
> > was
> >> created?
> >>
> >> Kind regards
> >>
> >> Georg
> >>
> >> __
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEAS
[R] Antwort: Re: Antwort: Re: Antwort: Re: Re: sink(): Cannot open file (SOLVED)
Hi Henrik, Jim, Sarah, Duncan,
Hi All,
I have tried the built-in solution using PowerShell:
$lockedFile="C:\Windows\System32\wshtcpip.dll"
Get-Process | foreach{$processVar = $_;$_.Modules | foreach{if($_.FileName
-eq $lockedFile){$processVar.Name + " PID:" + $processVar.id}}}
It did not show any processes.
Then I tried the solution using "RessourceMonitor". There I found two
processes:
rstudio.exe
rsession.exe
Right-clicking on rstudio.exe and selecting "Warteschlange analysieren" (=
analyse queue?) showed nothing. Right-clicking on rsession.exe and
selecting "Warteschlage" said:
"Mindestens ein Thread von rsession.exe wartet auf die Fertigstellung von
Netzwerk E/A". (= "At least one thread of "rsession.exe" is waiting for
finishing a network i/o operation").
Putting rsession.exe into the search field of the handles tap of
RessourceMonitor gave no results. No handles were identified.
I can not follow the suggestions where installation of software is
required due to security rules of the company I work for.
I had a look at different R versions on my machine:
1) R i386 3.2.2
2) R i386 3.2.4 (revised)
3) R i386 3.2.5
4) R x54 3.2.2
5) R x64 3.2.4 (revised)
6) R x64 3.2.5
I did
## capture all the output to a file.
zz <- file("C:/Temp/all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message")
sink()
unlink("C:/Temp/all.Rout")
on R i386 3.2.2 and R x64 3.2.2 directly without RStudio. In both cases
the file was locked.
Adding
close(zz)
solved the problem in both versions.
Encouraged by this I tired (successivly refered to as "complete code")
## capture all the output to a file.
zz <- file("C:/Temp/all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message")
sink()
unlink("C:/Temp/all.Rout")
close(zz)
on R i386 3.2.4 (revised) and R x64 3.2.4 (revised) without RStudio. Works
in both cases. The same with R i386 3.2.5 and R x64 3.2.5 each without
RStudio.
It did the same with RStudio altering the R version in the RStudio session
using "complete code". The results are:
R i386 3.2.2: OK
R. x64 3.2.2: OK
R i386 3.2.4 (revised): OK
R x64 3.2.4 (revised): OK
R i386 3.2.5: OK
R x64 3.2.5: OK
This got me lost. I had tried the complete code the last days a hundred
times. It never worked.
Then I restarted my machine powering up RStudio x64 3.2.5 using the
"complete code" and ... it worked.
I have no idea what was wrong the last days.
As far as I can say today the documentation of ?sink in R is currently
## capture all the output to a file.
zz <- file("all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message")
sink()
file.show("all.Rout")
and should be - in my opinion - supplemented with
close(zz).
Any thoughts?
Kind regards
Georg
Von:Henrik Bengtsson
An: g.maub...@weinwolf.de,
Kopie: Duncan Murdoch , "r-help@r-project.org"
Datum: 11.05.2016 21:48
Betreff:Re: [R] Antwort: Re: Antwort: Re: Re: sink(): Cannot open
file
Sounds like it would be helpful to find out exactly which process is
holding on to the file in order to figure out what's going on. From a
quick look, it seems that
http://superuser.com/questions/117902/find-out-which-process-is-locking-a-file-or-folder-in-windows
gives some useful info on how to track down the process that looks the
file.
/Henrik
On Wed, May 11, 2016 at 9:47 AM, wrote:
> Duncan,
>
> thanks for the hint.
>
> I have done it correctly in R fashion
>
> ## capture all the output to a file.
> zz <- file("C:/Temp/all.Rout", open = "wt")
> sink(zz)
> sink(zz, type = "message")
> try(log("a"))
> ## back to the console
> sink(type = "message")
> sink()
> unlink("C:/Temp/all.Rout")
>
> But the error persits.
