Hallo
you can extract POSIX object
tv <- as.POSIXct(index(dt_train))
and use cut together with aggregate
cut(tv, "hour")
aggregate(dt_train, list(cut(tv, "hour")), mean)
2014-10-06 21:00:00 9.807692
2014-10-06 22:00:00 8.67
Cheers.
Petr
čt 3. 10. 2024 v 17:25 odesílatel roslinazairimah
Hallo Francesca
If you had an object with correct setting, something like template
> dput(res)
structure(list(V1 = c("1", "2", "3", "4", "5", "6", "7", "8"),
V2 = c(2, 7, 10, 4, 9, 5, 2, 6)), class = "data.frame", row.names =
c("1",
"2", "3", "4", "5", "6", "7", "8"))
you could merge it with
Hi
what about
now <- Sys.time()
now
[1] "2024-08-22 11:24:23 CEST"
now.utc <- as.POSIXct(now, tz = "UTC")+2*3600
now.utc
[1] "2024-08-22 11:24:23 UTC"
Cheers.
Petr
čt 22. 8. 2024 v 11:00 odesílatel Iago Giné Vázquez
napsal:
> Hi,
>
> How should POSIXct time zone be changed without modifying
Hallo Nadja
Similar as Bert I do not know how the function works. From the help page -
synth.data - is used in example. Check if your data structure is consistent
with synth.data by comparing str(synth.data) and str( INVESTMENTVOLUME).
Names of columns should be in both cases match the names in da
Hallo James
Just a wild guess, are your problems connected with change of default
download method from wininet to libcurl?
Cheers
Petr
út 20. 2. 2024 v 18:24 odesílatel James Powell napsal:
>
[[alternative HTML version deleted]]
__
R-help@r
Hi
> all problems solved. thank you for your help!
> for the sake of completeness, here my solution:
> #1) read in data:
> daten<-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE,
sep="\t")
> daten<-as.data.frame(daten)
not needed, daten is already data frame
indxind[z2,z1]<-topf
> }
> #print(c(s,z1,z2,indxind[1,2])) ##counts s, z1 and z2 properly,
but
> gives always 8 for indxind[1,2]
> }
> #indxind[1:5,1:5] #empty matrix
> }
> #indxind[1:5,1:5] #empty matrix
> }
>
> #4) check:
Hi
>
> > Specifically, since it has only a single detection indicator column
> > (ceneq1), it implies that within any single sample either all the
analytes
> > were detected, or all were not. Not what I would expect.
>
> Don,
>
>I have been thinking about this and wondered whether the cast
I am not allowed to connect to Nabble from work, sorry.
Petr
>
> hi peter the pdf folder is in
> http://r.789695.n4.nabble.com/file/n4639434/aj.pdf
>
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/help-to-
> program-my-function-tp4639434p4639629.html
> Sent from the R
Hi
please cc your post also to Rhelp list, others may give you better/quicker
answer.
>
>
> HI peter
> there is the function that i want to programm (joint in pdf folder).
No pdf allowed.
> my data is > dataexpv Ti1 265.792 26 1579.523 26
2323.704
> 28 68.855 28 426.07
Hi
>
> hello, this is my script:
>
> #1) read in data:
>
daten<-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt',
> header=TRUE, sep="\t")
> daten<-as.matrix(daten)
If there is any column with nonnumeric values it will transfer all numeric
values from daten data.frame
Hi
Maybe it is time for you to read some basic stuff like R intro. It seems
to me that you expect R to behave like some other language you know but
probably your expectation is wrong.
