Re: [R] sapply, lattice functions
You're right. It's necessary for xyplot though to prevent grouping. On Mar 20, 2010 10:43 AM, Dieter Menne dieter.me...@menne-biomed.de wrote: Sundar Dorai-Raj-2 wrote: Or perhaps more clearly, histogram(~a1 + b1 + c1, data = aa, o... Why outer=TRUE? Looks same for me without: Dieter library(lattice) aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5)) histogram(~a1 + b1 + c1, data = aa) -- View this message in context: http://n4.nabble.com/sapply-lattice-functions-tp1618134p1676043.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz [[alternative HTML version deleted]] __ 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.
Re: [R] sapply, lattice functions
Or perhaps more clearly, histogram(~a1 + b1 + c1, data = aa, outer = TRUE) --sundar On Fri, Mar 19, 2010 at 3:50 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try this: histogram(~ values | ind, stack(aa)) On Fri, Mar 19, 2010 at 5:44 PM, Santosh santosh2...@gmail.com wrote: Dear R-gurus aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5)) sapply(aa,function(x) histogram(x,breaks=NULL)) or px - sapply(aa,function(x) histogram(x,breaks=NULL)) print(px,split=c(1,1,1,1),more=F) The above code does not seem to work. am I missing something? Thanks, Santosh [[alternative HTML version deleted]] __ 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. __ 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. [[alternative HTML version deleted]] __ 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.
Re: [R] Source code for the t-distribution
Here it is. https://svn.r-project.org/R/trunk/src/nmath/pt.c --sundar On Tue, Mar 9, 2010 at 4:24 AM, Ravi Kulkarni ravi.k...@gmail.com wrote: I have tried looking for the source code for the pt() function in https://svn.r-project.org/R/trunk/src/library/stats/ and am unable to find it there. Can someone please tell me where to find it? Thanks, Ravi Kulkarni -- View this message in context: http://n4.nabble.com/Source-code-for-the-t-distribution-tp1585875p1585875.html Sent from the R help mailing list archive at Nabble.com. __ 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. [[alternative HTML version deleted]] __ 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.
Re: [R] capturing errors in Sweave
Thanks, Berwin. That works just great!
--sundar
On Tue, Mar 2, 2010 at 12:57 AM, Berwin A Turlach
ber...@maths.uwa.edu.auwrote:
G'day Sundar,
On Mon, 1 Mar 2010 23:46:55 -0800
Sundar Dorai-Raj sdorai...@gmail.com wrote:
Thanks for the input, but I don't want try in the Sweave output. I
want the output to look just like it does in the console, as if an
uncaptured error really did occur.
I don't think that you will get around using try; and you will have
to work moderately hard to make the output appear as it does on the
console. Probably somewhere along the lines:
Sweave code start ++
Function-4a=
MySqrt - function(x) {
if (missing(x)) {
stop('x' is missing with no default)
}
if (!is.numeric(x)) {
stop('x' should only be numeric)
}
if (x 0) {
stop('x' should be non-negative)
}
return(sqrt(x))
}
@
echo=FALSE=
tmp - try(MySqrt())
@
eval=FALSE=
MySqrt()
@
echo=FALSE=
cat(tmp[1])
@
echo=FALSE=
tmp - try(MySqrt(a))
@
eval=FALSE=
MySqrt(a)
@
echo=FALSE=
cat(tmp[1])
@
echo=FALSE=
tmp - try(MySqrt(-2))
@
eval=FALSE=
MySqrt(-2)
@
echo=FALSE=
cat(tmp[1])
@
=
MySqrt(4)
@
+++ Sweave code end ++
Now what I would like to know is how to include easily warning messages
in my Sweave output without having to try whether Jean Lobry's [1] hack
still works. :)
HTH.
Cheers,
Berwin
[1]
https://www.stat.math.ethz.ch/pipermail/r-help/2006-December/121975.html
== Full address
Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr)
School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
The University of Western Australia FAX : +61 (8) 6488 1028
35 Stirling Highway
Crawley WA 6009e-mail: ber...@maths.uwa.edu.au
Australiahttp://www.maths.uwa.edu.au/~berwin
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Re: [R] capturing errors in Sweave
What I ended up using was: cat(unclass(tmp)) --sundar On Tue, Mar 2, 2010 at 8:58 AM, Berwin A Turlach ber...@maths.uwa.edu.auwrote: G'day Sundar, On Tue, 2 Mar 2010 01:03:54 -0800 Sundar Dorai-Raj sdorai...@gmail.com wrote: Thanks, Berwin. That works just great! You are welcome. I noticed by now that cat(tmp) is sufficient; the tmp[1] in cat(tmp[1]) was a left over from earlier attempts to get the output to look correct. Cheers, Berwin [[alternative HTML version deleted]] __ 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.
[R] capturing errors in Sweave
Hi,
I'm writing a manual using Sweave and I want to be able to print errors from
bad code. Here's an example:
Function-4a=
MySqrt - function(x) {
if (missing(x)) {
stop('x' is missing with no default)
}
if (!is.numeric(x)) {
stop('x' should only be numeric)
}
if (x 0) {
stop('x' should be non-negative)
}
return(sqrt(x))
}
MySqrt()
MySqrt(a)
MySqrt(-2)
MySqrt(4)
@
And I would like the output to be:
MySqrt - function(x) {
...
}
MySqrt()
Error in MySqrt() : 'x' is missing with no default
MySqrt(a)
Error in MySqrt(a) : 'x' should only be numeric
MySqrt(-2)
Error in MySqrt(-2) : 'x' should be non-negative
MySqrt(2)
[1] 1.414214
I.e. I want the Error statements to print in the tex file and not just make
Sweave to bomb.
Thanks,
--sundar
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Re: [R] capturing errors in Sweave
Thanks for the input, but I don't want try in the Sweave output. I want
the output to look just like it does in the console, as if an uncaptured
error really did occur.
--sundar
On Mon, Mar 1, 2010 at 11:42 PM, Sharpie ch...@sharpsteen.net wrote:
Sundar Dorai-Raj-2 wrote:
Hi,
I'm writing a manual using Sweave and I want to be able to print errors
from
bad code. Here's an example:
Function-4a=
MySqrt - function(x) {
if (missing(x)) {
stop('x' is missing with no default)
}
if (!is.numeric(x)) {
stop('x' should only be numeric)
}
if (x 0) {
stop('x' should be non-negative)
}
return(sqrt(x))
}
MySqrt()
MySqrt(a)
MySqrt(-2)
MySqrt(4)
@
And I would like the output to be:
MySqrt - function(x) {
...
}
MySqrt()
Error in MySqrt() : 'x' is missing with no default
MySqrt(a)
Error in MySqrt(a) : 'x' should only be numeric
MySqrt(-2)
Error in MySqrt(-2) : 'x' should be non-negative
MySqrt(2)
[1] 1.414214
I.e. I want the Error statements to print in the tex file and not just
make
Sweave to bomb.
Thanks,
--sundar
You can catch errors in R using the try() function:
foo - try( log(A), silent = TRUE )
if( class( foo ) == 'try-error' ){
print( unclass( foo ) )
}
[1] Error in log(\A\) : Non-numeric argument to mathematical function\n
There might be a more clever way to override the default error handling
that
would save you from wrapping everything in try()-- but this method should
work.
Hope it helps!