>
> Kind regards
>
> Georg
>
>
>
>
> Von:Duncan Murdoch
> An: John Sorkin , drjimle...@gmail.com,
> g.maub...@weinwolf.de,
> Kopie: r-help@r-project.org
> Datum: 10.05.2016 19:03
> Betreff:Re: [R] Antwort: Re: Re: sink(): Cannot open file
>
>
>
> On 10/05/2016 11:15 AM, John Sorkin wrote:
>> George,
>> I do not know what operating system you are working with, but when I
use
> sink() under windows, I need to specify a valid path which I don't see
in
> your code. I might, for example specify:
>>
>> sink("c:\myfile.txt")
>
> Note that the backslash should be doubled (so it isn't interpreted as an
> escape for the "m" that follows it), or replaced with a forward slash.
>
> Duncan Murdoch
>
>> R code goes here
>> sink()
>>
>> with the expectation that I would create a file myfile.txt that would
> contain the output of my R program.
>>
>> John
>>
>>
>> John David Sorkin M.D., Ph.D.
>> Professor of Medicine
>> Chief, Biostatistics and Informatics
>> University of Maryland School of Medicine Division of Gerontology and
> Geriatric Medicine
>> Baltimore VA Medical Center
>> 10 North Greene Street
>> GRECC (BT/18/GR)
>> Ba
[R] Antwort: Antwort: Re: Re: Antwort: Re: Re: sink(): Cannot open file
Hi Martin,
many thanks for following-up on my question.
I did it again:
## capture all the output to a file.
zz <- file("C:/Temp/all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
## back to the console
sink(type = "message")
sink()
close(zz)
This works.
I tried several other combinations of the commands, e.g.
## capture all the output to a file.
zz <- file("C:/Temp/all.Rout", open = "wt")
sink(zz)
sink(zz, type = "message")
try(log("a"))
close(zz)
Does not work.
As far as I have understood right now, I have to loosen the connection of
the streams with sink(zz, type = "message") and sink() before I can close
the file connection itself.
If I did it like in the last example the connection to the file is lost
and then the connection to the streams of sink() can not be recovered.
This will last until the R session is closed and opened again.
To me it looks like I need to learn more about the operation of R under
the hood.
Kind regards
Georg
Von:Martin Maechler
An: ,
Kopie: Sarah Goslee ,
Datum: 12.05.2016 10:40
Betreff:[R] Antwort: Re: Re: Antwort: Re: Re: sink(): Cannot open
file
> Hi Sarah,
> yes, I followed your suggestion.
I doubt that you followed it correctly. Sarah's advise is
usually really very sound -- and your code below is *not* :
> If I do exactly what is in the example of the documentation:
> sink("C:/Temp/sink-examp.txt")
> i <- 1:10
> outer(i, i, "*")
> sink()
> unlink("C:/Temp/sink-examp.txt")
> it does not write anything, i. e. no file is created in "C:/Temp/". The
> script is executed without an error or warning message.
Well, did you ever lookup what unlink() does ?
I save you the time : it does *REMOVE* a file.
So no wonder that you don't see any result after executing the
above R code block..
Martin
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[R] How to plot a bunch of dichotomous code variables in one plot using ggplot2
Hi All, I have a bunch of dichotomous code variables which shall be plotted in one graph using one of their values, this is "1" in this case. The dataset looks like this: -- cut -- var1 <- c(1,0,1,0,0,1,1,1,0,1) var2 <- c(0,1,1,1,1,0,0,0,0,0) var3 <- c(1,1,1,1,1,1,1,1,0,1) ds <- data.frame(var1, var2, var3) -- cut -- I would like to have a bar plot like this * * * * * * * * * * * * * * * * * * * * - var1 var2 var3 If this possible in R? If so, how can I achieve this? Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: RE: How to plot a bunch of dichotomous code variables in one plot using ggplot2
Hi Bob, Hi John, Hi readers, many thanks for your reply. I did barplot(colSums(dataset %>% select(FirstVar:LastVar))) and it worked fine. How would I do it with ggplot2? Kind regards Georg Von:"Fox, John" An: "g.maub...@weinwolf.de" , Kopie: "r-help@r-project.org" Datum: 05.10.2016 15:01 Betreff:RE: [R] How to plot a bunch of dichotomous code variables in one plot using ggplot2 Dear Georg, How about barplot(colSums(ds)) ? Best, John - John Fox, Professor McMaster University Hamilton, Ontario Canada L8S 4M4 Web: socserv.mcmaster.ca/jfox > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of > g.maub...@weinwolf.de > Sent: October 5, 2016 8:47 AM > To: r-help@r-project.org > Subject: [R] How to plot a bunch of dichotomous code variables in one plot > using ggplot2 > > Hi All, > > I have a bunch of dichotomous code variables which shall be plotted in one > graph using one of their values, this is "1" in this case. > > The dataset looks like this: > > -- cut -- > var1 <- c(1,0,1,0,0,1,1,1,0,1) > var2 <- c(0,1,1,1,1,0,0,0,0,0) > var3 <- c(1,1,1,1,1,1,1,1,0,1) > > ds <- data.frame(var1, var2, var3) > -- cut -- > > I would like to have a bar plot like this > > > > * > * > * > * > * * > * * > * * * > * * * > * * * > * * * > - > var1 var2 var3 > > If this possible in R? If so, how can I achieve this? > > Kind regards > > Georg > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Documenting a function using roxygen2
Hi All,
I began to document my functions using roxygen2. This is an example of a
function I would like to write for training and testing purposes:
t_simple_table <- function(variable,
useNA = TRUE,
print = FALSE) {
#' @title Create a simple table for one variable.