See inline
>
> HI
>
> >i have a problem please help me to solve it:
> http://r.789695.n4.nabble.com/file/n4
> > http://cran.r-project.org/web/packages/plan/plan.pdf
> >
> > -- Bert
> >
> > On Tue, Aug 7, 2012 at 1:28 AM, Petr PIKAL
wrote:
> >> Dear all
> >>
> >> I need to perform some process evaluation. Sorry for not posting data
and
> >&g
Probably a problem in your setting or envirenment or print function
Here is what I get
> x1<-64.90
> x2<-17.7025
> c(x1,x2)
[1] 64.9000 17.7025
> x<-c(x1,x2)
> x
[1] 64.9000 17.7025
>
Regards
Petr
>
> HI
>
> >i have a little problem please help me to solve it
>
> >this is the code in R:
>
>
p://cran.r-project.org/web/packages/plan/plan.pdf
>
> -- Bert
>
> On Tue, Aug 7, 2012 at 1:28 AM, Petr PIKAL
wrote:
> > Dear all
> >
> > I need to perform some process evaluation. Sorry for not posting data
and
> > code - I do not have any, I ask only for
Dear all
I need to perform some process evaluation. Sorry for not posting data and
code - I do not have any, I ask only for pointing me to correct direction.
Suppose I have several connected processes P1, P2, ..., Pn. Each process
takes some time and have some capacity (let say like preparing a
Hi
>
> Dear Jean
>
> Thanks a lot for your help.
>
> The reason I did not provide producible code is that my work started
with
> reading in some large csv files, e.g. the data is not created by myself.
> But the data is from the same data provider so I would expect to
receive
> data in exact
Hi
It is better to use dput for presenting data for others. You probably want
?merge.
Something like
merge(datuak, datuak2, by = "calee_id", all.x=TRUE)
However calee_id seems to be a floating point number and it may be rounded
so you shall beware of it.
Regards
Petr
> Thank you very muc
Hi
>
> Hi,
>
> I appreciate your help with the segmented function. I am relatively new
to
> R. I followed the introduction of the 'segmented'-package by Vito
Muggeo,
> but still it does not work.
> Here are the lines I wrote:
>
> data_test<-data.frame(x=c(1:10),y=c(1,1,1,1,1,2,3,4,5,6))
> lr_t
Hi
Your description is quite long but almost uninformative about what you
really want.
You do not say which values you want to sum but you say it is completely
equal which value you want to add to what and what shall be the final
vector length
Based on this I would just use simple "+"
x<-1:
Hi
OTOH I wonder why cbind gives error as OP told us
x <- data.frame(x = 1:5)
y <- data.frame(y = 6:15)
merge(x,y)
cbind(x,y)
Gives different results but without any error.
Regards
Petr
>
> On Thu, Aug 2, 2012 at 4:52 PM, Ayyappa Chaturvedula
> wrote:
> > Michael,
> > Thank you for this ,
Hi
one of possible options
f<- function(x, parts) split(x,rep(1:length(parts),parts))
f(9:1, c(3,2,4))
$`1`
[1] 9 8 7
$`2`
[1] 6 5
$`3`
[1] 4 3 2 1
You can also check if your parts vector agrees with x vector, if you want.
Regards
Petr
>
> Anyone got a neat way to chop a vector up into sma
Hi
did you find function dist? It seems that it can do directly what you
want.
Regards
Petr
>
> Hello everyone. Like others on this list, I'm new to R, and really not
much
> of a programmer, so please excuse any obtuse questions! I'm trying to
> repeat a function across all possible combina
Hi
Something like
aggregate(DF$data, list(quarters(DF$date), format(DF$date, "%Y")), mean)
Regards
Petr
>
> Hello everybody,
>
> I need to calculate seasonal means with temperature data for my work.
> I have 70 files coming from weather stations, which looks like this for
> example:
>
> sta
Hi
Well, this is help list for R not for Excel, maybe you shall contact
Microsoft guys. I believe that probably easiest would be to make a simple
macro in Excel.
If you want to do merging in R you shall go through help pages for
read.xls, merge, cbind, rbind and R data import/export manual.
R
Hi
I do not have direct answer. You shall probably search ggplot2 web.
