Hi!
With 'dplyr':
dt_count %>% mutate(STATUS=ifelse(STATUS %in%
c("Resolved","Closed"),"Resolved/Closed",STATUS)) %>% group_by(STATUS)
%>% summarise(n=sum(N))
Output:
1 Assigned 135
2 Cancelled 20
3 In Progress56
4 Pending75
5 Resolved/Closed 1180
HTH,
Ki
llo Kimmo,
>
> Yes. I did that and it worked fine -- as far as it goes. But it
> didn't cover what to do when using the same twitter account on a
> computer with a different user name -- which is what my question was
> about.
>
>
> On Wed, 08-Apr-2020 at 08:55AM
Hi!
Have you already read this:
https://cran.r-project.org/web/packages/rtweet/vignettes/auth.html
I think they explain rather well how to use Twitter tokens with
rtweet...
HTH,
Kimmo
ke, 2020-04-08 kello 17:19 +1200, Patrick Connolly kirjoitti:
> I'm using the rtweet package which makes use o
nha kirjoitti:
> Thanks for providing the code but I also needed the output sheet in
> .csv format with all the four columns corresponding to the value
> (Chrom,
> Start_pos, End_pos & Value ranging from what I specified earlier).
>
> Puja
>
> On Fri, Jan 31, 2020 at 10:23
Hi!
Oh, sorry, one "s" too much in my code. Here the correct one:
df$Value[ (df$Value>=0.2 & df$Value<=0.4) | df$Value>=0.7 ]
Best,
Kimmo
pe, 2020-01-31 kello 17:12 +0200, K. Elo kirjoitti:
> Hi!
>
> Let's assume your data is stored in a data frame call
Hi!
Let's assume your data is stored in a data frame called 'df'. So this
code should do the job:
df$Value[ (df$Value>=0.2 & df$Values<=0.4) | df$Value>=0.7 ]
Best,
Kimmo
pe, 2020-01-31 kello 09:21 -0500, pooja sinha kirjoitti:
> Hi All,
>
> I have a .csv file with four columns (Chrom, Start
Hi,
cannot reproduce, either, on my Linuxmint 19.3 + R 3.6.2.
Here the outputs:
--- snip ---
> test(mean, 1:10)
[1] 5.5
> test(NULL, 1:10)
NULL
Error in FUN(args) : could not find function "FUN"
> test(mean, list(x=1:10, na.rm=TRUE))
[1] NA
Warning message:
In mean.default(args) : argument is n
Hi Rajesh,
2019-05-05 10:23 +0530, Rajesh Ahir_GJ wrote:
> Hello R users,
>
> I am getting an error while running following code.
>
> library(ggplot2)
> ggplot(hourly_data1,aes(hour, power))+
> geom_boxplot(aes(fill=monthname),outlier.shape=NA) +
> facet_wrap(~monthname)
> ggplot(hourly_data1,ae
Hi Drake,
2019-05-04, 17:34 -0700, Drake Gossi wrote:
> Hello everyone,
>
> I'm trying to learn how to put together a citation network, and, in
> doing so, I'm playing around with a data set of my own making. I'm
> going back and forth between two .csv files. One has two columns and
> is simply l
Hi!
2019-03-25 kello 09:30 +0800, Steven Yen wrote:
> The second command is ugly. How can I print the 25 numbers into 2
> rows
> of ten plus a helf row of 5? Thanks.
Something like this?
x<-1:25; for (i in seq(1,length(x),10))
print(x[i:ifelse((i+9)>length(x),length(x),i+9)])
HTH,
Kimmo
_
Hi!
2019-02-27 22:51 -0500, Aimin Yan wrote:
> I have a question about assigning color based on the value of a
> matrix
>
> The following is my matrix.
>
> d
> lateRT earlyRT NAD ciLAD
> lateRT 1.0 0.00 0.006224017 0.001260241
> earlyRT 0.
Hi!
Not having a data chunk prevents me from testing abit, but maybe you
should take a look on:
?table
?xtabs
to start with.