#'
#' @description t_simple_table() creates absolute and relative
#' frequencies, cumulative sums and column sums for both as well as
#' overall statistics about valid N and missing values.
#'
#'
#' @param variable (vector, list, data.frame): variable the table is
#' created for.
#' @param useNA (logical): flag to include or exclude missing values
#' from the computation.
#' @param print (logical): flag to print/not print a table before
#' returning it as an object.
#'
#' @operation
#' Coerces the given variable to a factor.
#' If useNA = TRUE NA is also transformed to a valid value,
#' if useNA = FALSE it is disregarded in all operations.
#'
#' @return Returns a table with the following statistics:
#'
#' Frequencies Percent Cumulative
#' Percent
#' Valid . .
#' Missing . .
#' Total . 100
#' Categories
#' Cat 1 . ..
#' Cat 2 . ..
#' Cat 3 . ..
#' ... . . 100
#' Total . 100
#'
#' @errorhandling None
#'
#' @version "0.1"
#'
#' @created "2016-10-11"
#' @updated "2016-10-11"
#'
#' @status development
#'
#' @see Manderscheid: Sozialwissenschaftliche Datenanalyse mit R,
#' p. 79ff
#'
#' @author Georg
#'
#' @license GPL-2
# function body to be defined
}
Is this a correct header for a function?
How could I do better?
Kind regards
Georg
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[R] Visibility of libraries called from within functions
Hi All,
in my R programs I use different libraries to work with Excel sheets, i.
e. xlsx, excel.link.
When running chunks of code repeatedly and not always in the order the
program should run for development purposes I ran into trouble. There were
conflicts between the methods within these functions causing R to crash.
I thought about defining functions for the different task and calling the
libraries locally to there functions. Doing this test
-- cut --
f_test <- function() {
library(xlsx)
cat("Loaded packages AFTER loading library")
print(search())
}
cat("Loaded packages BEFORE function call ")
search()
f_test()
cat("Loaded packages AFTER function call -")
search()
-- cut --
showed that the library "xlsx" was loaded into the global environment and
stayed there although I had expected R to unload the library when leaving
the function. Thus confilics can occur more often.
I had a look into ?library and saw that there is no argument telling R to
hold the library in the calling environment.
How can I load libraries locally to the calling functions?
Kind regards
Georg
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[R] Antwort: Re: Visibility of libraries called from within functions
Hi Duncan,
many thanks for your reply.
Your suggestion of using requireNamespace() together with explicit
namespace calling using the "::" operator is what I was looking for:
-- cut --
f_test <- function() {
requireNamespace("openxlsx")
cat("Loaded packages AFTER loading library")
print(search())
xlsx::read.xlsx(file = "c:/temp/test.xlsx",
sheetName = "test")
}
cat("Loaded packages BEFORE function call ")
search()
f_test()
cat("Loaded packages AFTER function call -")
search()
-- cut --
When reading ?requireNamespace I did not really get how R operates behind
the scenes.
Using "library" attaches the namespace to the search path. Using
"requireNamespace" does not do that.
But how does R find the namespace then? What kind of list or directory
used R to to store the namespace and lookup the correct function or
methods of this namespace?