-Charlie
-
Charlie Sharpsteen
Undergraduate-- Environmental Resources Engineering
Humboldt State University
--
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http://n4.nabble.com/capturing-errors-in-Sweave-tp1574642p1574686.html
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Re: [R] How do I juxtapose two lattice graphs with common X axes such that the X axes line up?
Try googling latticeExtra x.same for some examples. Here's one:
http://www.mail-archive.com/r-help@r-project.org/msg39048.html
On Wed, Jan 20, 2010 at 9:44 AM, George Chen glc...@stanford.edu wrote:
Hello,
I would like to juxtapose two lattice graphs with common X axes such that
the X axes line up. I am using plot right now but the edges are not neat
and it would be nice if I could just draw 1 X axis and not both of them.
Here is my code:
upper-bwplot(SignalUsed~as.factor(AllNormalHitsNamesCount),data=NmlOverviewArray2,
xlab=,
ylab=Intensity of Individual Antibody Responses,
main=Intensity, Frequency, Distribution, Quantity of Normal
Antibody Responses,
box.ratio=1,
panel = function (AllNormalHitsNamesCount,...) {
panel.bwplot(...)
}
)
lower-barchart(as.vector(table(NmlOverviewArray2$AllNormalHitsNamesCount))
~as.factor(as.numeric(names(table(NmlOverviewArray2$AllNormalHitsNamesCount,
data=NmlOverviewArray2,
ylab=Number of Individual Antibody Responses,
xlab=Occurrence of Individual Antibody Responses
(Out of 45 Normals),
box.ratio=1)
plot (upper, newpage=TRUE, more=TRUE, position = c(0,.15,1,1))
plot (lower, newpage=FALSE, more=TRUE, position = c(0,0,1,.3))
George
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Re: [R] Dynamic arguments in rbind function
Use a list instead of assign then do.call(rbind, thelist).
import.files - c(a.txt, b.txt, c.txt, d.txt, e.txt)
imp - vector(list, length(import.files))
for (i in 1:length(import.files)) {
imp[[i]] - read.delim(import.files[i], sep = , header = TRUE)
}
combined - do.call(rbind, imp)
HTH,
--sundar
On Mon, Jan 4, 2010 at 4:31 PM, Steven Kang stochastick...@gmail.comwrote:
Hi, all
Basically, I have unknown number of data that need to be imported and
collapsed row-wisely.
The code below works fine, however the rbind function may require 50
arguments if there are 50 data files...
Thus, I would like to explore whether there are any methods in using
dynamic
objects (i.e from the resulting objects in the for loop) as an argument in
the *rbind* function.
setwd(.)
import.files - c(a.txt, b.txt, c.txt, d.txt, e.txt)
for (i in 1:length(import.files)) {
assign(paste(imp, i, sep = .), read.delim(eval(paste(.\\,
import.files[i], sep = )), header = TRUE))
}
combined - rbind(*imp.1, imp.2, imp.3, imp.4, imp.5, imp.6*)
Your expertise in resolving this issue would be greatly appreciated.
Steve
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Re: [R] makefile for sweave
Is texi2dvi in your PATH? What happens if you open a CMD window and
type texi2dvi at the prompt?
--sundar
On Thu, Nov 26, 2009 at 6:14 AM, Wolfgang Raffelsberger wr...@igbmc.fr wrote:
Dear all,
I can't get texi2dvi working right. Basically I'd like to convert a .lex to
.pdf without having to fiddle with the issue Sweave.sty not being in my
current directory (as this was sugested in other posts on this list).
When I'm in the R-Gui I can get the help via
?texi2dvi
(So I conclude its installed.)
However, when I try to use it to concert a .tex to .pdf I get trouble ...
For example :
A)
The file test02.r contains :
Sweave(Sweave_test01.rnw)
library(tools)
texi2dvi(Sweave_test01.tex, pdf =T)
Now, when I run on the linux command line :
R --vanilla -q test02.r
I get :
Sweave(Sweave_test01.rnw)
Writing to file Sweave_test01.tex
Processing code chunks ...
1 : term hide (label=chunk_ini)
2 : term verbatim eps pdf (label=Fig01)
3 : term tex (label=packageVersionInfo)
Loading required package: xtable
You can now run LaTeX on 'Sweave_test01.tex'
library(tools)
texi2dvi(Sweave_test01.tex, pdf =T)
Error in texi2dvi(Sweave_test01.tex, pdf = T) :
Running 'texi2dvi' on 'Sweave_test01.tex' failed.
Messages:
sh: texi2dvi: command not found
Execution halted
B)
In a previous message on this list I found the following command line(s)
suggested, but I my case it won't work
star5_R_test_ R CMD texi2dvi --help
/usr/local/lib64/R/bin/Rcmd: line 62: exec: texi2dvi: not found
similarly, when execute (as sugested) I get the same error message
star5_R_test_ R CMD texi2dvi -p Sweave_test01.tex
/usr/local/lib64/R/bin/Rcmd: line 62: exec: texi2dvi: not found
I don't understand how can a command can be present (= installed) and still
not being found as the error messages suggest ?
For completeness :
sessionInfo()
R version 2.10.0 (2009-10-26)
x86_64-unknown-linux-gnu
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] xtable_1.5-6 mouse4302probe_2.5.0 AnnotationDbi_1.8.1
[4] mouse4302cdf_2.5.0 MASS_7.3-3 fdrtool_1.2.5 [7]
limma_3.2.1 affyPLM_1.22.0 preprocessCore_1.8.0
[10] gcrma_2.18.0 affy_1.24.2 Biobase_2.6.0
loaded via a namespace (and not attached):
[1] Biostrings_2.14.3 DBI_0.2-4 IRanges_1.4.4 RSQLite_0.7-3
[5] affyio_1.14.0 splines_2.10.0 tools_2.10.0
Thank's in advance,
Wolfgang
Charles C. Berry a écrit :
On Tue, 8 Sep 2009, Welma Pereira wrote:
Hello, I have the following makefile. The problem is that the
bibliography
doesn t work. Any help would be appreciated! I really don t don t what to
do..:-(
# The sources of the report (tex, Rnw and other files (e.g. bib, idx))
TEX_CMPS = Report problem
RNW_CMPS = prop1 prop2 ExeExps
OTHER = Report.bib
# The name of the report to produce
all: Report.pdf
code: $(RNW_CMPS:=.R)
clean:
rm -f *.log *.dvi *~
# On what does the report depends?
Report.pdf: $(TEX_CMPS:=.tex) $(RNW_CMPS:=.tex) ${OTHER} makefile
TEXINPUTS=${TPUTS} pdflatex $
TEXINPUTS=${TPUTS} pdflatex $
IIRC
R CMD texi2dvi -p target.tex
takes care of finding sweave.sty and running latex thru all the iterations
needed to build cross-references and a usable pdf.
Try
R CMD texi2dvi --help
at the shell prompt.
HTH,
Chuck
rm *.log
# mv *.aux $(dir $)
# How to build the tex files from the Rnw (Sweave) files
%.tex: %.Rnw
echo library(utils); options(width=60); Sweave('$') | ${R_PRG}
--no-save --vanilla
mv $(notdir $*.tex) $(dir $)
# How to build the R code files from the Rnw (Sweave) files
%.R: %.Rnw
echo library(utils); Stangle('$') | ${R_PRG} --no-save --vanilla
%.bib:
TEXINPUTS=${TPUTS} pdflatex $
bibtex $
%.aux:
TEXINPUTS=${TPUTS} pdflatex $
bibtex $
%.idx:
TEXINPUTS=${TPUTS} pdflatex $
makeindex $
cheers!