Searching "legend" gave me about sixty results from which you probably
could learn how to modify legend(s) according to your wish.
e.g.
http://had.co.nz/ggplot2/docs/opts.html
Regards
Petr
>
> Dear all,
>
> I produced th
Hi
> Hello,
>
> It is not what happens.
>
> Function "convexhull" exists in both "siar" and "spatstat" packages. As
> you already load "spatstat", when you are loading "siar", the
> "convexhull" in "spatstat" is masked by the one in "siar".
>
> Thus, when you will run "convexhull" function,
Hi
> Hi all,
>
> I would like create a new column in a data.frame (a1) to store 0, 1 data
> converted from a factor as below.
>
> a1$h2<-NULL
> for (i in 1:dim(a1)[1]) {
> if (a1$h1[i]=="H") a1$h2[i]<-1 else a1$h2[i]<-0
> }
>
> My question: is it possible to remove the loop from a
Hi
>
> Dear list members,
>
> I am trying to create a subset of a data frame based on conditions in
two
> columns, and after spending much time trying (and search R-help) have
not
> had any luck. Essentially, I have a data frame that is something like
this:
>
> date<-as.POSIXct(as.characte
Hi
I am not at all an expert in regular expressions but
gsub("^[[:punct:]]+[[:digit:]]+:", "",m)
does the output you want. Maybe by chance :-)
Regards
Petr
>
> Hello,
> I would like to substitute a substring of characters defined by a
specific
> start and end sequence.
> i.e. in the example
Hi
In new ggplot2 version following works too
p + geom_boxplot(aes(fill = factor(cyl))) +
labs(fill = "Cylinders") + ylab("Miles per Gallon")+xlab("Number of
Cylinders")
Regards
Petr
>
> Yes you can do all of the things you want.
>
> Below is a start, to give you an idea of how to ap
>
> Hi again!
> I have a question about R.
> I have done gam in previous version of R with "mgcv" package and saved
the
> workspace. This workspace contains different models and I will do
prediction
> by these GAMs.
>
> However, I install new version of R. and use the same workspace. when I
ty
Hi
>
> hi,R-users:
> Now I plot some data with the name(aveobsdata) in column , How can I add
> the
> some number(e,g. the sample number) in each of the column?
> plot
>
(aveobsdata,type='h',lwd=line_width,col=line_col,main=titleinfo,xlab=xxlab,ylab=yylab,xaxt
> = "n")
We do not have aveobsdata
Hi
>
> Dear Sir,
>
> We are able to read dataset in the syn.data (species:yeast) using R.But
we
> want to read a new dataset using R.
> We are not able to do that please tell us the procedure of reading a new
> dataset of a new species.
Did you try it by similar way as you read the first datas
Hi
It seems to me that it can be done by ggplot2 package. However I do not
understand what is three boxplots one on top of another? How could you see
the bottom boxplot when it is twice overplotted?
library(ggplot2)
p <- ggplot(mtcars, aes(factor(cyl), mpg))
p + geom_boxplot(aes(fill = factor(
Hi
I use R for quite a long time and as I remember I did not use such assign
paste i loop yet. Insted of such construct with polluting environment with
plenty of objects named something(i)somethingelse it is always advisable
to use lists.
When you want to shorten variables to some common lengt
Hi
>
> Thanks for your reply Jon.
>
> I need to actually do more than print the name of the variable (I just
made
> the example simpler). I need to manipulate var_1, var_2 etc. but setting
> values of NA to 0.
Why? R has pretty strong system for handling NAs. The only exception AFAIK
is cums
Hi
>
> How could I select the rows of a dataset that have the maximum value in
> one variable and to do this nested in another variable. It is a
dataframe
> in long format with repeated measures per subject.
> I was not successful using aggregate, because one of the columns has
You could do
>
> I do now know how to navigate through the table.
> But my question is, what kind of graphical and numerical summaries can I
> make with keeping in mind that I am interested in the average working
hour
> per employee?