But as already suggested by other users, a small data set would be of
great help :)
HTH,
Kimmo
su, 2019-01-06 kello 13:49 -0500, Rachel Thompson kirjoitti:
> Hi Rich,
>
Hi!
Maybe this would do the trick:
--- snip ---
library(reshape2) # Use 'reshape2'
library(dplyr)# Use 'dplyr'
datatransfer<-data %>% mutate(letter2=letter) %>%
dcast(id+letter~letter2, value.var="weight")
--- snip ---
Or did I misunderstood something?
Best,
Kimmo
2019-01-06, 13:16
Hi!
Have you tried to use 'fromJSON' with the parameter 'simplifyDataFrame'
set to TRUE?
See:
https://cran.r-project.org/web/packages/jsonlite/vignettes/json-aaquickstart.html
-> Section "Data Frames" explains how this affects the data frame
structure. IMHO this should solve your problem...
Be
Hi!
Maybe not the most elegant solution, but a workaround is to have a
function:
> save2<-function(y, ...) { save(y,...)}
> save2(x1,x2,file="test.RData")
The point is to include the variables to be "renamed" as parameters (in
my example: y). The function will use the parameter variable names wh
Hi!
How about this:
--- snip --
for (i in 1:(length(split_str)-1)) {
assign(paste("DF",i,sep=""),DF[
c((which(DF$name==split_str[i])+1):(which(DF$name==split_str[i+1])-1)),
])
}
--- snip ---
'assign' creates for each subset a new data.frame DFn, where n ist a
count (1,2,...).
But note: i
Hi!
Seems to be an encoding problem. This worked for me (have not full-
checked the output, though):
fromJSON(encodeString(zWebObj))
HTH,
Kimmo
2018-05-08 12:49 -0700, David Winsemius wrote:
> >
> > On May 8, 2018, at 10:08 AM, Evans, Richard K. (GRC-H000) > k.ev...@nasa.gov> wrote:
> >
> >
Hi!
Not just an gmail issue. After my last reply I have gotten tons of
spams from "Samantha Smith". Keep hitting my "rank as spam"-button in
the hope that my MUA could learn :)
Best,
Kimmo
ti, 2018-04-17 kello 19:34 +, Ding, Yuan Chun kirjoitti:
> No, I do not use gmail, still got dirty spam
Hi!
An alternative with 'assign':
for ( i in 71:75) {
setwd(paste("C:/Awork/geneAssociation/removed8samples/neuhausen", i,
sep=""))
temp.df<-read.csv("seg.pr3.csv", head=T)
temp.df$id<-paste0("sn",i,sep="")
assign(paste0("seg",i,sep=""),temp.df)
}
rm(temp.df,i) # Clean up
HTH,
Kimmo
Hi!
2017-09-18 07:13 -0500, Therneau, Terry M., Ph.D. wrote:
> This question likely has a 1 line answer, I'm just not seeing
> it. (2, 3, or 10 lines is
> fine too.)
>
> For a vector I can do group <- match(x, unqiue(x)) to get a vector
> that labels each
> element of x.
Actually, you get a v
Hi!
Maybe this helps:
http://r.789695.n4.nabble.com/Error-in-normalizePath-path-with-McAfee-td2532324.html
Best,
Kimmo
15.12.2016, 08:18, Amelia Marsh via R-help wrote:
Hi
I had installed R studio Desktop 1.0.44. However whenever I wanted to write any
command, before I could complete, I was
Hi!
18.10.2016, 14:38, Abhinaba Roy wrote:
Hi R helpers,
I have json inputs from an app which I want to convert to dataframes. Below
are the two inputs. Can someone help me in converting these to dataframes
[...]
IMHO, the best way is to use the package 'jsonlite', see:
* https://cran.r-pro
Dear all,
I am currently working a research project on social media interaction.
As a part of this project, mostly for teaching purposes, I should
develop a R-based approach for real-time visualisation of streamed data
(from Twitter).
My idea is simple (and working :) ): A Python-script stre
Hi,
what is the exact problem? I tried you code and it works fine...
Best,
Kimmo
24.08.2016, 10:07, Serpil ŞEN wrote:
Dear Authorized Sir / Madam,
I need your help on clustering with R.