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Charles C. Berry (858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
__
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wolfgang Raffelsberger, PhD
Laboratoire de BioInformatique et Génomique Intégratives
IGBMC, 1 rue Laurent Fries, 67404 Illkirch Strasbourg, France
Tel (+33) 388 65 3300 Fax (+33) 388 65 3276
wolfgang.raffelsberger (at) igbmc.fr
Re: [R] Why F value and Pr are not show in summary() of an aov() result?
It's hard to read your code, so I won't comment on your specific
example. So when all else fails read the documentation for
?summary.aov:
They have columns ‘Df’, ‘Sum Sq’, ‘Mean
Sq’, as well as ‘F value’ and ‘Pr(F)’ if there are non-zero
residual degrees of freedom.
So if you do df.residual(afit), is it 0?
--sundar
On Sun, Nov 22, 2009 at 7:19 AM, Peng Yu pengyu...@gmail.com wrote:
I have the following code. I'm wondering why summary() doesn't show F
value and Pr?
Rscript multi_factor.R
a=3
b=4
c=5
d=6
e=7
A=1:a
B=1:b
C=1:c
D=1:d
E=1:e
X=matrix(nr=a*b*c*d*e,nc=5)
colnames(X)=LETTERS[1:5]
for(i_a in 1:a-1) {
+ for(i_b in 1:b-1) {
+ for(i_c in 1:c-1) {
+ for(i_d in 1:d-1) {
+ for(i_e in 1:e-1) {
+ X[(((i_a * b + i_b) * c + i_c) * d + i_d) * e + i_e + 1, ]
= c(i_a+1, i_b+1, i_c+1, i_d+1, i_e+1)
+ }
+ }
+ }
+ }
+ }
Y=matrix(nr=a*b*c*d*e,nc=1)
for(i in 1:(a*b*c*d*e)) {
+ fa=X[i,'A']
+ fb=X[i,'B']
+ fc=X[i,'C']
+ fd=X[i,'D']
+ fe=X[i,'E']
+
+ Y[i,1]= fa +fb +fc +fe +fa*fb +fa*fc +fb*fc +fa*fe +fc*fe
+fa*fb*fc +fa*fc*fe + rnorm(1)
+ }
aframe = data.frame(
+ A=as.factor(X[,'A'])
+ , B=as.factor(X[,'B'])
+ , C=as.factor(X[,'C'])
+ , D=as.factor(X[,'D'])
+ , E=as.factor(X[,'E'])
+ ,Y)
afit=aov(Y ~ A * B * C * D * E, aframe)
summary(afit)
Df Sum Sq Mean Sq
A 2 1512240 756120
B 3 453324 151108
C 4 2549895 637474
D 5 2 0.3693
E 6 1451057 241843
A:B 6 33875 5646
A:C 8 189839 23730
B:C 12 56024 4669
A:D 10 7 1
B:D 15 25 2
C:D 20 18 1
A:E 12 107574 8964
B:E 18 21 1
C:E 24 180413 7517
D:E 30 16 1
A:B:C 24 4167 174
A:B:D 30 37 1
A:C:D 40 42 1
B:C:D 60 63 1
A:B:E 36 30 1
A:C:E 48 13298 277
B:C:E 72 62 1
A:D:E 60 79 1
B:D:E 90 87 1
C:D:E 120 122 1
A:B:C:D 120 140 1
A:B:C:E 144 131 1
A:B:D:E 180 145 1
A:C:D:E 240 225 1
B:C:D:E 360 398 1
A:B:C:D:E 720 713 1
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Re: [R] denoting max value in ylim
ylim = c(0, max(log10(D10$Part.P))) Make sure you remove any 0s or NAs before computing the max though. --sundar On Fri, Nov 20, 2009 at 6:12 AM, helene frigstad helenefrigs...@hotmail.com wrote: Hi, is there any way to set the ylim range from zero to whatever is the max value in that dataset? I am plotting many similar plots to the one below, and would like to avoid having to find the max value each time. plot (D10$Part.P ~ D10$Klorofyll,pch=16,log = xy, xlab = (Chla), ylab = (POP), ylim = c (0, ???)) abline(m3, untf=TRUE, lty=1,col=blue) text(5, 0.05,paste(round(glm(D10$Part.P ~ D10$Klorofyll, data = D10, family = Gamma(link = identity))$coef, 2),collapse = )) thank you very much for your time and help. Best regards, Helene Frigstad -- View this message in context: http://old.nabble.com/denoting-max-value-in-ylim-tp26441590p26441590.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] lapply() not converting columns to factors (no error message)
Works for me: x - read.csv(url(http://dc170.4shared.com/download/153147281/a5c78386/Testvcomp10.csv?tsid=20091116-075223-c3093ab0;)) names(x) x[2:13] - lapply(x[2:13], factor) levels(x$P1L55) [1] 0 1 is.factor(x$P1L96) [1] TRUE sessionInfo() R version 2.10.0 (2009-10-26) i386-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.0 tools_2.10.0 On Mon, Nov 16, 2009 at 4:50 AM, A Singh aditi.si...@bristol.ac.uk wrote: Sorry, my file is at: http://www.4shared.com/file/153147281/a5c78386/Testvcomp10.html -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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. __ 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.
Re: [R] lapply() not converting columns to factors (no error message)
Could it be you have factor redefined in your workspace? Have you tried it in a clean directory? I.e. a directory where no .RData exists? On Mon, Nov 16, 2009 at 5:07 AM, A Singh aditi.si...@bristol.ac.uk wrote: Oh, strange! I thought it might be a problem with the 'base' package installation, because the same thing's worked for me too before but won't do now. I tried to reinstall it (base), but R says its there already which I expected it to be anyway. I don't quite know where the issue is. Very odd. --On 16 November 2009 04:59 -0800 Sundar Dorai-Raj sdorai...@gmail.com wrote: Works for me: x - read.csv(url(http://dc170.4shared.com/download/153147281/a5c78386/Testvc omp10.csv?tsid=20091116-075223-c3093ab0)) names(x) x[2:13] - lapply(x[2:13], factor) levels(x$P1L55) [1] 0 1 is.factor(x$P1L96) [1] TRUE sessionInfo() R version 2.10.0 (2009-10-26) i386-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.0 tools_2.10.0 On Mon, Nov 16, 2009 at 4:50 AM, A Singh aditi.si...@bristol.ac.uk wrote: Sorry, my file is at: http://www.4shared.com/file/153147281/a5c78386/Testvcomp10.html -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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. -- A Singh aditi.si...@bristol.ac.uk School of Biological Sciences University of Bristol __ 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.