Rather vague question without data or code, so rather vague answer.
For nu
Hi
>
> I have the graph plotted with x axis(-50 to 250) and y axis (-50 to
500).I
> need the x axis values(-50 to 250) with spacing between two tick marks
as
> 1or 0.1.The graph should be wider to get enough resolution.
For the first case you shall have some special display as you will need a
Hi
both na.omit and complete cases works for me smoothly when NA is not a
valid level in factor.
If this is the case, as it seems to be, you need reset your factor levels
so that NA is not a valid level.
ex10s$dg <- factor( ex10s$dg )
both commands shall work than.
Regards
Petr
>
> Removi
hmaltz äż®ĺ®‰çŞ Departamento de PortuguĂŞs / Department
of
> Portuguese Faculdade de CiĂŞncias Sociais e Humanas / Faculty of Social
> Science and Humanities - UM
澳门大ĺ¦ç¤ľäĽšç§‘ĺ¦ĺŹŠäşşć–‡ĺ¦é™˘č‘ˇčŻçł»
> http://www.umac.mo/fsh/ciela/staff/Marcia_Schmaltz.html (+853) 6231-2114
Hi
I can read the example you provided without much problem.
dput(head(test))
structure(list(n = 0:5, X = c(NA, NA, NA, NA, NA, NA), start = c(11185L,
39530L, 40544L, 109684L, 114629L, 118841L), X.1 = c(NA, NA, NA,
NA, NA, NA), dur = c(1L, 2L, 1L, 1L, 0L, 1L), X.2 = c(NA, NA,
NA, NA, NA, NA),
Hi
>
> Dear all
> I have a 'character' vector containing mixed formats (thanks Excel!)
> and I'd like to translate it into a default "%Y-%m-%d" Date vector.
> x <- c("1/3/2005", "13/04/2004", "2/5/2005", "2/5/2005", "7/5/2007",
>"22/04/2004", "21/04/2005", "20080430", "13/05/2003", "20080
Hi
>
> I have a table like the following:
>
> TABLE NO. 1
> ID TIME
> 1325 0
> 1325 0
> . .
> . .
> . .
> TABLE NO. 1
> ID TIME
> 1325 0
> 1325 0
> . .
> . .
> . .
> TABLE NO. 1
> ID TIME
> 1325 0
> 1325 0
> . .
> . .
> . .
> TABLE NO. 1
> ID TIME
>
Hi
> Hi,
>
> Yes, the columns are related: V1 is related to V6, V2 is related to V7
and
> so on. The columns V1,V2,V3,V4,V5 contains the number of employees (in a
> filling team). The columns V6,V7,V8,V9,V10 contains the number of worked
> hours of the filling team.
You shall rather include y
Hi
>
> Hello,
>
> I am trying to fit a model to some "death over time" data that does not
> fit the criteria for the usual LD50 type models (the counts are too
> large). I am using a simple linear model in an attempt to plot a nice
line
> on a scatter plot and calculate some LD values to use
Hi
What is your intention?
basically one output column can be made by
cumsum(bd)
Then you can shuffle bd by let say
bd <- sample(bd, 64)
and repeat cumsum for new bd.
bd<-scan("blap.txt")
output<-matrix(0,64,10)
for (i in 1:10) {
bd<-sample(bd, 64)
cs<-cumsum(bd)
output[,i]<-cs
}
If you in
>
> I'm not sure what you are trying to prove with that example - the
loopless
> versions are massively faster, no?
In some languages loops are integral part of programming habits. In R you
can many things do with whole objects without looping - vectorisation
approach. See R-Inferno from Patri
Hi
>
> On May 22, 2012, at 4:08 AM, HAOLONG HOU wrote:
>
> > Dear list,
> >
> > The name of R-language is too short and is not friendly to search
> > engines.
