Hi,
There is also the R package "highlight" [1,2] serving the same purpose
but based on information from the R parser, not a static list.
highlight can render in html, tex, and directly for the console (with
the help of the xterm256 package [3]).
Romain
[1] http://cran.r-project.org/web/pa
Hi,
If you can first convert the image to ppm format, the pixmap package
has an addlogo function that can do just what you want,
x <- read.pnm(system.file("pictures/logo.ppm", package="pixmap")[1])
plot(1:10,1:10)
addlogo(x, px=c(2, 4), py=c(6, 8), asp=1)
One could probably get inspiration from
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Peng Yu
> Sent: Friday, December 04, 2009 2:19 PM
> To: r-h...@stat.math.ethz.ch
> Subject: Re: [R] grep() exclude certain patterns?
>
[snip]
> Here is another bad example. Se
I am sure that you mentioned before that your are using 2.7.1, and possibly
even why, but with the number of posts to this list each day and the number of
different posters, I cannot keep track of what version everyone is using (well,
I probably could, but I am unwilling to put in the time/effor
Thanks David, it worked!
On Fri, Dec 4, 2009 at 10:36 PM, David Winsemius wrote:
>
> On Dec 4, 2009, at 10:17 PM, Xin Ge wrote:
>
> @ Francisco: Thanks, it worked.
>>
>>
>> @ All: I'm able to change the colors of legend using following code:
>>
>> par.settings=simpleTheme(col=c(451,26,652)),
>>
On Dec 4, 2009, at 10:17 PM, Xin Ge wrote:
@ Francisco: Thanks, it worked.
@ All: I'm able to change the colors of legend using following code:
par.settings=simpleTheme(col=c(451,26,652)),
key=list(space="right", cex=.96,
text=list(c("A","B","C")),
rectangles=list(size=1.7, border="white"
If your group sizes are not too large, I would use jittered stripcharts.
They're more informative than boxplots and much less subject to
misinterpretation. One warning, I'm not fond of the default pch=0.
-Peter Ehlers
DispersionMap wrote:
What ways are there to plot categorical vs numerical dat
@ Francisco: Thanks, it worked.
@ All: I'm able to change the colors of legend using following code:
par.settings=simpleTheme(col=c(451,26,652)),
key=list(space="right", cex=.96,
text=list(c("A","B","C")),
rectangles=list(size=1.7, border="white", col = c(451,26,652)))
*Q. Using the foll
Try this which assumes that comma does not appear in any of the strings.
You can substitute a different non-appearing character if a comma does
appear in any of the strings:
strsplit(gsub("string,", "string", paste(vec, collapse = ",")), ",")[[1]]
It first runs all the strings together into a sin
On Dec 4, 2009, at 8:42 PM, Jill Hollenbach wrote:
Hi all,
I would like to combine elements of a vector:
vec <- c("astring", "b", "cstring", "d", "e")
vec
[1] "astring" "b" "cstring" "d" "e"
such that for every element that contains "string" at the end, it is
combined with th
On Fri, 4 Dec 2009, David Winsemius wrote:
On Dec 4, 2009, at 5:49 PM, Hien Nguyen wrote:
Dear Dr. Winsemius,
Thank you very much for your reply.
I have tried many possible combinations (even with the model of only 2
predictors) but it produces the same message. With more than 4000
observ
On Fri, Dec 4, 2009 at 7:51 PM, Charles C. Berry wrote:
> On Fri, 4 Dec 2009, Peng Yu wrote:
>
>> On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
>>>
>>> The invert argument seems a likely candidate, you could also do perl=TRUE
>>> and use negations within the pattern (but that is probably overk
Hi all,
I would like to combine elements of a vector:
vec <- c("astring", "b", "cstring", "d", "e")
> vec
[1] "astring" "b" "cstring" "d" "e"
such that for every element that contains "string" at the end, it is
combined with the next element, so that I get this:
> res
[1] "ast
On Fri, 4 Dec 2009, Peng Yu wrote:
On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
The invert argument seems a likely candidate, you could also do perl=TRUE and
use negations within the pattern (but that is probably overkill for your
original question).