Kind regards
Georg
Von:Duncan Murdoch
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 13.10.2016 10:43
Betreff:Re: [R] Visibility of libraries called from within
functions
On 13/10/2016 4:18 AM, g.maub...@weinwolf.de wrote:
> Hi All,
>
> in my R programs I use different libraries to work with Excel sheets, i.
> e. xlsx, excel.link.
>
> When running chunks of code repeatedly and not always in the order the
> program should run for development purposes I ran into trouble. There
were
> conflicts between the methods within these functions causing R to crash.
>
> I thought about defining functions for the different task and calling
the
> libraries locally to there functions. Doing this test
>
> -- cut --
>
> f_test <- function() {
> library(xlsx)
> cat("Loaded packages AFTER loading library")
> print(search())
> }
>
> cat("Loaded packages BEFORE function call ")
> search()
>
> f_test()
>
> cat("Loaded packages AFTER function call -")
> search()
>
> -- cut --
>
> showed that the library "xlsx" was loaded into the global environment
and
> stayed there although I had expected R to unload the library when
leaving
> the function. Thus confilics can occur more often.
>
> I had a look into ?library and saw that there is no argument telling R
to
> hold the library in the calling environment.
>
> How can I load libraries locally to the calling functions?
You can detach at the end of your function, but that's tricky to get
right: the package might have been on the search list before your
function was called. It's better not to touch the search list at all.
The best solution is to use :: notation to get functions without putting
them on the search list. For example, use
xlsx::write.xlsx(data, file)
If you are not sure if your user has xlsx installed, you can use
requireNamespace() to check.
Duncan Murdoch
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and provide commented, minimal, self-contained, reproducible code.
[R] Antwort: Re: Antwort: Re: Visibility of libraries called from within functions
Von:Duncan Murdoch
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 13.10.2016 12:34
Betreff:Re: Antwort: Re: [R] Visibility of libraries called from
within functions
On 13/10/2016 6:21 AM, g.maub...@weinwolf.de wrote:
> Hi Duncan,
>
> many thanks for your reply.
>
> Your suggestion of using requireNamespace() together with explicit
> namespace calling using the "::" operator is what I was looking for:
>
> -- cut --
>
> f_test <- function() {
> requireNamespace("openxlsx")
> cat("Loaded packages AFTER loading library")
> print(search())
> xlsx::read.xlsx(file = "c:/temp/test.xlsx",
> sheetName = "test")
> }
Not sure if that's a typo in your message or a real error, but you
require "openxlsx" and then use "xlsx".
It's a typo!
>
> cat("Loaded packages BEFORE function call ")
> search()
>
> f_test()
>
> cat("Loaded packages AFTER function call -")
> search()
>
> -- cut --
>
> When reading ?requireNamespace I did not really get how R operates
behind
> the scenes.
>
> Using "library" attaches the namespace to the search path. Using
> "requireNamespace" does not do that.
>
> But how does R find the namespace then? What kind of list or directory
> used R to to store the namespace and lookup the correct function or
> methods of this namespace?
R has an internal list of packages that are loaded. Functions in them
are only visible to user code if the package is *also* on the search
list, or if the package name prefix is used with ::.
Can I have a look at this internal list like I can do with search() for
pachages or ls() for objects?
If xlsx is loaded, xlsx::read.xlsx will just use it; if it is not
loaded, the package will be loaded to make the call. So you don't need
the requireNamespace call if you can be sure that xlsx will be found.
You would normally use its return value (FALSE if the package is not
found) to test whether it will be safe to make the xlsx::read.xlsx call.
Got it!
Duncan Murdoch
>
> Kind regards
>
> Georg
>
>
>
>
> Von:Duncan Murdoch
> An: g.maub...@weinwolf.de, r-help@r-project.org,
> Datum: 13.10.2016 10:43
> Betreff:Re: [R] Visibility of libraries called from within
> functions
>
>
>
> On 13/10/2016 4:18 AM, g.maub...@weinwolf.de wrote:
>> Hi All,
>>
>> in my R programs I use different libraries to work with Excel sheets,
i.
>> e. xlsx, excel.link.
>>
>> When running chunks of code repeatedly and not always in the order the
>> program should run for development purposes I ran into trouble. There
> were
>> conflicts between the methods within these functions causing R to
crash.