I have symmetric distance matrix which i created usign ClustalOmega program.
and used this R codes for cl
Hi!
22.06.2016, 22:00, chalabi.el...@yahoo.de wrote:
Dear Kimmo,
I already used df$speed[training] in df.xyf but I get this error:
Error in xyf(Xtraining,factor(df$speed[training]),grid=somgrid(5, :
NA/NaN/Inf in foreign function call (arg 1)
Please check for zeros (0) and NAs in df$spee
Hi again!
21.06.2016, 21:33, chalabi.el...@yahoo.de wrote:
Hi Kimmo, Thanks for your reply. I think now my problem is that I
don't understand what does factor(df.classes[training]) do?
Sorry, my mistake, should habe been 'df$speed'. Please try the following:
--- snip ---
set.seed(7) training
4 3 3 2 1 4 1 2 3 ..
and I want to cluster my data based on speed, to see the coming costumer's
protocols fall into which speed group and I think I need to bring this speed
column in Y element of xyf
On Thursday, June 16, 2016 2:29 PM, K. Elo wrote:
Hi!
Some sample data could help us t
Hi!
Some sample data could help us to help you...
But have you read '?xyf' in order to ensure that your 'Y' is what 'xyf'
expects it to be?
What kind of error messages do you get?
Regards,
Kimmo
16.06.2016, 15:13, ch.elahe via R-help wrote:
Is there any answer?
Hi all, I have a df and I
Hi Juho!
01.06.2016, 14:40, Juho Kiuru wrote:
Hi all, I am new to R and TwitteR and would love to get some advice from
you.
I managed to get list of tweets containing word 'innovation' tweeted in
Helsinki with following script:
searchTwitter('innovation', n=1, geocode='60.1920,24.9458,30mi
Hi!
Many thanks to Duncan and Jim for their quick replies.
27.05.2016, 01:08, Jim Lemon wrote:
Hi Kimmo,
par(mar=c(5,7,4,2))
dotchart(kedf$x)
mtext(kedf$Group.2,side=2,at=1:6,line=0.5,
las=2,cex=log(abs(kedf$Freq))+1)
Jim
This 'dotchart' solution worked fine and I got what I wanted :) Howev
Dear R-helpers!
I have a data frame storing data for word co-occurrences, average
distances and co-occurence frequency:
Group.1Group.2 x Freq
1 deutschland achtziger 2.001
2 deutschlandalt 1.254
3 deutschland anfang -2.001
4 deutschlandansehen 1.00
Hi,
thanks to Duncan and Jeroen to quick replies. I was actually my thinking
error :) I suppoed 'fromJSON' to cope with a multi-line file or a list,
but this seems not to be the case. So I first read the file with
'readLines' into a list and processed all items with 'fromJSON' within a
for-lo
Hi!
You can download the example file with this link:
https://www.dropbox.com/s/tlf1gkym6d83log/example.json?dl=0
BTW, I have used a JSON validator and the problem seems to related to
wrong/missing EOF.
--- snip ---
Error: Parse error on line 1:
...:"1436705823768"} {"created_at":"Sun J
-
Hi!
I have collected 500.000+ tweets with a Python script using 'tweepy',
which stored the data in JSON format. I would like to use R for data
analysis, but have encountered problems when trying to import the data
file with 'jsonlite'.
Here what I have tried:
> data.df<-fromJSON("example.js
Hi!
25.08.2015, 18:17, Sam Albers wrote:
Hi all,
This is a process question. How do folks efficiently identify column
numbers in a dataframe without manually counting them. For example, if I
want to choose columns from the iris dataframe I know of two options. I can
do this:
str(iris)'data.
Hi!
10.06.2015, 13:20, khatri wrote:
My date column is in following format : "%m/%d H:M:S"
There is not mention of year in the data.So how can I read this using
strptime function.
I have tried strptime(dates,"%m/%d H:M:S") but this is returning NA.
Thanks
How about:
strptime(dates,"%m/%d %H:%
Hi!