I don't see 'invert' in the R versi
On Fri, 4 Dec 2009, Peng Yu wrote:
On Fri, Dec 4, 2009 at 12:18 PM, Charles C. Berry wrote:
On Fri, 4 Dec 2009, Peng Yu wrote:
On Tue, Dec 1, 2009 at 4:04 PM, Charles C. Berry
wrote:
On Tue, 1 Dec 2009, Peng Yu wrote:
Could somebody recommend some textbook how to compute contrast when
t
Hello Xin,
Take a look at the examples under ?print.trellis
Using your original example, you could use:
require(lattice)
p1=barchart(yield ~ variety | site, data = barley,
groups = year, layout = c(1,6),
ylab = "Barley Yield (bushels/acre)",
scales = list(x = list(abb
I try it again and it works.
Thank you.
TTsai wrote:
>
> Hello,
>
> I have problem running WinBUGS from R.
> The following example works in WinBUGS but it does not work in R through
> package R2WinBUGS.
> Does anyone know what the problem is?
>
> x <- c(0.2, 1.1, 1, 2.2, 2.5, 2.9, 2.9, 3.6,
Dear R users,
i would like to have expression on my plot written in bold
italic font and use following:
text(0.01,70,expression(bolditalic(r^2==0.67)),pos= 4)
However, only the letter r appears bold and italic, but not
the whole expression. How should i change it?
Thank you for your help,
Alla.
_
Hello,
I've written a brush for R for the SyntaxHighlighter JavaScript
library. It allows you to display R code on a web page with the proper
syntax highlighting. It's available here:
http://demitri.com/code
Comments and suggestions for improvement are welcome!
Cheers,
Demitri
__
Maybe, atleast for the most used functions, there should be a section
in the .Rd file
with name "for newbies"?
Kjetil
On Fri, Dec 4, 2009 at 6:18 PM, Peng Yu wrote:
> On Fri, Dec 4, 2009 at 3:06 PM, Peng Yu wrote:
>> On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
>>> The invert argument seem
That should be:
transform(DF, Time = ave(1:nrow(DF), Subject, FUN = function(ix)
with(DF[ix,], if (any(Marker == 1)) Time - Time[Marker == 1] else Time)))
On Fri, Dec 4, 2009 at 6:36 PM, Gabor Grothendieck
wrote:
> Try this:
>
> transform(DF, Time = ave(1:nrow(DF), Subject, FUN = function(ix
On Fri, 2009-12-04 at 15:18 -0600, Peng Yu wrote:
> On Fri, Dec 4, 2009 at 3:06 PM, Peng Yu wrote:
> > On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
> >> The invert argument seems a likely candidate, you could also do perl=TRUE
> >> and use negations within the pattern (but that is probably o
On 04/12/2009 3:46 PM, Bert Gunter wrote:
?help (see argument help_type)
?options
This has been asked before. My understanding is that there was a licensing
issue with Microsoft's compiled html help; so what you need to do is specify
There was nothing new there: CHM is discouraged (or at
Try this:
transform(DF, Time = ave(1:nrow(DF), Subject, FUN = function(ix)
if (any(Marker[ix] == 1)) Time - Time[Marker == 1] else Time))
On Fri, Dec 4, 2009 at 5:47 PM, Dennis Fisher wrote:
> Colleagues,
>
> R 2.9.0 on all platforms
>
> I have a dataset that contains three columns of int
You could try using merge:
>
d<-data.frame(Subject=rep(11:13,each=3),Time=101:109,Marker=c(0,1,0,
0,0,0, 0,0,1))
> d
Subject Time Marker
1 11 101 0
2 11 102 1
3 11 103 0
4 12 104 0
5 12 105 0
6 12 106 0
Dennis Fisher wrote:
>
> Colleagues,
>
> R 2.9.0 on all platforms
>
> I have a dataset that contains three columns of interest: ID's, serial
> elapsed times, and a marker. Representative data:
> Subject TimeMarker
> 1 100.5 0
> 1
On Dec 4, 2009, at 5:49 PM, Hien Nguyen wrote:
Dear Dr. Winsemius,
Thank you very much for your reply.