>>
>> I thought about defining functions for the different task and calling
> the
>> libraries locally to there functions. Doing this test
>>
>> -- cut --
>>
>> f_test <- function() {
>> library(xlsx)
>> cat("Loaded packages AFTER loading library")
>> print(search())
>> }
>>
>> cat("Loaded packages BEFORE function call
")
>> search()
>>
>> f_test()
>>
>> cat("Loaded packages AFTER function call
-")
>> search()
>>
>> -- cut --
>>
>> showed that the library "xlsx" was loaded into the global environment
> and
>> stayed there although I had expected R to unload the library when
> leaving
>> the function. Thus confilics can occur more often.
>>
>> I had a look into ?library and saw that there is no argument telling R
> to
>> hold the library in the calling environment.
>>
>> How can I load libraries locally to the calling functions?
>
> You can detach at the end of your function, but that's tricky to get
> right: the package might have been on the search list before your
> function was called. It's better not to touch the search list at all.
>
> The best solution is to use :: notation to get functions without putting
> them on the search list. For example, use
>
> xlsx::write.xlsx(data, file)
>
> If you are not sure if your user has xlsx installed, you can use
> requireNamespace() to check.
>
> Duncan Murdoch
>
>
>
>
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Reshaping geographic data
Hi All, I need to reshape an ESRI shape file: http://arnulf.us/PLZ and resp http://www.metaspatial.net/download/plz.tar.gz I found an instruction for T-SQL Server: https://blog.oraylis.de/2010/05/german-map-spatial-data-for-plz-postal-code-regions/ How can I do this using R? Kind regards Georg -- cut -- Here's my code so far: download.file( url = "http://www.metaspatial.net/download/plz.tar.gz";, destfile = "C:/temp/plz.tar.gz") untar(tarfile = "C:/temp/plz.tar.gz", exdir = "C:/temp", compressed = "gzip") __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Storing long string with white space in variable
Hi All,
I would like to store a long string with white space in a variable:
-- cut --
# Create README.md
readme <- "---
title: "Your project title here"
author: "Author(s) name(s) here"
date: "Current date here"
output: html_document
---
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, cache = FALSE)
```
# Project Context
# Goals
# Approach
# Reference to main program
´´´{r}
source("main_program.R")
´´´
# Information on used system and configuration
```{r}
cat("Gathering system information ...\n)
sessionInfo()
```
"
cat(readme, file = "README.md")
-- cut --
I am looking for an equivalent to Pythons """ """ long string feature.
I searched the web and found this:
http://stackoverflow.com/questions/6329962/split-code-over-multiple-lines-in-an-r-script
https://stat.ethz.ch/pipermail/r-help/2006-October/115358.html
But this is not the solution to the problem.
How can I store long strings with white space in a variable?
Kind regards
Georg
PS: This is a template for a project folder for each project. I would like
to create it with R script instead of distributing it as a template file.
This way one needs only the R script to setup a project like this:
#---
# Module: t_setup_project_directory.R
# Author: Georg Maubach
# Date : 2016-10-19
# Update: 2016-10-19
# Description : Setup a directory structure for a new analytics
# project
# Source System : R 3.3.0 (64 Bit)
# Target System : R 3.3.0 (64 Bit)
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
#1-2-3-4-5-6-7--
t_version = "2016-10-19"
t_module_name = "t_setup_project_directory.R"
t_status = "development"
cat(
paste0(
"\n",
t_module_name,
" (Version: ",
t_version,
", Status: ",
t_status,
")",
"\n",
"\n",
"Copyright (C) Georg Maubach 2016
This software comes with ABSOLUTELY NO WARRANTY.",
"\n",
"\n"
)
)
library(svDialogs)
# If do_test is not defined globally define it here locally by
un-commenting it
t_do_test <- FALSE
# [ Function Defintion
]