How about trying this:
data[ data$col1!=data$col2 & !is.na(data$col3), ]
col1 col2 col3
2a1 ST001
3b2 ST002
HTH, Kimmo
28.05.2014 15:35, jeff6868 wrote:
> Hi everybody,
>
> I have a little problem in my R-code which seems be easy to solve, but I
> wasn't able to find t
Hi!
28.09.2012 09:13, Bhupendrasinh Thakre wrote:
Statement I tried :
b <- unclass(Sys.time())
b = 1348812597
c_b <- rnorm(1,2,1)
Do you mean this:
--- code ---
> df<-data.frame("x"=0,"y"=0)
> colnames(df)
[1] "x" "y"
> colnames(df)[2]<-paste("b",unclass(Sys.time()),sep="_")
> colnames(df)
Hi!
28.09.2012 08:41, Atte Tenkanen wrote:
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
How about this:
--- code ---
g1<- c(2,1,1,4,6,3)
g2<- c(2,1,1,6,4
Hi!
On Thu, 2 Aug 2012, Prof Brian Ripley wrote:
You may bave a micro-packaged distribution (some form of SuSE as I recall):
is there a separate R-devel RPM? (Fedora had one an one time.)
Thanks, this made the trick. After having R-base-devel installed I
succeeded in intalling the 'igraph' p
Hi!
On Thu, 2 Aug 2012, Uwe Ligges wrote:
R.h should be part if your R installation, given the output above
probably in /usr/lib64/R/include ?
But there is no such file "R.h" in my system???
~$ locate \/R.h
~$
Or with 'find':
# find / -name R.h
#
Any ideas?
Kind regards,
Kimmo
__
Hi!
I want to use R for network analysis and have tried to install the
'igraph' package. Unfortunately, the installation is aborted by an error:
--snip--
gcc -std=gnu99 -I/usr/lib64/R/include -DNDEBUG -I/usr/local/include
-DUSING_R -I. -Ics -Iglpk -Iglpk/amd -Iglpk/colamd -I/usr/local/includ
Hi!
I recently posted a similar question (entitled "Adding mean line to a
lattice density plot"). Have not got any usable solution forcing my to
fall back to the use of the normal 'plot' function. The problem was the
same as yours: using panel.abline simply did not work, the position of
the m
Hi!
Maybe not the most elegant solution, but works:
for(i in seq(1,length(data)-(length(data) %% 3), 3)) {
ifelse((length(data)-i)>3, { print(sort(data)[ c(i:(i+2)) ]);
print(mean(sort(data)[ c(i:(i+2)) ])) }, { print(sort(data)[
c(i:length(data)) ]); print(mean(sort(data)[ c(i:length(data))
Hi,
some sample data would be *very* helpful...
Kind regards,
Kimmo
16.03.2012 15:44, statquant2 wrote:
Hello I am looking for a special plot.
Let's suppose I have *100 days and
*each day I have a (1D)
distribution of the same variable.
I would like
HI again,
thanks for the replies. Unfortunately, due to my deadlines, I had no
time to test Petr's suggestion. I will test it later. I had to figure
out an alternate solution, so I decided to use "normal" plotting functions:
> with(subset(ISGFINC2, as.numeric(SGENDER)==1), plot(density(PV1CIV
Hi,
thank you, Pascal, for your quick reply. Unfortunately your suggestion
is not working. Please have a look on the attachment, I have added
manually the mean lines I am trying to plot. The problem with 'abilne'
seems to be that the argument 'v' is relative to the graph area, not the
x-axis
Hi!