I have tried many possible combinations (even with the model of only
2 predictors) but it produces the same message. With more than 4000
observations, I think 14 predictors might not be
Dear Dr. Winsemius,
Thank you very much for your reply.
I have tried many possible combinations (even with the model of only 2
predictors) but it produces the same message. With more than 4000
observations, I think 14 predictors might not be too many.
Although my dependent variable (Pin) is
Colleagues,
R 2.9.0 on all platforms
I have a dataset that contains three columns of interest: ID's, serial
elapsed times, and a marker. Representative data:
Subject TimeMarker
1 100.5 0
1 101 0
1
I installed RF on Linux OpenSuSe 11.1 and while it did install and did run a
model I had created on Windows correctly, it gave me a lot of "uninitialized"
warnings. I don't know if these are significant and so am a little concerned
even though my model ran. Any thoughts?
Thanks
R version 2
Hi All,
I'm trying par(mfrow(c(1,2))) with barchart(), but its not working. Can I
display two or more barcharts on a same page using some other function? I'm
using following code --- where barchart() part is taken from help manual.
library(lattice)
par(mfrow=c(1,2))
barchart(yield ~ variety | sit
Thanks Greg for your suggestion!
On Thu, Dec 3, 2009 at 11:50 AM, Greg Snow wrote:
> A bar graph including both stacked and grouped bars will put lots of pretty
> colors on the page and probably be eyecatching, but is unlikely to be the
> most effective way to convey the actual meaning of the da
Hello,
I am not too sure if anyone tried this out. but I ma fitting a mixture model
via the EM algorithmn.
But then I tried using CAMAN package and I seem to get different results.
Has anyone else had this problem? For example, I am fitting the Inness data
from the text Medical Applications of Fini
Hi all,
I fit a Tobit model to Fluid milk consumption (dependent variable) data
using survreg. 1) The square root of the dependent variable was used to
correct for
heteroscedasticity since it provides the best fit. 2) The output before
transformation of numeric variables in scores (by "Make.Z")
__
R-help@r-project.org mailing list
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.
Jorge,
It really helps. I appreciate your help a lot!
Chen
On Fri, Dec 4, 2009 at 12:04 AM, Jorge Ivan Velez [via R] <
ml-node+948281-1243760...@n4.nabble.com
> wrote:
> Try this:
>
> a <- c(9,3,5)
> b<-c(3,4,1)
> cbind(a, b)
> # a b
> # [1,] 9 3
> # [2,] 3 4
> # [3,] 5 1
>
> cov(cbind(a,
Thanks a lot, Jorge. It works well!
On Fri, Dec 4, 2009 at 12:00 AM, Jorge Ivan Velez [via R] <
ml-node+948277-873392...@n4.nabble.com
> wrote:
> Hi aegea,
>
> Here is one:
>
> m <- structure(c(6, 5, 20, 7, 8, 25, 14, 8, 9, 10, 11, 12, 13, 14,
> 1, 2, 3, 4, 5, 6, 7), .Dim = c(7L, 3L))
>
> m1 <-
On 2009.12.03 23:52:15, Yoseph Zuback wrote:
> Hi Frank,
>
> I'm trying to repair heteroscedastic variables using the hccm. A
> statistician in my department gave an incomplete solution that included:
>
>
> OLS1$coefficients/(sqrt(hccm(OLS1)))
>
> Trying to solve my problem I get different resu
What does a status value of -3 mean when I do a regression with RF and use the
getTree function?
left daughter right daughter split var split point status prediction
12 311 4.721000e+03 -3 15.8489576
24 5 5 6.500
On Fri, Dec 4, 2009 at 12:18 PM, Charles C. Berry wrote:
> On Fri, 4 Dec 2009, Peng Yu wrote:
>
>> On Tue, Dec 1, 2009 at 4:04 PM, Charles C. Berry
>> wrote:
>>>
>>> On Tue, 1 Dec 2009, Peng Yu wrote:
>>>
Could somebody recommend some textbook how to compute contrast when
there are inte
Hello everyone,
I'm having a problem performing reshape() on a large data frame. The
operation is fairly trivial but it makes R run out of memory.