Did you try it? Even with plain Google
something R will usually lead to quite good hit. Try e.g.
linear model R
Regards
Petr
> > Do
And as followup
> system.time(d<-1000*1001/2)
user system elapsed
0.020.000.02
> identical(a,b,d)
[1] TRUE
>
Regards
Petr
>
> Hi
>
> >
> > For loops are really, really slow in R. In general, you want to avoid
> them
>
> I strongly disagree. ***Proper*** use of loopi
Hi
>
> For loops are really, really slow in R. In general, you want to avoid
them
I strongly disagree. ***Proper*** use of looping is quite convenient and
reasonably fast.
Consider
> system.time( {
+ a=0
+ for (i in 1:1000) {
+ a <-a+i
+ }
+ a
+ })
user system elapsed
10.220.0
Hi
>
> Dear All,
>
> The function I wrote can run well with the small data, but with the
large
> data, the function runs very very slowly. How can I correct it? Thank
you
Your function does not run slowly, it does not run at all.
> l(10,20)
Error in l(10, 20) : object 's' not found
No s obj
Hi
You can do it by hand and remove row/col with max number of NA values.
rem<-which.max(colSums(is.na(M)))
M1<-M[-rem, -rem]
rem<-which.max(colSums(is.na(M1)))
M2<-M1[-rem, -rem]
M2
1 2 3 4 5 7 8 10 11 12
10 143 92 134 42 123 40 107 49 93
2 143 0 77 6 99 46 47 114
Hi
>
> I have a dataset called "raw-data" . I am trying to use the following
code -
>
>
> col_name<-names(raw_data)
> for (i in 1:(length(names(raw_data))-2))
> {
> tbl=table(raw_data$Pay.Late.Dummy, raw_data$col_name[i])
>
> chisqtest<-chisq.test(tbl)
> }
>
>
> Say the 1st column of my
Hi
You did not provide data but I can see some problems in your code. See
inline.
>
> I'm failing to get a for loop working. I'm sure it's something simple,
and I
> have found some posts relating to it, but I'm just not understanding why
> this isn't working.
>
> I have a data frame and woul
Hi
I had not see any answer yet but maybe there is nobody who wants to touch
the elusive object of "outlier". Neither me, but here are some ideas how
one can proceed.
First of all its always up to you what is considered an outlier and how
will you deal with them.
I usually call an outlier an
Hi
>
> I want to use the standard error values in the summary that is produced
using
> coeftest, but I am getting an error code- any ideas?
See what is structure of coeftest object by
str(coeftest(lmodT_WBHO))
and from this you shall deduct how to select second column.
Regards
Petr
>
> >
>
> x[is.na(z)] <- NA
>
> This might send you a nasty bug if x and z are different lengths
> though -- just a head's up.
Another option
x*!is.na(z)*z
Regards
Petr
>
> Michael
>
> On Wed, May 16, 2012 at 12:55 PM, Mintewab Bezabih
> wrote:
> > Dear R users,
> >
> > I was wondering how I ca
Hi
>
> On 2012-05-15 08:36, Melissa Rosenkranz wrote:
> > Here is an R problem I am struggling with:
> > My dataset is organized like this...
> >
> > subject sessionvariable_x variable_y
> > 01 11
> > 01 1
> Something along the lines of
>
> dat2 <- ifelse( dat1==1 , "yes", "no")
Another option is in this case
dat2 <- c("no", "yes")[dat1+1]
Regards
Petr
>
> should do it.
>
> John Kane
> Kingston ON Canada
>
>
> > -Original Message-
> > From: s1010...@student.hsleiden.nl
> > Sent:
Hi
>
> > score<-read.csv("http://users.stat.umn.edu/~chen2285/hw/ACT.csv";)
> score<-read.csv("http://users.stat.umn.edu/~chen2285/hw/ACT.csv";)
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") : unable to connect to 'users.stat.umn.edu'
Hi
>
> how do I plot only the data below 10? everything is white for the 0-10
and
> 10-90 is black ..