Re: [R] Trellis settings get lost when printing to pdf
Did you make the changes before or after starting the device: library(lattice) ## before doesn't change the settings on the device: trellis.par.set(plot.symbol = list(col = red)) trellis.device(pdf, file = tmp.pdf) xyplot(1 ~ 1) dev.off() ## after does trellis.device(pdf, file = tmp.pdf) trellis.par.set(plot.symbol = list(col = red)) xyplot(1 ~ 1) dev.off() I never do things like this, though. I would suggest creating a theme instead and supplying it to xyplot (or whatever plot you're using) using par.settings: my.theme - list(plot.symbol = list(col = red)) trellis.device(pdf, file = tmp.pdf) xyplot(1 ~ 1, par.settings = my.theme) dev.off() HTH, --sundar 2009/11/13 Joel Fürstenberg-Hägg joel_furstenberg_h...@hotmail.com: Hi all, I've got some problems when changing the trellis settings for the lattice plots. The plots look exactly as I want them to when calling show.settings() as well as when plotting them in the graphical window. But when printing to a pdf file, none of the settings are used!? Does anyone know what might have happened? Because the when changing the trellis settings, these should remain in the new state until you close R right..? # Change settings for the boxplot appearance new.dot=trellis.par.get(box.dot) new.rectangle=trellis.par.get(box.rectangle) new.umbrella=trellis.par.get(box.umbrella) new.symbol=trellis.par.get(plot.symbol) new.strip.background=trellis.par.get(strip.background) new.strip.shingle=trellis.par.get(strip.shingle) new.dot$pch=| new.dot$col=black new.rectangle$col=black new.rectangle$fill=grey65 new.umbrella$col=black new.umbrella$lty=1 # Continous line, not dotted new.symbol$col=black new.strip.background$col=grey87 # Background colour in the upper label new.strip.shingle$col=black # Border colour around the upper label trellis.par.set(box.dot=new.dot, box.rectangle=new.rectangle, box.umbrella=new.umbrella, plot.symbol=new.symbol, strip.background=new.strip.background, strip.shingle=new.strip.shingle) Best regards, Joel _ Nya Windows 7 - Hitta en dator som passar dig! Mer information. http://windows.microsoft.com/shop [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] tapply function
you must have missing values in data. Try tapply(data, group, mean, na.rm = TRUE) If that's not the case, read the bottom of this email about the posting guide. HTH, --sundar On Tue, Nov 3, 2009 at 5:28 AM, FMH kagba2...@yahoo.com wrote: Hi, I tried to use tapply function to find the mean of the data in each group as the following command, but the result are NA, as there are several missing values in each group. tapply(data,group,mean) Could someone please advice me the way to ignore the missing data in order for the fucntion to run successfully? Thanks Fir __ 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. __ 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.
Re: [R] Issue with %in% - not matching identical rows in data frames
?%in% says x and table must be vectors. You supplied data.frames. So %in% is coercing your today.sequence to a vector using as.character(today.sequence) Perhaps you should paste the columns together first: x - do.call(paste, c(sequence, sep = ::)) table - do.call(paste, c(today.sequence, sep = ::)) x[7] %in% table I'm not sure if this is what you want/need, but it does match your example. HTH, --sundar On Tue, Nov 3, 2009 at 7:53 AM, Kaushik Krishnan kaushik.s.krish...@gmail.com wrote: Hi folks I have two data frames. I know that the nth (let's say the 7th) row in the first data frame (sequence) is there in the second (today.sequence). When I try to check that by doing 'sequence[7,] %in% today.sequence', I get all FALSE when it should be all TRUE. I'm certain I'm making some trivial mistake. Any solutions? The code to recreate the data frames and see for yourself is: sequence - structure(list(DATE = structure(c(14549, 14549, 14553, 14550, 14557, 14550, 14551, 14550), class = Date), DATASET = c(1L, 2L, 1L, 2L, 2L, 3L, 3L, 4L), REP = c(1L, 0L, 2L, 2L, 3L, 0L, 1L, 0L), WRONGS_ABS = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), WRONGS_RATIO = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), DONE = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), class = data.frame, row.names = c(NA, -8L)) today.sequence - structure(list(DATE = structure(c(14551, 14550), class = Date), DATASET = 3:4, REP = c(1L, 0L), WRONGS_ABS = c(0L, 0L), WRONGS_RATIO = c(0L, 0L), DONE = c(0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), row.names = 7:8, class = data.frame) sequence[7,] #You should see '2009-11-03 3 1 0 0 0' today.sequence #You can clearly see that sequence [7,] is the first row in today.sequence sequence[7,] %in% today.sequence #This should show 'TRUE TRUE TRUE TRUE TRUE TRUE'. Instead # it shows 'FALSE FALSE FALSE FALSE FALSE FALSE' Thanks -- Kaushik Krishnan (kaushik.s.krish...@gmail.com) __ 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. __ 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.
Re: [R] How can I run a function to a piece of text?
Based solely on what you told us, this can be done using eval(parse(text=...)) cmd - sprintf(mean(%s), script) eval(parse(text = cmd)) However, with more context, there may be a better solution. See, for example, install.packages(fortunes) library(fortunes) fortune(parse()) HTH, --sundar On Fri, Oct 16, 2009 at 11:39 AM, Javier PB j.perez-barbe...@macaulay.ac.uk wrote: Dear users, I got really stuck trying to apply a function to a piece of code that I created using different string functions. To make things really easy, this is a wee example: x-c(1:10) script-x, trim = 0, na.rm = FALSE ##script created by a number of paste() and rep() steps mean(script) ##function that I want to apply: doesn't work Is there any way to convert the script class so that the function mean() can read it as if it were text typed in the console? Thanks and have a superb day Javier -- View this message in context: http://www.nabble.com/How-can-I-run-a-function-to-a-piece-of-text--tp25930315p25930315.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] Error in family$family : $ operator is invalid for atomic vectors
Check to see if you have an old workspace being loaded. You might have an object called 'family' which you might need to remove. --sundar On Oct 11, 2009 12:15 PM, romunov romu...@gmail.com wrote: Thank you Jorge and Barry for your input. I've fiddled around a bit and as a result, am even more confused. If I start R console via Notepad++ (I use Npp2R) and execute the model1, it goes through just fine. Here is the sessionInfo() for this working session: sessionInfo() R version 2.9.2 (2009-08-24) i386-pc-mingw32 locale: LC_COLLATE=Slovenian_Slovenia.1250;LC_CTYPE=Slovenian_Slovenia.1250;LC_MONETARY=Slovenian_Slovenia.1250;LC_NUMERIC=C;LC_TIME=Slovenian_Slovenia.1250 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.9.2 And if I run R normally, via an icon from the desktop (Rgui.exe) it gives the aforementioned error. Here is the sessionInfo() for non-working session. Is it possible that grid, reshape, plyr, ggplot2 and proto could be causing this? If so, how can I prevent them from loading automatically or unloading from a live session? sessionInfo() R version 2.9.2 (2009-08-24) i386-pc-mingw32 locale: LC_COLLATE=Slovenian_Slovenia.1250;LC_CTYPE=Slovenian_Slovenia.1250;LC_MONETARY=Slovenian_Slovenia.1250;LC_NUMERIC=C;LC_TIME=Slovenian_Slovenia.1250 attached base packages: [1] stats graphics grDevices utils datasets grid methods [8] base other attached packages: [1] reshape_0.8.3 plyr_0.1.9proto_0.3-8 loaded via a namespace (and not attached): [1] ggplot2_0.8.3 Cheers, Roman On Sun, Oct 11, 2009 at 6:32 PM, Jorge Ivan Velez jorgeivanve...@gmail.comwrote: Hi Romain, It works for me: model1 - glm(as.vector(x) ~dept*sex*admit,poisson) model1 ... [[alternative HTML version deleted]] __ 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.