The data frame has the following structure:
ID DATE1 DATE2VALTYPE
VALUE
'abcd1233' 2009-11-12
At long last pgfSweave has finally made its way to CRAN.
The pgfSweave package is about speed and style of graphics. For speed,
the package provides capabilities for “caching” graphics generated
with Sweave on top of the caching funcitonality of cacheSweave. For
style the pgfSweave package facilit
Hi,
when I create huge pdf files (width is 6meters) with R I cannot open
them in Adobe Acrobat reader (I tried version 9 and some lower). I use
pdf() of grDevices version 2.8.1 and CairoPDF from Cairo version
1.4-4. When I add the (perhaps since pdf version 1.7 pagesizes of more
than 200in are poss
On Fri, Dec 4, 2009 at 3:06 PM, Peng Yu wrote:
> On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
>> The invert argument seems a likely candidate, you could also do perl=TRUE
>> and use negations within the pattern (but that is probably overkill for your
>> original question).
>
> I don't see '
On Dec 4, 2009, at 4:03 PM, Jorge Ivan Velez wrote:
Hi Austin,
What version of R are you using? It works for me for R 2.10.0
Patched on Win
XP Pro:
R> a <- matrix(c(1, 2, 3, 4), nrow = 2)
R> a
# [1] 1 3
# [2] 2 4
R> rmask <- c(TRUE, FALSE)
R> a[rmask,]
# [1] 1 3
Doesn't work on a Mac (ad
On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
> The invert argument seems a likely candidate, you could also do perl=TRUE and
> use negations within the pattern (but that is probably overkill for your
> original question).
I don't see 'invert' in the R version (2.7.1) that I use. Here is the
Hi Austin,
What version of R are you using? It works for me for R 2.10.0 Patched on Win
XP Pro:
R> a <- matrix(c(1, 2, 3, 4), nrow = 2)
R> a
# [1] 1 3
# [2] 2 4
R> rmask <- c(TRUE, FALSE)
R> a[rmask,]
# [1] 1 3
HTH,
Jorge
On Fri, Dec 4, 2009 at 3:52 PM, Austin Huang <> wrote:
> One problem I'
Check out `drop'
?drop
a <- matrix(c(1, 2, 3, 4), nrow = 2)
rmask <- c(TRUE, FALSE)
b <- a[rmask, , drop=FALSE]
colSums(b)
This will maintain the matrix structure.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Pro
b <- a[rmask,,drop=FALSE]
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
> One problem I've been having is the special case in which only one
> row/column remains and the variable gets converted into a vector when
> entries are removed by logical masking. This is a problem because
> subsequent
> code may rely on matrix operations (apply, colsums, dim, etc) For example:
On Dec 4, 2009, at 3:52 PM, Austin Huang wrote:
One problem I've been having is the special case in which only one
row/column remains and the variable gets converted into a vector when
entries are removed by logical masking. This is a problem because
subsequent
code may rely on matrix operati
One problem I've been having is the special case in which only one
row/column remains and the variable gets converted into a vector when
entries are removed by logical masking. This is a problem because subsequent
code may rely on matrix operations (apply, colsums, dim, etc) For example:
> a <- ma
?help (see argument help_type)
?options
This has been asked before. My understanding is that there was a licensing
issue with Microsoft's compiled html help; so what you need to do is specify
options(help_type = "html")
in your startup process. There are a variety of ways to do so, explained
Phil Spector's solution is the best way to get your
triplicating job done.
As for why the result of your apply call is a 6 by
4 matrix, read the help file for apply where it talks
about how it 'simplifies' the result. For FUN's that
do not return a scalar the simplification algorithm
may be surpr
The invert argument seems a likely candidate, you could also do perl=TRUE and
use negations within the pattern (but that is probably overkill for your
original question).