What data below 10? I do not see any. You posted some mails before but I
do not keep all mails from R. Only those which helped me somehow.
Basically
x <- sample(1:100, 100, raplace=TRUE)
x[x
Hi
what is wrong with
heatmap(as.matrix(test), col=my.colors(25))
with test from your dput
Regards
Petr
> The heat map generated the correct result:
>
> library(gplots)
> arq <-read.table("l")
> matrix_l <-data.matrix(arq)
> my.colors <-
> colorRampPalette(c
>
("gray0","gray10","gray20","gr
Hi
as already mentioned your data can not be deciphered. Use
dput(table1) for sending usable data.
>From what you describe probably
?aggregate can be used.
But without suitable data you hardly get any advice.
Regards
Petr
>
>
> Hi R user,I am struggling to figure out on how I can calculate
Hi
>
> >arq <-read.table("file")
> > arq_matrix <-data.matrix(arq)
Are you sure that arg_matrix is numeric? Did you check it somehow?
> >dput(arq)
You forgot to include dput(arg) result. Without that only you know what
arg is.
> > arq_heatmap <- heatmap(arq_matrix, Rowv = NA, Colv = NA,col
Hi
>
>
> I might be silly but if I was going to type in dput() then how should I
> send the data over here?
> dput(zdrz20)
outputs tou your console
structure(list(sklon = c(95, 95, 40, 40, 40, 40, 20, 20, 20,
20, 20, 20, 20), ot = c(15, 4, 10, 15, 4, 1.5, 1.5, 4, 10, 15,
4, 10, 15), doba
Hi
I wold vote aggregate
> aggregate(my.df[,-1], list(pathway=my.df$pathway), mean, na.rm=T)
pathway cond.one cond.two cond.three
1pw.A 0.5 0.6NaN
2pw.B 0.4 0.90.1
3pw.C NaN 0.2NaN
>
Regards
Petr
>
> Esteemed UseRs,
>
> Thi
.OO#. .OO#.
rocks...1k
>
---
> Sent from my phone. Please excuse my brevity.
>
> Petr PIKAL wrote:
>
> >Hi
> >
> >Or you can use base function
> >
> >write.table(varxls, file="some output file.xls", sep = "\t", row.na
>
> Hi, Dear all,
>
> Could you please tell me how to select specified column in dataset and
how
> to do
my.data[, x]
my.data[, "aa"]
my.data$aa
> the concentration-time profiles in R?
What is that?
Besides you could help youself a lot reading an intro to R which I believe
is in doc direct
Hi
Or you can use base function
write.table(varxls, file="some output file.xls", sep = "\t", row.names =
F)
Regards
Petr
>
> As it says, you need to supply a file name to be outputted to.
>
> write.xls(, file = "abc.xls")
>
> Michael
>
> On Fri, May 4, 2012 at 11:31 AM, PaulJr wrote:
Hi
One option for substantial distinguishable range of colours is jet.colors
from matlab package.
Regards
Petr
> Hi,
>
> Thanks for the help. Extending the palette to 16 or 20 would be a big
> help. The largest number of files I've had to handle in a single group
is
> 42 and I wouldn't ex
Hi
I would convert it to propper date format and then you can extract
anything.
dat<-strptime("12/31/11 23:45", format="%m/%d/%y %H:%M")
as.Date(dat)
[1] "2011-12-31"
format(dat, "%H:%M")
[1] "23:45"
Regards
Petr
>
> Hello there, I was wondering if you could help me with a quick R issue.
>
Hi
you maybe can use mapply
If you have 2 lists
xl<-list(x, x+5)
xl
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8 9 10
> yl<-list(9,10)
> yl
[[1]]
[1] 9
[[2]]
[1] 10
and this function
fff<- function(xl,yl) (xl-yl)/yl
mapply(fff, xl, yl)
[,1] [,2]
[1,] -0.889 -0.4
[2,] -0.778
> Hi!