Re: [R] How to speed up R with version 2.9.2?
Another possibility is a very large .RData file in the directory where you're starting R. You can try Rgui --no-restore (I don't have windows, so I'm not sure if this an option with RGui, though I know it is with R.) --sundar On Fri, Oct 2, 2009 at 8:50 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Its under 5 seconds on my Vista laptop. Do you have any startup files? If Rgui --vanilla is much faster then your startup files are the problem. On Fri, Oct 2, 2009 at 11:45 AM, FMH kagba2...@yahoo.com wrote: Thank you for your answer. I'm using Win XP with 2GB RAM in memory. Cheers Fir - Original Message From: stephen sefick ssef...@gmail.com To: FMH kagba2...@yahoo.com Cc: r-help@r-project.org Sent: Fri, October 2, 2009 4:38:10 PM Subject: Re: [R] How to speed up R with version 2.9.2? You're fine, but please do read the posting guide. What OS etc. Where you doing anything else on the computer? Is this a RAM limitation? I have 2.9.2 running on two flavours of linux, mac os x and windows all 2.9.2 and there doesn't seem to be a problem. regards, Stephen On Fri, Oct 2, 2009 at 10:35 AM, FMH kagba2...@yahoo.com wrote: Dear All, I'm sorry if my question does not suit with this R group. I have recently installed R software with version 2.9.2, but i found the program took almost 1 minute as soon as it was opened, before it can be used. However, the previous version 2.9.1 only take few seconds after the menu bar was clicked. This circumstance has caused me to wait for couple of minutes as several R windows were opened simultaneously. Are there any ways to speed up on this 2.9.2 version? Could someone give some hints, please? Thank you Fir __ 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. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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. __ 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. __ 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.
Re: [R] Skipping missing files when importing data
Try ?file.exists.
if (file.exists(fxxx)) {
read.table(fxxx)
} else {
cat(\, fxxx, \ is missing\n, sep = )
}
HTH,
--sundar
On Sun, Sep 20, 2009 at 9:28 PM, jiangrm jian...@gmail.com wrote:
Trying to import a bunch of data files named like f001, f002, f999. Some
of the files may be
missing and the missing files vary from time to time.
Used for loop and read.table. When it reaches the missing file (say f100), it
shows:
Error in file(file, r) : cannot open the connection
In addition: Warning message:
In file(file, r) :
cannot open file 'f100': No such file or directory
and the program stops.
How can I skip the missing ones and keep the program running? Guess either
checking the validity of
filenames, or ignore the error message may work. Which functions should be
used or any better ideas?
-RJ
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Re: [R] xyplot: Can I identify groups but not use them for regression?
I think this ought to work for you:
library(lattice)
set.seed(42)
d - data.frame(year = c(rep(2007,12), rep(2008,12)),
treatment = rep(LETTERS[1:3], each = 4, times = 2))
d$cover - rnorm(nrow(d))
d$variable - rnorm(nrow(d))
xyplot(variable ~ cover | year, d,
panel = function(x, y, ...) {
panel.superpose(x, y, ...)
panel.lmline(x, y, ...)
},
groups = treatment)
HTH,
--sundar
On Fri, Sep 18, 2009 at 3:42 PM, Seth W Bigelow sbige...@fs.fed.us wrote:
I wish to identify groups representing different treatments, but to plot
them and do a regression using a continuous variable (cover)
ignoring the groupings.
d$year - NA
d$year -c(rep(2007,12), rep(2008,12))
d$treatment - c(rep(A,4),rep(B,4),rep(C,4), rep(A,4), rep(B,4),
rep(C,4))
d$cover - rnorm(24)
d$variable - rnorm(24)
xyplot(variable ~ cover | year, d,
type=c(p,r),
groups=treatment
)
As it stands, a different regression line is plotted for each treatment.
Oh, and how do I display the actual numeric value of year (e.g., 2007)
in the strip, rather than the word year?
--Seth
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
1731 Research Park Drive, Davis California
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Re: [R] Lattice graph tweaking
A reproducible example would be nice. Try grid = FALSE for the first question, though I'm unaware which lattice plot you are using where the default is TRUE. So I can't guarantee that will even work. For your second question, add par.settings = list(strip.background = list(col = white)) to your call (e.g. xyplot(..., par.settings = ...)). To see what other parameters you can set in par.settings, try: str(trellis.par.get()) HTH, --sundar On Tue, Aug 25, 2009 at 6:14 AM, Wallis, Daviddavid.walli...@imperial.ac.uk wrote: To: silwood-r Subject: Removing lattice graph gridlines and editing label box colour Hi, Is it possible to remove the background gridlines from a lattice graph (ie graph made up of multiple individual graphs with annoying blue grid in the backgroun)? Also, Is it possible to change the colour of the individual graph label boxes? - ie the default pink boxes above the individual graphs Thanks [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] Help with nls and error messages singular gradient
Hi, Michael,
I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:
con - textConnection(time bod
11 0.47
22 0.74
33 1.17
44 1.42
55 1.60
67 1.84
79 2.19
8 11 2.17)
mydata - read.table(con, header = TRUE)
close(con)
beta - c(3, -0.1) # your initial values
beta - c(2.4979, -2.02456) # SPSS answer
mycurve - function(x) {
beta[1]/(1 - exp(beta[1] * x))
}
curve(mycurve, from = 1, to = 11,
ylim = range(mydata$bod, mycurve(mydata$time)))
points(mydata$time, mydata$bod)
You might want to look at package nls2 which allows a brute force grid
search to find some starting values. Or rethink the equation you're
trying to fit.
HTH,
--sundar
On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmainmpearm...@google.com wrote:
Hi All,
I'm trying to run nls on the data from the study by Marske (Biochemical
Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
(1988).
Data is as follows, (stored as mydata)
time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17
I then run the following;
#Plot initial curve
plot(mydata$time, mydata$bod,xlab=Time (in days),ylab=biochemical oxygen
demand (mg/l) )
model - nls(bod ~ beta1/(1 - exp(beta2*time)), data =
mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
The start values are recommended, (I have used these values in SPSS without
any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
but return the error message,
Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
list(beta1 = 3, : singular gradient
Can anyone offer any advice?
Thanks in advance
Mike
--
Michael Pearmain
Senior Analytics Research Specialist
“Statistics are like women; mirrors of purest virtue and truth, or like
whores to use as one pleases”
Google UK Ltd
Belgrave House
76 Buckingham Palace Road
London SW1W 9TQ
United Kingdom
t +44 (0) 2032191684
mpearm...@google.com
If you received this communication by mistake, please don't forward it to
anyone else (it may contain confidential or privileged information), please
erase all copies of it, including all attachments, and please let the sender
know it went to the wrong person. Thanks.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with nls and error messages singular gradient
Another alternative is to use SSlogis which is very similar to the
model you're fitting except with one additional parameter:
Asym - 3
xmid - 0
scal - 10
model - nls(bod ~ SSlogis(time, Asym, xmid, scal), data = mydata)
summary(model)
plot(bod ~ time, mydata)
newdata - data.frame(time = seq(1, 11, length = 100))
lines(newdata$time, predict(model, newdata))
HTH,
--sundar
On Tue, Aug 25, 2009 at 10:05 AM, Sundar Dorai-Rajsdorai...@gmail.com wrote:
Hi, Michael,
I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:
con - textConnection(time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17)
mydata - read.table(con, header = TRUE)
close(con)
beta - c(3, -0.1) # your initial values
beta - c(2.4979, -2.02456) # SPSS answer
mycurve - function(x) {
beta[1]/(1 - exp(beta[1] * x))
}
curve(mycurve, from = 1, to = 11,
ylim = range(mydata$bod, mycurve(mydata$time)))
points(mydata$time, mydata$bod)
You might want to look at package nls2 which allows a brute force grid
search to find some starting values. Or rethink the equation you're
trying to fit.