Could you explain to us the process that you use to search for answers to your
questions before posting? You have been ask
I *think* this is from from 'StatsRUs' - how about
as.data.frame(lapply(df,function(x)rep(x,n)))
hth, david freedman
pengyu.ut wrote:
>
> I want to duplicate each line in 'df' 3 times. But I'm confused why
> 'z' is a 6 by 4 matrix. Could somebody let me know what the correct
> way is to dupli
df[rep(1:nrow(df),each=3),]
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
I want to duplicate each line in 'df' 3 times. But I'm confused why
'z' is a 6 by 4 matrix. Could somebody let me know what the correct
way is to duplicate each row of a data.frame?
df=expand.grid(x1=c('a','b'),x2=c('u','v'))
n=3
z=apply(df,1
,function(x){
result=do.call(rbind,rep(list(x
use !grepl
On Fri, Dec 4, 2009 at 2:43 PM, Peng Yu wrote:
> On Fri, Dec 4, 2009 at 11:54 AM, Duncan Murdoch
> wrote:
> > On 04/12/2009 12:52 PM, Peng Yu wrote:
> >>
> >> The external grep program has an option -v to select non-matching
> >> lines. I'm wondering if how to exclude certain pattern
What ways are there to plot categorical vs numerical data in R.
I have two columns: one with categorical data in 5 categories a,b,c,d,e, and
a numerical column with integers between 1 and 100.
I have used a boxplot with a,b,c,d,e on the x-axis and an increasing
numerical scale on the y-axis. T
Gerrit Draisma wrote:
>
> Hallo,
> I have a dataset with one or two columns with character data
> and the rest with numeric data.
> Using latex.table from the quantreg package produced a table,
> but I cannot set the decimals.
> For instance:
> ---
> > x<-data.frame(Name=c("Jan","Piet","Jan"),
I just upgraded from 2.8.1 to 2.10 on Windows Vista. BIG MISTAKE
apparently because now when I type:
> help(functionname)
or
?functionname
I get only a small text window giving some very basic info on the topic, e.g.:
base-package package:baseR Documentation
The
On Fri, Dec 4, 2009 at 11:54 AM, Duncan Murdoch wrote:
> On 04/12/2009 12:52 PM, Peng Yu wrote:
>>
>> The external grep program has an option -v to select non-matching
>> lines. I'm wondering if how to exclude certain patterns in grep() in
>> R?
>>
>
> ?grep
I don't see which argument to use.
__
The boxplot function has an 'at' argument that you can use to specify where to
plot the boxes, could you just use this to group the boxplots?
Or the lattice package can put each group into its own panel to show the
grouping.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Ce
This is historical and compatibility. The story as I heard it is that on the
computer where the original S language was developed had a key that produced a
left pointing arrow and that was used as the assignment operator (later <- was
added as an alternative for other computer systems). When k
Exactly, that's waht I want. Thank you very much!
Lisa
Phil Spector wrote:
>
> Lisa -
> I think this is what you're looking for:
>
> myfunction = function(...)do.call(cbind,list(...))
>
>
> - Phil Spector
>St
On Dec 4, 2009, at 2:04 PM, Bradley W. Settlemyer wrote:
Hello
I need to fit a distribution to a histogram data set. I have read
Ricci's guide to distribution fitting, and am ready to begin
experimenting with the techniques it mentions, but I am uncertain
how to get my data in the form
Michael, thanks a lot, it works!
Now I have to study the dataset to mining some interesting rules.
Just one more question, I saw ways to find rules that contains some itens.
But, Is there a method to find rules that doesn't have a item?
Thanks again! You were very helpfull!!!
2009/12/4 Michael
Hello
I need to fit a distribution to a histogram data set. I have read
Ricci's guide to distribution fitting, and am ready to begin
experimenting with the techniques it mentions, but I am uncertain how to
get my data in the format he uses.
My problem is that my data is binned. So for
Thanks Baptiste,
I think you nailed it.
baptiste auguie wrote:
Hi,
I think the size mismatch occurs because of a different default for
the fontsize (and grid.points has a size of 1 character by default).
Compare the following two examples,
# default
grid.newpage()
pushViewport(viewport(x=unit
Lisa -
I think this is what you're looking for:
myfunction = function(...)do.call(cbind,list(...))
- Phil Spector
Statistical Computing Facility
Department of Statistics
Dear all
Is it normal that R ignores options("width"=100) at start-up? Although
li...@debian-liv:~$ cat /usr/lib/R/etc/Rprofile.site | grep width
options(width = 100)
, R will start with
[Previously saved workspace restored]
> options()$width
[1] 80
Am I doing something wrong?