>
> I'm trying to do a lm() test on three objects. My problem is that R
protests
> and says that the variable lengths differ for one of the objects
> (Sweden.GDP.gap). But I have double checked that the number of
observations
You shall check it again. BTW How did you checked it?
> are th
Hi
If you used shorter names for your objects you will get probably more
readable advice
Is this what you wanted?
truncated_dataframe[truncated_dataframe$CLAIM_NO %in%
setdiff(truncated_dataframe$CLAIM_NO, truncated_list$CLAIM_NO),]
Regards
Petr
> Hi there,
> I've tried the noted solutions
Hi
Isn't possible just to change values in etc/Rconsole file?
## Colours for console and pager(s)
# (see rw/etc/rgb.txt for the known colours).
background = White
normaltext = NavyBlue
usertext = Red
highlight = DarkRed
You can customise it and keep your setting somewhere. With each new
ins
Hi
>
> Dear R-community,
>
> I am using R (V 2.14.1) on Windows 7. I have a dataset which consists of
19
> variables for 91 individuals or rows. Two of my variables are Age
> (adult/chick, with no NA values) and Sex (0 for females/1 for females,
with
> quite a few NA values). The sex of many
Hi
what about
rbind(a,a)
[,1] [,2] [,3] [,4]
[1,]1234
[2,]5678
[3,]1234
[4,]5678
Regards
Petr
>
> > a <- matrix(1:8, 2, 4, byrow=TRUE)
> > a
> [,1] [,2] [,3] [,4]
> [1,]1234
> [2,]5678
>
Hi
Without data it is only a guess.
>
> I found out something strange when I used the same thing on another data
> file.
>
> In a excel file I have same data too and there I asked in a certain
column
> what values where above the 7.5. Result: 206.
> Now I have done the same thing in R and I g
Hi
> On Fri, Apr 20, 2012 at 4:42 PM, Jeff Newmiller
> wrote:
> > If you read the help, it talks about compiling vectors into matrices,
or
> scalars into vectors. It does not say anything about combining matrices.
> >
> > For the error about 14 elements, you should keep in mind that matrices
Hi
Your question is rather cryptic. Why the output shall be 3? What has
unique count to do with match function?
Maybe you want something what is described in switch help.
See
?switch
Regards
Petr
>
> Hi
>
> My code looks like this
>
> I have two parameters x and par1. X contains values an
Hi
>
> Hi,
>
> I am working on dataframe and column names are multiwords but when i
read
> it it become one word with space relplaced by "." How can i keep
normall
> spacing reading file.
>
> for e.g.
>
> input file
>
> Data Test Data Out
> 35
> 5
Hi
>
> Hi,
>
>
> I m working on bar chart.
>
> *Input file:*
> index -5 1
> index -4 3
> index -3 2
> index -2 10
> index -1 7
> index 0 2
> index 1 1
>
> barplot(t(as.matrix(i[3])), ylab= "value", main = "testdata",
beside=TRUE,
>
col=c("burlywood1"),horiz=TRU
Hi
>
> Hi there
>
> is it possible that pdfs generated using the pdf() function with default
> settings leads to loss of information? I was plotting copy number
changes
> from Agilent 180k data in form of rectangles (rect()) while each
rectangle
> represents one region of copy number change
dify
for instance this
h.lab <- h$counts
h.lab[seq(2,16,2)]<-NA
hist(x, labels=as.character(h.lab))
prints every second label
Regards
Petr
>
> thanks
>
> From: Petr PIKAL
> To: carol white
> Cc: "r-h...@stat.math.ethz.ch"
> Sent: Thursday, April 12, 20
Hi
>
> Hello,
> Is it possible to selectively display labels on a histogram?
What labels?