t_setup_project_directory <- function() {
#-
# Setup a directory structure for a new analytics
#
# Args:
# None.
#
# Operation:
# The user can create or select a directory for the projects files.
# The function then places all sub directories in this project
# folder.
# The function saves a RData file with objects containing the path
# to project directory and its sub folders.
#
# Returns:
# Nothing.
#
# Error handling:
# None.
#
# See also:
# ./.
#-
# Get and/or create project directory
v_project_dir <- svDialogs::dlgDir()$res
# Define names for sub directories
data <- "data" # data to be loaded into or
# saved from R
documentation <- "documentation"# explanatory material for results
# (e. g. knitR documents)
fundamentals <- "fundamentals" # background knowledge
input <- "data/input" # input data eventually manually
# revised for import
meta <- "data/meta"# meta data (e. g. lookup tables)
output<- "data/output"
raw <- "data/raw" # a copy of all input data never
# touched for safety reasons and
# not read by R
program <- "program" # all scripts and runnable files
modules <- "program/modules" # project specific packages, files
# or functions in separate files as
# well as all other sub routines to
# be sourced or loaded
results <- "results" # container for all resulring data
# in an aggregated form
graphics <- "results/graphics"
tables<- "results/tables"
presentations <- "results/presentations"
temp <- "temp"
v_paths_relative <- list(
project = v_project_dir,
documentation = documentation,
fundamentals = fundamentals,
input = input,
meta = meta,
output= output,
raw = raw,
program = program,
modules = modules,
graphic = graphics,
table = tables,
presentation = prese
[R] openxlsx Error: length of rows and cols must be
Hi All, when using -- cut -- number_style <- openxlsx::createStyle( numFmt = "COMMA" ) openxlsx::addStyle( wb = xlsx_workbook, sheet = "Kundenliste", style = number_style, rows = 2:nrow(customer_list), cols = 4:5 ) --cut -- I get the error Error in openxlsx::addStyle(wb = xlsx_workbook, sheet = "Kundenliste", : Length of rows and cols must be equal. The customer_list can be of any arbritrary length due to subgroup definitons. I do not see why the argument "rows" and "cols" should be of the same length. This would mean that number formatting can only be done for rectangular areas. What do I need to change to format my numbers in the given area correctly? Kind regards Georg __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Different results when converting a matrix to a data.frame
Hi All, I build an empty dataframe to fill it will values later. I did the following: -- cut -- matrix(NA, 2, 2) [,1] [,2] [1,] NA NA [2,] NA NA > data.frame(matrix(NA, 2, 2)) X1 X2 1 NA NA 2 NA NA > as.data.frame(matrix(NA, 2, 2)) V1 V2 1 NA NA 2 NA NA -- cut -- Why does data.frame deliver different results than as.data.frame with regard to the variable names (V instead of X)? Kind regards Georg [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] for loop is looping only once
Hi All,
I need to execute a loop on variables to compute several KPIs.
Unfortunately the for loop is executed only once for the last KPI given.
The code below illustrates my current solution but is not completely
necessary to spot the problem. I just give an idea what I am doing
overall. Looks much but isn't if copied and run in RStudio. The problem
occurs in function f_create_kpi_table() in lines 150 to 157:
for (item in length(kpis)) # This loop runs only once!
{
print(kpis[[item]])
ds_kpi <- f_compute_kpi(
years= years,
kpi = kpis[[item]],
kpi_base = kpi_bases[[item]])
print(ds_kpi)
Here is the complete example code with example data:
- cut --
dataset <-
structure(
list(
to_2012 = c(
85,
822,
891,
700,
386,
127,
938,
381,
871,
254,
793,
0,
934,
217,
163,
755,
607,
794,
477
),
to_2013 = c(
289,
0,
963,
243,
608,
47,
0,
941,
998,
775,
326,
0,
0,
470,
248,
439,
212,
0,
0
),
to_2014 = c(0, 0, 71, 0, 0, 434, 0, 282, 0,
0, 405, 0, 0, 642, 0, 0, 0, 47, 299),
to_2015 = c(
705,
134,
659,
0,
609,
807,
783,
0,
0,
304,
141,
500,
0,
0,
764,
790,
851,
0,
802
),
kpi1_2013 = c(0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0,
0, 0, 0, 1, 1),
kpi1_2014 = c(1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0,
1, 1, 0, 1, 1, 1, 0, 0),
kpi1_2015 = c(0, 0, 0, 1, 0, 0, 0, 1,
1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0),
kpi1_2016 = c(0, 1, 0, 1, 0,
1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1),
kpi2_2013 = c(1, 0,
1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0),
kpi2_2014 = c(0,
0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1),
kpi2_2015 = c(1,
1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1),
kpi2_2016 = c(1,
0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0)
),
.Names = c(
"to_2012",
"to_2013",
"to_2014",
"to_2015",
"kpi1_2013",
"kpi1_2014",
"kpi1_2015",
"kpi1_2016",
"kpi2_2013",
"kpi2_2014",
"kpi2_2015",
"kpi2_2016"
),
row.names = c(NA, 19L),
class = "data.frame"
)
f_compute_kpi <- function(
years,
kpi,
kpi_base)
{
print(years)
print(kpi)
print(kpi_base)
ds_result <- data.frame()
for (year in years) {
current_year <- year
previous_year <- year - 1
result <- sum(dataset[dataset[[paste0(kpi,
"_",
current_year)]] == 1 ,
paste0(kpi_base,
"_", previous_year)],
na.rm = TRUE)
ds_result <- rbind(ds_result, result)
}
ds_result <- t(ds_result)
rownames(ds_result) <- kpi
colnames(ds_result) <- years
invisible(ds_result)
}
f_create_kpi_table <- function(
years,
kpis,
kpi_bases)
{
print(length(kpis))
#-- Problematic loop --
for (item in length(kpis)) # This loop runs only once!