Liviu
> sessionI
Thank you for your reply. But this is not what I want.
For example, I have several variables, like
arg1 <- c(1, 2, 3, 5, 6, 6)
arg2 <- c(3, 1, 5, 5, 7, 8)
arg3 <- c(8, 10, 4, 0, 9, 1)
arg4 <- c(11, 22, 30, 5, 61, 22)
…
I just want to bind some of these variables based on the arguments assigned
On Dec 4, 2009, at 1:27 PM, Gray Calhoun wrote:
Hi Muhammad,
Load the data from all of the files into an array (probably using a
for loop), then call apply on the resulting array: ie
I like that idea a lot and it does not need an explicit loop:
apply( array(c(File1, File2, File3), dim=c(3,3
Hi Muhammad,
Load the data from all of the files into an array (probably using a
for loop), then call apply on the resulting array: ie
## replace the first line with code to load your data
> rr <- array(c(rep(1:3,3), rep(4:6,3), rep(7:9,3)), c(3,3,3))
> apply(rr, c(1,2), sd)
[,1] [,2] [,3]
Pascale,
If you do want an nls fit with the associated error structure
assumptions, check ?SSlogis.
fm <- nls(y ~ SSlogis(x, Asy, xmid, scal))
summary(fm)
xx <- seq(123, 248, length = 101)
yy <- predict(fm, list(x = xx))
plot(x, y)
lines(xx, yy)
-Peter Ehlers
Gabor Grothendieck wrote:
Hi Everyone,
I am new to R and just trying my hand at plotting a network from my dataset. I
am trying to teach myself network analysis, so this is probably an easy
question.
Using the package network, I have successfully plotted my network. I simply
want to add labels to my vertices, with th
Muhammad -
Here's one way:
files = c('File1','File2','File3')
themats = lapply(files,read.table)
ans = matrix(0,3,3)
for(i in 1:3)for(j in 1:3)ans[i,j] = sd(sapply(themats,function(x)x[i,j]))
Here's another
files = c('File1','File2','File3')
themats = lapply(files,read.table)
ans =
outer
On Fri, 4 Dec 2009, Peng Yu wrote:
On Tue, Dec 1, 2009 at 4:04 PM, Charles C. Berry wrote:
On Tue, 1 Dec 2009, Peng Yu wrote:
Could somebody recommend some textbook how to compute contrast when
there are interactions terms? "Applied Linear Regression Models"
(book) mentioned contrast, but I
An alternative to str is the TkListView function in the TeachingDemos package.
You still get the long listing, but it is in a separate window that you can
control the scrolling on by hand. For more complicated lists/objects it
provides a tree structure so that you can look at only the detailed
Hi,
try ?do.call
do.call(cbind, replicate(3, 1:10, simplify=FALSE))
HTH,
baptiste
2009/12/4 Lisa :
>
> Hello, All,
>
> I want to write a function to do some works based on the arguments. For
> example, bind some variables (arguments) as this:
>
> myfunction <- function(arg1, arg2, arg3, …)
> {
This reminds me of a quote I saw once (I think it may have been in one of those
Murphy's Laws calendars), my parahpase:
If you make someone think that they are thinking,
They will love you for it.
If you make them actually think,
They will hate you for it.
This explains why peopl
Hello, All,
I want to write a function to do some works based on the arguments. For
example, bind some variables (arguments) as this:
myfunction <- function(arg1, arg2, arg3, …)
{
x <- cbind(arg1, arg2, arg3, …)
}
myfunction(arg1, arg2, arg3, …)
The function can automatically determine t
Hello R-users,
I would like to know how to find the standard deviation for each element
in a set of matrices.
Given the following files,
File1File2File3
1 1 1 4 4 4 7 7 7
2 2 2 5 5 5 8 8 8
3 3 3 6 6 6 9 9 9
I want to calculate the standard deviation for every
On 04/12/2009 12:52 PM, Peng Yu wrote:
The external grep program has an option -v to select non-matching
lines. I'm wondering if how to exclude certain patterns in grep() in
R?