I have used the following command:
densityplot(~PV1CIV, groups=SGENDER, data=ISGFINC2,
lwd=2, col=1, lty=c(1,2), pch=c("+","o"),
key=list(text=list(lab=levels(ISGFINC2$SGENDER), col=1),
space="bottom", columns=2, border=T, lines=T, lwd=2,
lty=c(1,2), col=1), ref=T, plot.points=F)
to
Hi,
19.12.2011 11:12, Ville Iiskola wrote:
Error in if (abs(x - oldx)< ftol) { : missing value where TRUE/FALSE
needed
The reason for this error is in the row 563. There the choice has
value 1 and Ie has missing value. If the choice has value 0 and Ie
has missing values, then there is no erro
Hello,
three commands might do the job (NOTE: df=your data frame,
obser=Observation, var1=Variable 1 [TYPE: string], var1flag=Variable 1
flag [TYPE: string])
1. df$var1flag<-NA
2. df$var1flag[ is.na(as.numeric(df$var1)) ]<-df$var1[
is.na(as.numeric(df$var1)) ]
3. df$var1<-as.numeric(df$var1
Dear Jim,
17.03.2011 20:54, Jim Silverton wrote:
I have a matrix say:
23 1
12 12
00
0 1
0 1
0 2
23 2
I want to count of number of distinct rows and the number of disinct element
in the second column and put these counts in a column. SO at the end of the
day I should have:
c(1, 1
Hi!
21.02.2011 08:50, Schmidt, Lindsey C (MU-Student) wrote:
> What is plot.new? How can I fix this data or add plot.new so it works?
?plot.new
?plot
plot(u[h],v[h],type="l",asp=1) seems to work for me...
HTH,
Kimmo
__
R-help@r-project.org mailing li
Hi!
21.02.2011 08:50, Schmidt, Lindsey C (MU-Student) wrote:
> What is plot.new? How can I fix this data or add plot.new so it works?
?plot.new
?plot
plot(u[h],v[h],type="l",asp=1) seems to work for me...
HTH,
Kimmo
__
R-help@r-project.org mailing li
Dear Katharina,
08.02.2011 11:21, Katharina Lochner wrote:
> The following appeared on my console:
>
>> install.packages("psych")
>
> Warnung in install.packages("psych") :
>
> 'lib = "C:/PROGRA~1/R/R-212~1.1/library" ist nicht schreibbar
>
Apparently You do not have permissions to write to
Dear Romezo,
a solution maybe not that elegant and effective, but seems to work:
test_calculate<-function() {
tarrow<-1
TARGET<-data.frame("Thesis"=0, "Day"=0,"A"=0,"B"=0,"C"=0)
for (i in c(unique(my.data$Thesis))) {
for (j in c(unique(my.data$Day[ my.data$Thesis==i ]))) {
TEMPDF
Hi!
Do you mean something like this (df is your original data frame):
--- cut here ---
df1<-df
df1[[1]]<-paste("R",df[[1]],sep="_")
colnames(df1)<-c("SERIES","YEAR","value")
df1$value[ df1$YEAR==2009 ]<-5
for (i in c(2009:2007)) { df1$value[ df1$YEAR==(i-1) ]<-( df1$value[
df1$YEAR==i ]-df$DELTA
Hi Jens!
24.03.2010 14:48, koj wrote
>
> Hi Kimmo, thank you for your answer, but this is not the thing I am searching
> for. Unfortunately, I have not described the problem very good. But just in
> this moment I have a good idea: I use add=TRUE and paint two plots. And so I
> am sure that I can
Hi Jens!
23.03.2010 17:18, koj wrote:
The problem is: I want to group the data. I want to have ten groups. The
first two bars should be [1,1] and [2,1] together in one bar and in the
second bar of the first obervation should be [1,2] and [2,2] (stacked with
beside =TRUE). Therefore the first obs
Hi!
22.02.2010 19:53, xin wei wrote:
>
> hi, Kevin and K.Elo:
> thank you for the suggestion. Can you be more specific on these? (like how
> exactly get into x-switch or man ssh). I am totally ignorant about linux and
> SSH:( Memory limitation forces me to switch from windows to Linux
> c
Hi!
22.02.2010 17:45, xin wei wrote:
>
> thank you for reply. I just type: hist(x) from SSH terminal, expecting a
> histogram to pop up like what i got under windows.instead I got the
> following error msg:
>
> Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma,
> d$colortype,
Hi!
Right, my solution did not take into accound paired negative values
summing up to zero.