{
print(kpis[[item]])
ds_kpi <- f_compute_kpi(
years= years,
kpi = kpis[[item]],
kpi_base = kpi_bases[[item]])
print(ds_kpi)
}
# This for loop is executed only once for kpi2 instead of
# as many times as given kpis in length(kpis), i. e.
# kpi1 AND kpi2.
# Why?
# What do I do wrong?
}
-- cut --
What do I need to change to get the loop work correctly and loop over two
elements instead of one when calling the function
f_create_kpi_table(years = 2013:2016, kpis = c("kpi1", "kpi2"), kpi_bases
= c("to", "to"))
Kind regards
Georg
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[R] Antwort: Re: for loop is looping only once [SOLVED]
Hi Ulrik,
oh no! What a mistake did I make. But I definitely did not see the
failure.
Many thanks for helping me.
Kind regards
Georg
Von:Ulrik Stervbo
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 17.11.2016 12:24
Betreff:Re: [R] for loop is looping only once
Hi Georg,
Your for loop iterates over just one value, to get it to work as you
intend use for(item in 1:length(kpis)){}
HTH
Ulrik
On Thu, 17 Nov 2016 at 12:18 wrote:
Hi All,
I need to execute a loop on variables to compute several KPIs.
Unfortunately the for loop is executed only once for the last KPI given.
The code below illustrates my current solution but is not completely
necessary to spot the problem. I just give an idea what I am doing
overall. Looks much but isn't if copied and run in RStudio. The problem
occurs in function f_create_kpi_table() in lines 150 to 157:
for (item in length(kpis)) # This loop runs only once!
{
print(kpis[[item]])
ds_kpi <- f_compute_kpi(
years= years,
kpi = kpis[[item]],
kpi_base = kpi_bases[[item]])
print(ds_kpi)
Here is the complete example code with example data:
- cut --
dataset <-
structure(
list(
to_2012 = c(
85,
822,
891,
700,
386,
127,
938,
381,
871,
254,
793,
0,
934,
217,
163,
755,
607,
794,
477
),
to_2013 = c(
289,
0,
963,
243,
608,
47,
0,
941,
998,
775,
326,
0,
0,
470,
248,
439,
212,
0,
0
),
to_2014 = c(0, 0, 71, 0, 0, 434, 0, 282, 0,
0, 405, 0, 0, 642, 0, 0, 0, 47, 299),
to_2015 = c(
705,
134,
659,
0,
609,
807,
783,
0,
0,
304,
141,
500,
0,
0,
764,
790,
851,
0,
802
),
kpi1_2013 = c(0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0,
0, 0, 0, 1, 1),
kpi1_2014 = c(1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0,
1, 1, 0, 1, 1, 1, 0, 0),
kpi1_2015 = c(0, 0, 0, 1, 0, 0, 0, 1,
1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0),
kpi1_2016 = c(0, 1, 0, 1, 0,
1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1),
kpi2_2013 = c(1, 0,
1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0),
kpi2_2014 = c(0,
0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1),
kpi2_2015 = c(1,
1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1),
kpi2_2016 = c(1,
0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0)
),
.Names = c(
"to_2012",
"to_2013",
"to_2014",
"to_2015",
"kpi1_2013",
"kpi1_2014",
"kpi1_2015",
"kpi1_2016",
"kpi2_2013",
"kpi2_2014",
"kpi2_2015",
"kpi2_2016"
),
row.names = c(NA, 19L),
class = "data.frame"
)
f_compute_kpi <- function(
years,
kpi,
kpi_base)
{
print(years)
print(kpi)
print(kpi_base)
ds_result <- data.frame()
for (year in years) {
current_year <- year
previous_year <- year - 1
result <- sum(dataset[dataset[[paste0(kpi,
"_",
current_year)]] == 1 ,
paste0(kpi_base,
"_", previous_year)],
na.rm = TRUE)
ds_result <- rbind(ds_result, result)
}
ds_result <- t(ds_result)
rownames(ds_result) <- kpi
colnames(ds_result) <- years
invisible(ds_result)
}
f_create_kpi_table <- function(
years,
kpis,
kpi_bases)
{
print(length(kpis))
#-- Problematic loop --
for (item in length(kpis)) # This loop runs only once!