HTH,
--sundar
On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmainmpearm...@google.com wrote:
Hi All,
I'm trying to run nls on the data from the study by Marske (Biochemical
Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
(1988).
Data is as follows, (stored as mydata)
time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17
I then run the following;
#Plot initial curve
plot(mydata$time, mydata$bod,xlab=Time (in days),ylab=biochemical oxygen
demand (mg/l) )
model - nls(bod ~ beta1/(1 - exp(beta2*time)), data =
mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
The start values are recommended, (I have used these values in SPSS without
any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
but return the error message,
Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
list(beta1 = 3, : singular gradient
Can anyone offer any advice?
Thanks in advance
Mike
--
Michael Pearmain
Senior Analytics Research Specialist
“Statistics are like women; mirrors of purest virtue and truth, or like
whores to use as one pleases”
Google UK Ltd
Belgrave House
76 Buckingham Palace Road
London SW1W 9TQ
United Kingdom
t +44 (0) 2032191684
mpearm...@google.com
If you received this communication by mistake, please don't forward it to
anyone else (it may contain confidential or privileged information), please
erase all copies of it, including all attachments, and please let the sender
know it went to the wrong person. Thanks.
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Re: [R] how to check if ... is empty
Try
dots - list(...)
if (length(dots) == 0) {
## do something
}
On Thu, Jul 16, 2009 at 6:46 AM, Thomas Roth (geb.
Kaliwe)hamstersqu...@web.de wrote:
Hi,
I was wondering what would be the best way to check if the three dots
argument contains any arguments (i.e. does ... contain any arguments or not?
)
#Example
test = function(x,y, ...)
{
#Wanted R-Code
# if(empty(...))
# do some calculation
plot(x,y,...)
}
Thanks
Thomas Roth
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Re: [R] Lattice group colors?
Look at show.settings() and str(trellis.par.get()). This will show you what the default settings are. The group colors are set by the superpose.* elements (e.g. superpose.line is for group lines). To set them, I usually create a list and pass it to par.settings. For example, my.theme - list(superpose.line = list(col = c(darkred, darkblue), lty = 1:2)) xyplot(y ~ x, data = mydata, groups = g, par.settings = my.theme, auto.key = list(points = FALSE, lines = TRUE)) HTH, --sundar On Mon, Jun 22, 2009 at 6:30 AM, Fredrik Karlssondargo...@gmail.com wrote: Dear list, I have been struggling to find how I would go about changing the bakground colors of groups in a lattice barchart in a way so that the auto.key generated also does the right thing and pick it up for the key. I have used the col argument (which I guess is sent to par()) in a way so that the colors are like I would want them, but now I guess I need to know which part part of the theme I need to change in order to make this change permanent for all the plots, and all the keys. I am of course thankful for all the help I can get. (Also, how does one know these things about the theme variables? Is there some documentation somewhere?) /Fredrik -- Life is like a trumpet - if you don't put anything into it, you don't get anything out of it. __ 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. __ 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.
Re: [R] reading in multiple files
You could try: do.call(rbind, lapply(list.files(path/to/files, full = TRUE), read.csv)) And add more arguments to lapply if the files are not csv, have no header, etc. --sundar On Tue, Jun 9, 2009 at 11:18 AM, Erin Hodgesserinm.hodg...@gmail.com wrote: Dear R People: I have about 6000 files to be read in that I'd like to go to one matrix. There are two columns, 1 line in each file. Is there a way to bring them in to produce one matrix or data frame, please? Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodg...@gmail.com __ 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. __ 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.
Re: [R] find a sequence of characters in a vector
use gregexpr and paste aze - paste(c(a, z, e), collapse = ) sequence - paste(c(a,z,e,r,t,a,z,a,z,e,c), collapse = ) gregexpr(aze, sequence, fixed = TRUE) [[1]] [1] 1 8 attr(,match.length) [1] 3 3 HTH, --sundar On Fri, Jun 5, 2009 at 6:22 AM, Ptit_Bleuptit_b...@yahoo.fr wrote: Hello, I'm just looking for an easy way to find the positions of a complete sequence in a bigger vector. For example : c(a,z,e) in c(a,z,e,r,t,a,z,a,z,e,c) and the result should be 1 8 that is the positions of the beginning of the complete sequence. I tried with %in%, match, is.element but all I get is, for example which(c(a,z,e) in c(a,z,e,r,t,a,z,a,z,e,c)) 1 2 3 meaning that each character is in the bigger vector. It must be easy, except for me. Sorry. If you have a solution, thanks in advance to share it. Have a good week-end, Ptit Bleu. -- View this message in context: http://www.nabble.com/find-a-sequence-of-characters-in-a-vector-tp23888063p23888063.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] error message re: max(i), but code and output seen O.K.
This error is thrown if the argument to max is either NULL or length zero: [~] Rscript -e max(NULL) [1] -Inf Warning message: In max(NULL) : no non-missing arguments to max; returning -Inf [~] Rscript -e max(numeric(0)) [1] -Inf Warning message: In max(numeric(0)) : no non-missing arguments to max; returning -Inf HTH, --sundar On Wed, May 20, 2009 at 11:23 AM, Kirsten Miles sirole@gmail.com wrote: I have a researcher who is consistently get the warning message: In max(i) : no non-missing arguments to max; returning -Inf Best as I can tell the code is working properly and the output is as expected. I would like some help in understanding why he is getting this error message and what its implications are. I have his code. Sincerely, Kirsten Miles Support Specialist Research Computing Lab Charles L. Brown Science and Engineering Library kd...@virginia.edu [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] what is wrong with this code?