This should work in all cases:
df[, which(colSums(df!=0)!=0)]
Kind regards,
Kimmo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r
Dear Anna,
19.02.2010 08:17, Anna Carter wrote:
> (1) If the dataset contains some variables having all the entries = 0
> and while analysing I want to delete those pericular columns, how do
> acheive this. i.e.
Let's suppose 'df' is your data frame, then:
subset(df, select=which(colSums(df)!=0)
Hi!
29.01.2010 12:49, soeren.vo...@eawag.ch wrote:
> Hello,
>
> I read the help as well as the examples, but I can not figure out why
> the following code does not produce the *given* row names, "x" and "y":
>
> x <- 1:20
> y <- 21:40
> rbind(
> x=cbind(N=length(x), M=mean(x), SD=sd(x)),
> y
Hi!
Let's suppose the values for the x axis are stored in 'values'.
barplot(values, col=c(rep("Red",3),rep(1,length(values)-8),rep("Blue",5)))
HTH,
Kimmo
vikrant kirjoitti:
> Suppose I need to draw a Grouped bar plot with 100 values on the X axis. Now
> my question is If I need to highlight sup
Hi!
Jean-Baptiste Combes wrote:
> Hello,
>
> I use R 2.10, and I am new in R (I used to use SAS and lately Stata), I am
> using XP.
>
> I have a data which has a data.frame format called x.df (read from a csv
> file). I want to take from this data observations for which the variable
> "Code" sta
Hi!
Wacek Kusnierczyk wrote:
> m = matrix(1:4, 2)
>
> apply(m, 1, cat, '\n')
> # 1 2
> # 3 4
> # NULL
>
> why the null?
Could it be the return value of 'cat'. See ?cat, where:
---snip ---
Value
None (invisible NULL).
---snip ---
Kind regrads,
Kimmo
__
Hi!
mathallan wrote:
> How can I from the summary function, decide which glm (fit1, fit2 or fit3)
> fits to data best? I don't know what to look after, so I would please
> explain the important output.
Start with the AIC value (Akaike Information Criterion). The model
having the lowest AIC is the
Hi,
Crosby, Jacy R wrote:
>
> i.e. I'd like to have aov(Phen1~L1) use only Pat1-Pat4,and Pat 10.
> Similarly, aov(Phen1~L2) should use Pat1, 6, and 10.
> Etc.
>
> Is this something I can do in the aov function, or do I need to modify my
> dataset before running aov? In either case, I
Hi,
minben wrote:
> I am a new R-language user. I have set up a data frame "mydata",one of
> the colume of which is "skill". Now I want to select the observations
> whose "skill" value is equal to 1,by what command can I get it?
Try this:
mydata1<-mydatasubset(mydata, skill==1)
Maybe You should
Hi,
> https://stat.ethz.ch/mailman/listinfo/r-help
and there You'll find the section:
"To unsubscribe from R-help, get a password reminder, or change your
subscription options enter your subscription email address:"
Hope this helps,
Kimmo
__
R-help@r
Hi Yannick,
yannick misteli wrote:
> I have a vector x with certain ID numbers in it and want to create a
> subset from my dataset Y with this vector i.e extract only the data with
> the given IDs from vector x.
>
> example:
> x
> [1] 10066924 10207314 10257322 10334594 10348247
>
> and now I wa
Hi,
kayj wrote:
> Hi,
>
> I do have a data set with some missing values that appear as blanks. I want
> to fill these blanks with an NA. How can this be done? Thanks for your help
Something like this?
> my.data<-data.frame("var"=c(1,2,5,"",66,4,3,"",67,5,3,2,1,4,32,56,23),
stringsAsFactors=F)
Hi Ivo,
ivo welch wrote:
> alas, for some odd reason, my R boxplots have some fat black dots, not
> just the hollow outlier plots. Is there a description of when R draws
> hollow vs. fat dots somewhere?
> [and what is the parameter to change just the size of these dots?]
Have you tried the comma
ivo welch kirjoitti:
> dear R experts:
>
> I am playing with boxplots for the first time. most of it is
> intuitive, although there was less info on the web than I had hoped.