Like that?
x<-rnorm(1)
hist(x)
hist(x, axes=F, xlab="bla", ylab="ble", main="bleble")
axis(1, at=c(-4, -1, 1, 4))
Regards
Petr
>
> Thanks
>
> Carol
>
>[[alternative HTML version deleted]]
>
Hi
>
> Any help ?
Call 999
Regards
Petr
>
> --
> View this message in context: http://r.789695.n4.nabble.com/number-of-
> warnings-tp4550325p4551760.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailin
Hi
see inline
>
> Hello,
>
> I've got a small dataset on box turtle shell measurements that I would
> like to perform a detrended correspondence analysis on. I thought that
it
> would be interesting to examine the morphometrics for each species in
the
> area of overlap and in areas where n
Hi
>
> You'll need to save them manually to avoid name conflicts --
save.image()
> is the function to do so but you need to give a file name.
Or it is necessary have separate folder for each R session.
Regards
Petr
>
> Michael
>
> On Apr 10, 2012, at 7:41 AM, ya wrote:
>
> > Hi guys,
>
Hi
>
> i have a matrix like
>
> x1 x2 x3
> y1 2 34 5656
> y2 34 434 342
> y3 234 43 34
>
> i want to plot these values like here
> http://www.almob.org/content/2/1/12/figure/F5?highres=y
>
> The rainbow function could calculate a colour for each value.
> But how van i generate t
y is rather undeveloped so if you
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
e.g. **how** did you read your file to R.
Regards
Petr
> >
> > Regards,
> >
> > Lämarao
> >
Hi
>
> Hi R-listers,
>
> 1) I am having trouble understanding why the means I have calculated
from
> Aeventexhumed (A, B, and C) are different from the means showing on the
> boxplot I generated (see attached). I have added the script as to how
my
> data is organized.
Maybe the difference i
Thanks,
anyway, using build-in R features is preferable for colours
with(data, plot(axis1, axis2, col= c("red", "blue",
"green")[as.numeric(data$Region)]))
legend("topright", legend=levels(data$Region), fill= c("red", "blue",
"green"))
although sometimes can be preferable to get advantage of
Hi
Your post is rather screwed.
>
> [R] how to do piecewise linear regression in R?
Maybe segmented?
Regards
Petr
>
>
> Dear all,
> I want to do piecewise CAPM linear regression in R:
> RRiskArbâ’Rf = (1â’δ)[αMktLow+βMktLow(RMktâ’Rf)] + δ[αMkt
High
> +βMkt High(RMkt â’Rf )]
>
Hi
> Hello,
>
> I am a new user of R and I am trying to use the data I am reading from a
> spreadsheet.
> I installed the xlsReadWrite package and I am able to read data from
this
> files, but how can I assign the colums into values?
> E.g:
> as I read a spreadsheet like this one:
Maybe with
Hi
>
> I have a bivariate plot of axis2 against axis1 (data below). I would
like
> to use different size, type and color for points in the plot for the
point
> coming from different region. For some reasons, I cannot get it done.
Below
> is my code.
>
> col <- rep(c("blue", "red", "darkgreen"
Hi
It seems to me that probably split.screen or layout is preferable if you
want specify graph for identification. But I am not an expert in this and
after some testing identification does not work well with splitted screen.
So you are probably out of luck.
Regards
Petr
>
> Please forgive m
(dates))
dates[ which( c(distance<365, F) > c(distance<91,F) &
apply(within_one_year,2,sum) >=3)[1]]
}
You shall get some improvement, however I am still struggling to evaluate
how many consecutive dates are within one year.
>
> Best,
> Felix
>
>
>
Hi
Can you please be more specific? Based on this input, what do you want as
a result?
> set.seed(111)
> dates = as.Date(sort(rnorm(10,3000,100)), origin = "2000-1-1")
> dates
[1] "2007-08-01" "2007-10-21" "2007-12-08" "2007-12-15" "2008-01-29"
"2008-02-14" "2008-02-16" "2008-03-01"
[9] "2008
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