You're missing a ) off end of the first line. You should consider using an editor (e.g. ESS/Emacs) that does parentheses matching. I found this in less than 5 sec (less time than I'm taking to write you a note) by cut and pasting in Emacs. --sundar On Tue, May 19, 2009 at 12:52 PM, deanj2k dl...@le.ac.uk wrote: dlogl - -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta))) d2logl - (n/theta^2) - sum((-2y/theta^3)*(1-exp(y/theta))/(1+exp(y/theta))) - sum(((2*y/theta^4)*exp(y/theta))/((1+exp(y/theta))^2)) returns the error message: Error: unexpected symbol in: dlogl - -(n/theta)-sum((y/(theta)^2)*((1-exp(y/theta))/(1+exp(y/theta))) d2logl do you know what i have done wrong -- View this message in context: http://www.nabble.com/what-is-wrong-with-this-code--tp23623227p23623227.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] convert large integers to hex
Thanks for both answers. In the end I decided to use Gabor's bc package. Thanks, --sundar On Thu, May 7, 2009 at 5:10 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: There is an interface between R and bc -- not on CRAN but available from its home page here: http://r-bc.googlecode.com source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;) bc(obase = 16; 123456789123456789, retclass = character) [1] 1B69B4BACD05F15 On Wed, May 6, 2009 at 9:59 PM, jim holtman jholt...@gmail.com wrote: You can use the 'bc' command (use Cygwin if on Windows); /cygdrive/c: bc bc 1.06 Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. This is free software with ABSOLUTELY NO WARRANTY. For details type `warranty'. x=6595137340052185552 obase=16 x 5B86A277DEB9A1D0 You can call this from R. On Wed, May 6, 2009 at 3:26 PM, Sundar Dorai-Raj sdorai...@gmail.comwrote: Hi, I'm wondering if someone has solved the problem of converting very large integers to hex. I know about format.hexmode and as.hexmode, but these rely on integers. The numbers I'm working with are overflowing and losing precision. Here's an example: x - 6595137340052185552 # stored as character as.integer(x) # warning about inaccurate conversion format.hexmode(as.numeric(x)) # warnings about loss of precision as.hexmode(x) # more warnings and does not do what I expected I'm planning on writing a function that will do this, but would like to know if anybody already has a solution. Basically, I would like the functionality of format.hexmode on arbitrarily large integers. Thanks, --sundar __ 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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ 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. __ 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.
[R] convert large integers to hex
Hi, I'm wondering if someone has solved the problem of converting very large integers to hex. I know about format.hexmode and as.hexmode, but these rely on integers. The numbers I'm working with are overflowing and losing precision. Here's an example: x - 6595137340052185552 # stored as character as.integer(x) # warning about inaccurate conversion format.hexmode(as.numeric(x)) # warnings about loss of precision as.hexmode(x) # more warnings and does not do what I expected I'm planning on writing a function that will do this, but would like to know if anybody already has a solution. Basically, I would like the functionality of format.hexmode on arbitrarily large integers. Thanks, --sundar __ 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.
Re: [R] setting trellis auto.key color values
Set the colors in graph.sets and not auto.key. graph.sets - list(axis.text = list(cex = 0.65), par.ylab.text = list(cex = 1.25), par.xlab.text = list(cex = 1.25), superpose.polygon = list(col = 3:5)) Then remove the col = 3:5 from auto.key and barchart. Also, you can simplify your code by removing gator_IR$ and including data = gator_IR. I.e. barchart(MEAN ~ Hydro | as.factor(IR_ID), data = gator_IR, layout = c(4, 1), groups = Rain, ylim = c(0, 1), ...) HTH, --sundar On Tue, May 5, 2009 at 8:32 AM, steve_fried...@nps.gov wrote: I'm working with Lattice graphics and I would like very much to color code the auto.key fill color with the same corresponding colors that I use in the panels. I've looked on the web for clues, and on the CRAN-R help sites searching on trellis auto.key color and variations, unfortunately the responses there are not very specific. Would someone please explain I have the Book, if you can point me to the pages that explain this I'd appreciate that too graph.sets - list(axis.text = list(cex = 0.65), par.ylab.text = list(cex = 1.25), par.xlab.text = list(cex = 1.25)) barchart(gator_IR$MEAN ~ gator_IR$Hydro | as.factor(gator_IR$IR_ID),layout=c(4,1),col = c(3:5), groups=gator_IR$Rain, ylim=c(0,1), par.settings = graph.sets, main = Alligator Nesting Performance, ylab= Mean HSI, auto.key = list(top, columns=3, col=c(3:5)) this script creates the graph nicely, with the qualification that the color for the key titles are the correct, but the filling colors are default pastels. Where do I change them ? Windows XP with R 2.8.1 Thank you. Steve Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ 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. __ 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.
Re: [R] evaluate a expression
Hi, Ning, Try: eval(parse(text = expr)) HTH, --sundar On Sat, May 2, 2009 at 5:39 AM, Ning Ma pnin...@gmail.com wrote: Hi, I am new to R. Can anyone tell me how to evaluate an expression stored in a string? such as: expr - 3*5 I want to get the result 15. Thanks in advance. __ 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. __ 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.
Re: [R] Flipping axes of qqnorm
Try (re)reading ?qqnorm. Use datax = TRUE. --sundar On Sun, Apr 26, 2009 at 4:37 PM, Chris_d dewhurstch...@gmail.com wrote: Hi all, I have just started using R to produce qqnorm plots. I am trying to switch the x and y axes so that the theoretical values are plotted on the y axis and my data on the x axis. Can anyone help me with this? Thanks -- View this message in context: http://www.nabble.com/Flipping-axes-of-qqnorm-tp23248007p23248007.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] Cannot clean infinite values
Use ?is.infinite inf - is.infinite(data) data[inf] - 0.3 * sign(data[inf]) On Sun, Apr 26, 2009 at 5:44 PM, Nigel Birney na...@cam.ac.uk wrote: Hello all, I have to import numeric data from file but found it contains Infinite values which need to be eliminated. I tried to replace them in this way: data[which(data==-Inf)] - -0.3 data[which(data==+Inf)] - 0.3 But, somehow, the Infinite values stayed there. Any suggestions? regards, N. -- View this message in context: http://www.nabble.com/Cannot-clean-infinite-values-tp23248409p23248409.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] Setting lattice par parameters
Because you're not calling trellis.par.set correctly. It should be: trellis.par.set(par.ylab.text = list(cex = 0.65), par.xlab.text = list(cex = 0.65)) However, I usually do things like this: my.theme - list(par.ylab.text = list(cex = 0.65), par.xlab.text = list(cex = 0.65)) barchart(..., par.settings = my.theme) so the settings are only changed for the current plot and not globally. HTH, --sundar On Thu, Apr 23, 2009 at 6:25 AM, steve_fried...@nps.gov wrote: Hello I'm plotting a large suite of barcharts and need to modify the size of the text for both the yaxis and xaxis labels. I've tried using the following: trellis.par.set(list(par.ylab.text = list(cex = 0.65)), trellis.par.set(list = par.xlab.text = list(cex = 0.65 On inspection, however after I invoke this line, trellis.par.get(par.ylab.text) trellis.par.get(par.xlab.text) It looks like the parameters have not been modified. Do I need to do something different than this approach ? Thank you very much in advance. Steve Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ 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. __ 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.
Re: [R] question on using lattice panel plots
Try: z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30), each = 50),2)) z - as.data.frame(z) z - cbind(z, runif(300)) names(z) - c(Method, sigma, Error) z$sigma - factor(z$sigma, c(5, 10, 30)) library(lattice) sigma - as.numeric(levels(z$sigma)) sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s))) sigmaExpr - as.expression(sigmaExprList) bwplot(Error~Method | sigma, data = z, horiz = F, xlab = Method, strip = strip.custom(var.name = sigmaExpr, strip.levels = FALSE, strip.names = TRUE), layout = c(3,1)) HTH, --sundar On Thu, Apr 16, 2009 at 12:43 PM, Ranjan Maitra mai...@iastate.edu wrote: Hi, I think this question is best explained using the following self-contained toy example: ## cut code here and paste to R window z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30), each = 50),2)) z - as.data.frame(z) z - cbind(z, runif(300)) names(z) - c(Method, sigma, Error) library(lattice) bwplot(Error~Method | sigma, data = z, horiz = F, xlab = Method, layout = c(3,1)) ## end code Now the question: I would like the panels to be in the order of sigma, i. e. 5, 10, 30 and not 10, 30 and 5 as is currently the case. Is this possible? Not to seek too much indulgence, but to ask anyway, I wonder if it is possible to have a Greek sigma = 5, a Greek sigma = 10 and a Greek sigma = 30. (Sort of what we would get using expression(sigma == 5), expression(sigma == 10), expression(sigma == 10) on base R figures). Please let me know if my question is not clear. Many thanks for any suggestions and help and best wishes! Ranjan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question on using lattice panel plots
Sorry, that should be:
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr - as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
horiz = F, xlab = Method,
strip = function(which.given, which.panel, var.name,
strip.levels = FALSE,
strip.names = TRUE, ...) {
strip.default(which.given, which.panel,
var.name = sigmaExpr[which.panel],
strip.levels = FALSE,
strip.names = TRUE, ...)