{
print(kpis[[item]])
ds_kpi <- f_compute_kpi(
years= years,
kpi = kpis[[item]],
kpi_base = kpi_bases[[item]])
print(ds_kpi)
}
# This for loop is executed only once for kpi2 instead of
# as many times as given kpis in length(kpis), i. e.
# kpi1 AND kpi2.
# Why?
# What do I do wrong?
}
-- cut --
What do I need to change to get the loop work correctly and loop over two
elements instead of one when calling the function
f_create_kpi_table(years = 2013:2016, kpis = c("kpi1", "kpi2"), kpi_bases
= c("to", "to"))
Kind regards
Georg
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[R] openxlsx: No Formatting of Numbers (TEXT ONLY)
Hi All,
Dear Readers,
I am using openxlsx to export data to Microsoft Excel 2013, 32-Bit, German
Version:
--- schnipp ---
library("openxlsx")
dataset <- structure(
list(
a = c(1126039.81, 45636.44, 14847.41),
b = c(1194447.5,
88310.53, 18699.68),
c = c(1560307.73, 34203.73, 24755.99),
d = c(1068790.67,
67581.86, 12378.55)
),
.Names = c("a", "b", "c", "d"),
row.names = c(NA,
3L),
class = "data.frame"
)
xlsx_workbook <- openxlsx::createWorkbook()
openxlsx::addWorksheet(
wb = xlsx_workbook,
sheetName = "Numbers")
openxlsx::writeData(
wb = xlsx_workbook,
sheet = "Numbers",
x = dataset,
rowNames = TRUE,
colNames = TRUE,
startRow = 2,
startCol = 2,
borders = c("surrounding")
)
myStyle <- openxlsx::createStyle(numFmt = "###.###.##0")
openxlsx::addStyle(wb = xlsx_workbook,
sheet = "Numbers",
style = myStyle,
rows = 1:1,
cols = 10:10,
gridExpand = TRUE,
stack = TRUE)
openxlsx::saveWorkbook(
wb = xlsx_workbook,
file = "C:/temp/openxlsx_example.xlsx",
overwrite = TRUE
)
--- schnipp ---
The problem with this is, that it does not apply the number formats to the
Excel cell on the sheet. Also, sometimes the boarder of the data on the
Excel sheet is delete. I could not find out yet what the cause for this
behaviour is.
My sessionInfo() output is:
R version 3.3.2 (2016-10-31)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
locale:
[1] LC_COLLATE=German_Germany.1252
[2] LC_CTYPE=German_Germany.1252
[3] LC_MONETARY=German_Germany.1252
[4] LC_NUMERIC=C
[5] LC_TIME=German_Germany.1252
attached base packages:
[1] tools stats graphics grDevices utils
[6] datasets methods base
other attached packages:
[1] tidyr_0.5.1stringr_1.1.0 reshape2_1.4.1
[4] openxlsx_3.0.0 dplyr_0.5.0
loaded via a namespace (and not attached):
[1] lazyeval_0.2.0 plyr_1.8.4 magrittr_1.5
[4] R6_2.2.0 assertthat_0.1 DBI_0.4-1
[7] tibble_1.1 Rcpp_0.12.5stringi_1.1.1
I do not want to round the numbers in R, cause my clients would like to
use them as they are in further calculations.
How can I export a dataframe to Excel, print a border around the complete
table/dataset (not the single cells) and format the numbers like
123.456.789 (thousand delimiter dot ".", all numbers without decimals)?
Kind regards
Georg
__
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