>
> alas, for some odd reason, my R boxplots have some fat black dots, not
> just the hollow outlier plots. Is there a d
Hi,
is there a command or parameter for reducing the plotting area with
lattice? What I am looking for is an option similar to 'mai' or 'mar'
from the graphs package.
Background: I have plotted several charts with horizontal stacked bars
and now I would like to add info about percentages of each
Hi,
as mentioned in my previous posting, I run R on a linux machine. So a
possible function for printing (in linux) could look like this:
copy2lpr<-function(..., PRINTER="lpr") {
LPR<-pipe(PRINTER,"w")
capture.output(..., file=LPR)
close(LPR)
}
This seems to work... An allows the user to c
] 0.2730090246 0.4462614490 0.2382041477 0.9826505063 0.1556554718
0.3746872961
[7] 0.6108254879 0.6617410595 0.6694177436 0.4650380281 0.0414420397
0.2307212995
[13] 0.5338913775 0.9186298891 0.0006410333 0.8046684864 0.6205502201
0.5352788521
[19] 0.4255279053 0.7711444888
On Wed, Aug 27, 2008 at 7:44
Dear R-helpers,
I am desperately looking for a solution for how to print out the console
output to a standard printer. For example, I would like to print out the
summary.lm() output, the output of different ftable-functions etc. I use
R on a linux machine.
The only ways so far have been to c
Hi,
cimfasy_rwl[ is.na(cimfasy_rwl) ] <-0
Or did I understood Your right?
HTH,
Kimmo
Alfredo Alessandrini wrote:
I'm trying to replace NA with 0 value...
I've write a loop, but don't work...
Where's the problem?
cimfasy_rwl
1991 0.92 0.72 0.50 1.29 0.54 1.22
199
Dear John,
thanks for Your quick reply.
John Fox wrote:
Dear Kimmo,
MCA is a rather old name (introduced, I think, in the 1960s by
Songuist and Morgan in the OSIRIS package) for a linear model
consisting entirely of factors and with only additive effects --
i.e., an ANOVA model will no interac
Hi!
Is there any possibilities to do multiple classification analysis (MCA)
in R? (MCA examines the relationships between several categorical
independent variables and a single dependent variable, and determines
the effects of each predictor before and after adjustment for its
inter-correlat
Hi,
Monica Pisica wrote:
> - There is no perfect “beginner” book.
How about
- Crawley, Michael (2007). The R book, Wiley & Sons.
- Maindonald, John & John Braun (2007): Data Analysis and Graphics Using
R (2nd edition), Cambridge University Press.
As a political scientist (with programming exp
Prof Brian Ripley kirjoitti viestissään (06.05.2008):
> Does starting R --vanilla help?
>
> I am wondering if you have another corrupt copy of lattice somewhere.
>
The latter was the problem, many thanks for this! I use Rkward as GUI
and obviously some packages have been installed into the user d
Hi,
thanks for the quick reply :)
Prof Brian Ripley kirjoitti viestissään (06.05.2008):
> Try installing again by
>
> install.packages("lattice", .Library)
>
> (from an account with suitable privileges).
Tried (as root) - not working :(
> If that still fails, we need to see the output produced
Hi,
My problem is simple: since having updated the lattice package, I cannot
load lattice anymore. If I type in the command 'library(lattice)' the
loading fails with the following message:
--- cut here ---
Error in library.dynam(lib, package, package.lib) :
shared library 'lattice' not found
Hi again,
Deepayan Sarkar wrote (21.4.2008):
> Write your own panel function (which may or may not be a simple
> exercise depending on your level of expertise in R). You could use
> panel.barchart as a starting point. Basically, you need to insert
> some calls to panel.text() (or something equival
Dear all,
I use the barchart-function (lattice) for plotting stacked barcharts.
The data is a summary table (data frame) of likert-scale-evaluations
(strongly agree, agree...strongly disagree) to different issues
constructed as follows (L1=precentage of "strongly agree" evaluations,
L4=precent
Henrique Dallazuanna wrote (1.4.2008):
> You can try this:
>
> x <- data.frame()
> for(i in LETTERS[1:5]) x[1:10, i] <- rnorm(10)
> x
Or this:
--- cut here ---
df<-data.frame(0) [obsolet, if df already exists]
for (i in 1:10) { df<-data.frame(cbind(df,0)); names(df)
[ncol(df)]<-as.character(i) }
Hi,
the problem were a couple of overlength labels, indeed. After having
removed them, I was able to import the data without any problems.