?grep
Duncan Murdoch
__
R-help@r-project.org mailing list
https://sta
The external grep program has an option -v to select non-matching
lines. I'm wondering if how to exclude certain patterns in grep() in
R?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide h
On Dec 4, 2009, at 11:55 AM, Allen L wrote:
Dear R forum,
I want to replace all the elements in a data frame (dd) which match
the
character "x" with "0".
What's the most elegant way of doing this (there must be an easy way
which
I've missed)? I settled on the following loop:
for(i in 5:
Here a way of doing it:
for (i in 5:12){
# convert to character so you can substitute 'x'
a <- as.character(dd[,i])
a[a == 'x'] <- '0' replace with zero
dd[,i] <- as.numeric(a)
}
On Fri, Dec 4, 2009 at 11:55 AM, Allen L wrote:
>
> Dear R forum,
> I want to replace all the eleme
On Tue, Dec 1, 2009 at 4:04 PM, Charles C. Berry wrote:
> On Tue, 1 Dec 2009, Peng Yu wrote:
>
>> Could somebody recommend some textbook how to compute contrast when
>> there are interactions terms? "Applied Linear Regression Models"
>> (book) mentioned contrast, but I cannot extend it to the case
Dear R forum,
I want to replace all the elements in a data frame (dd) which match the
character "x" with "0".
What's the most elegant way of doing this (there must be an easy way which
I've missed)? I settled on the following loop:
>for(i in 5:12){# These are the column of dd I am
Hi,
I think the size mismatch occurs because of a different default for
the fontsize (and grid.points has a size of 1 character by default).
Compare the following two examples,
# default
grid.newpage()
pushViewport(viewport(x=unit(0.5, "npc"), y=unit(0.5, "npc")))
lplot.xy(data.frame(x=0.55,y=0.5
Stephanie Cooke wrote:
Is there a place to find the code for R functions like lsoda? Thanks
Yes, all code is in the source version of the package that contains that
function.
Best,
Uwe Ligges
__
R-help@r-project.org mailing list
https://stat
On Dec 4, 2009, at 7:07 AM, StRose, Suzanne wrote:
Hi again,
Apologies for the last email I forwarded. Apparently, my data was
not legible. So, I have a question regarding the construction of 3D
graphs in ‘R’, BUT these graphs also need to illustrate movement
(with time) of the prostate
A question that has come up a few times on r-help is how to rename columns
of a data.frame. There are several ways to do this by hand (see the list
archives). There is also a 'rename' function in the reshape package.
I often use the 'transform' function shortly after reading in a data file
and w
In developing a machine learner to classify sentences in plain text
sources of scientific documents I have been using the caret
package and
following the procedures described in the vignettes. What I miss
in the
package -- but quite possibly I am overlooking it! -- is functions
t
Sorry, but i was out for two days!
I just want to call R scripts from java code. So i think i need to review
the jri and the REngine API.
I will tell you whatever i find.
Thank you so much, because i was really lost!
2009/12/4 Romain Francois
> On 12/03/2009 08:43 PM, Gene Leynes wrote:
>
>>
>>
On Dec 4, 2009, at 6:55 AM, Pazur, Robert wrote:
Dear all,
is it possible to perform logstic type of geographical weighted
regression in R software?
We even have a fortune for such questions:
> library(fortunes)
> fortune("Yoda")
Evelyn Hall: I would like to know how (if) I can extract some
Dear R-users,
For the past few days, I have been trying to find the reason why some of
my plots were showing symbols of different sizes, while I thought I was
using the same .cex arguments everywhere. The problem is exemplified by
the following example code where the xyplot and grid.points fun
Lauren,
I think (?) you mean to say that you wish to create a factor and control the
range of values assigned to each level.
The "breaks" argument can specify either the number of intervals you desire
OR the values you wish to use to define levels of your factor. See '?cut'
and the following ex
On Dec 4, 2009, at 9:02 AM, khaz...@ceremade.dauphine.fr wrote:
Hi
I want to generate from invers gamma distribution truncated but when
I use
the rinvegamma() function, the system can not find this function.
Is ther someone to help me please?
Check your spledding.
thanks
soleiman
__
1 - 100 of 171 matches
Mail list logo