Thanks anyway for Your help.
Happy Easter,
Kimmo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mail
Hi,
Albrecht Kauffmann wrote (18.03.2008):
> Dear R-helpers,
>
> if I want to read a .dta-file generated by stata 9.0 with read.dta
> (foreign), I get the message
> "not a stata version 5-8 .dta-file". I'm using R-2.6.2 and the latest
> version of the foreign package. Has someone any hint?
Seems
Hi again,
many thanks for Your answers. So, if I understood Wei right:
Zhao, Wei (Cancer Center) wrote (12.3.2008):
> I had a similar problem when read one of my spss.sav with long
> variable label. But when I read another spss.sav with short label the
> same way, I don't have problem.
> Seems li
Dear all,
I have tried to import a SPSS file in R, but always get the following
message:
--- cut here ---
Error in read.spss("spss-data.sav", :
error reading system-file header
In addition: Warning message:
In read.spss("spss-data.sav", :
spss-data.sav: Variable Y6B_A indicates variable la
Hi,
if the columns always follow the same order (indx,var#, var#_lab ...),
then You could use the column numbers and do the following:
1) CN<-colnames(df)[#] (#=colnum of 'var#')
2) df$NEW<-(here the expression to create the new variable)
3) colnames(df)[ncol(df)]<-c(paste(CN,"new",sep="_"))
Th
Hi,
sorry, the correct commands should look like this:
> plot(example.df$StartDate[ (row(example.df)%%5)==0 ], example.df$DSR1[
(row(example.df)%%5)==0 ], type="p", ylim=c(0.3,0.9))
> points(example.df$StartDate[ (row(example.df)%%5)==0 ],
example.df$DSR2[ (row(example.df)%%5)==0 ], type="p", pc
Hi,
this might also work for You:
> points(example.df$StartDate[ (row(example.df)%%5)==0 ],
example.df$DSR2[ (row(example.df)%%5)==0 ], type="p", pch=3)
> points(example.df$StartDate[ (row(example.df)%%5)==0 ],
example.df$DSR2[ (row(example.df)%%5)==0 ], type="p", pch=3)
Kind regads,
Kimmo
Je
Hi,
Hans Ekbrand wrote (19.2.2008):
> I tried the following small code snippet which I copied from the
>
> "Introduction to R":
> > for (i in 2:length(meriter)) { table(meriter[[1]], meriter[[i]]) }
Try:
for (i in 2:length(meriter)) { print(table(meriter[[1]],
meriter[[i]])) }
Kind regards,
Kim
Hi,
joseph wrote (15.2.2008):
> Thanks. I have another question:
> In the following data frame df, I want to replace all values in col1
> that are higher than 3 with NA. df= data.frame(col1=c(1:5, NA),col2=
> c(2,NA,4:7))
My suggestion:
x<-df$col1; x[ x>3 ]<-NA; df$col1<-x; rm(x)
-Kimmo
__
Hi Mohamed,
mohamed nur anisah wrote (8.2.2008):
> Dear lists,
>
> I'm in my process of learning of writing a function. I tried to
> write a simple functions of a matrix and a vector. Here are the
> codes:
>
> mm<-function(m,n){ #matrix function
> w<-matrix(nrow=m, ncol=n)
> for
Hi,
I have a small problem when using barchart. I have the following data:
letters a
6f 18
1a 15
10 j 12
9i 12
4d 9
5e 6
The data is from a survey and summaries the alternatives selected in one
question. The idea is to have a bar chart illustr
Hi,
I am quite new to R (but like it very much!), so please apologize if
this is a too simple question.
I have a large data frame consisting of data from a survey. There is,
for example, information about age and education (a numeric value from
1-9). Now I would like to extract the total amoun
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