},
layout = c(3,1))
Not sure how to do this with strip.custom.
--sundar
On Thu, Apr 16, 2009 at 1:20 PM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Try:
z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30),
each = 50),2))
z - as.data.frame(z)
z - cbind(z, runif(300))
names(z) - c(Method, sigma, Error)
z$sigma - factor(z$sigma, c(5, 10, 30))
library(lattice)
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr - as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
horiz = F, xlab = Method,
strip = strip.custom(var.name = sigmaExpr,
strip.levels = FALSE, strip.names = TRUE),
layout = c(3,1))
HTH,
--sundar
On Thu, Apr 16, 2009 at 12:43 PM, Ranjan Maitra mai...@iastate.edu wrote:
Hi,
I think this question is best explained using the following
self-contained toy example:
## cut code here and paste to R window
z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30),
each = 50),2))
z - as.data.frame(z)
z - cbind(z, runif(300))
names(z) - c(Method, sigma, Error)
library(lattice)
bwplot(Error~Method | sigma, data = z, horiz = F, xlab = Method,
layout = c(3,1))
## end code
Now the question:
I would like the panels to be in the order of sigma, i. e. 5, 10, 30
and not 10, 30 and 5 as is currently the case. Is this possible?
Not to seek too much indulgence, but to ask anyway, I wonder if it is
possible to have a Greek sigma = 5, a Greek sigma = 10 and a Greek
sigma = 30. (Sort of what we would get using expression(sigma == 5),
expression(sigma == 10), expression(sigma == 10) on base R figures).
Please let me know if my question is not clear.
Many thanks for any suggestions and help and best wishes!
Ranjan
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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.
Re: [R] Changing the y-axis units of a lattice histogram
Try: library(lattice) histogram( ~ height | voice.part, data = singer, type = c, scales = list(y = list(at = seq(0, 20, 5), labels = seq(0, 200, 50 HTH, --sundar On Fri, Apr 3, 2009 at 2:01 PM, Judith Flores jur...@yahoo.com wrote: Hello, I need to multiply the number of counts in the y-axis of a lattice histogram by a constant factor, such that the plot would represent a different type of variable plotted in the y-axis, not counts. Can this be done? Thank you, Judith __ 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. __ 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.
Re: [R] labeling panels in lattice plots
Try converting year to a factor xyplot(min + max + ave ~ month | factor(year), data = rain.stats, ...) Also, notice the inclusion of the data argument. HTH, --sundar On Tue, Mar 31, 2009 at 6:28 AM, steve_fried...@nps.gov wrote: I am using windows XP with R 2.8.1 I am generating a lattice plot of annual rain patterns using the following function: xyplot(rain.stats$min+ rain.stats$max + rain.stats$ave ~ rain.stats$month |rain.stats$year, lty = 1, data = rain.stats, type = c(l,l, l), col = c(red, blue, green), distribute.type = TRUE, main = Annual Monthly Minimum, Average and Maximum PPT) Each panel is labels with rain.stats$year instead of the actual value referenced by this variable. the data.frame I'm using has the following form (year values range from 1965 to 2000 and month from 1 to 12). dim(rain.stats) 432 6 rain.stats[1:10,] year month X2 min max ave 1 1965 1 1 0. 1.9196 0.3650112 2 1966 1 1 2.1483 4.9615 3.5247034 3 1967 1 1 0.4038 3.9145 1.7133045 4 1968 1 1 0.2033 3.2119 1.1844769 5 1969 1 1 1.2533 5.6226 2.9505545 6 1970 1 1 0.9142 3.8861 2.6248453 7 1971 1 1 0.1191 1.6109 0.6570289 8 1972 1 1 0.2309 2.9380 0.9259674 9 1973 1 1 0.9471 3.6342 1.9019848 10 1974 1 1 0.1739 9.0225 1.0672980 The plot illustrates exactly what I'm after with the exception of the panel labels. I'm following an example in the Lattice Book by Deepayan Sarkar (which I find very informative - thanks), but I'm not getting the results as per the example. Can anyone offer a solution? Thanks in advance. Steve Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 steve_fried...@nps.gov Office (305) 224 - 4282 Fax (305) 224 - 4147 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can not get a prediction interval from Predict
?predict.glm has no interval argument. Perhaps you're thinking of ?predict.lm, which is different. To get intervals in glm, I've used: example(predict.glm) pr - predict(budworm.lg, se.fit = TRUE) family - family(budworm.lg) lower - family$linkinv(pr$fit - qnorm(0.95) * pr$se.fit) upper - family$linkinv(pr$fit + qnorm(0.95) * pr$se.fit) Note that these are confidence limits and not prediction limits. The latter would require more thought. You could also try RSiteSearch(glm interval). HTH, --sundar On Tue, Mar 31, 2009 at 9:58 AM, Taylor Davis taylor.da...@klasresearch.com wrote: I am trying to get a prediction interval from a glm regression. With newdat being my set of values to be fitted, and glmreg the name of my regression, I am using the following code. predict(glmreg, newdat, se.fit = TRUE, interval = confidence, level = 0.90) The problem is that I am only getting the standard error and the fitted value, not a prediction interval. Any help would be great! Thanks so much. ___ TAYLOR DAVIS | KLAS taylor.da...@klasresearch.com | 801.734.6279 __ 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. __ 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.
Re: [R] use R Group SFBA April meeting reminder; video of Feb k
Could be that you have some sort of ad filter in your browser that's blocking the video? It appears just fine for me in Firefox 3. On Mon, Mar 30, 2009 at 3:55 PM, Ted Harding ted.hard...@manchester.ac.uk wrote: On 30-Mar-09 22:13:04, Jim Porzak wrote: Next week Wednesday evening, April 8th, Mike Driscoll will be talking about Building Web Dashboards using R see: http://www.meetup.com/R-Users/calendar/9718968/ for details to RSVP. Also of interest, our member Ron Fredericks has just posted a well edited video of the February kickoff panel discussion at Predictive Analytics World The R and Science of Predictive Analytics: Four Case Studies in R with * Bo Cowgill, Google * Itamar Rosenn, Facebook * David Smith, Revolution Computing * Jim Porzak, The Generations Network and chaired by Michael Driscoll, Dataspora LLC see: http://www.lecturemaker.com/2009/02/r-kickoff-video/ Best, Jim Porzak It could be very interesting to watch that video! However, I have had a close look at the web page you cite: http://www.lecturemaker.com/2009/02/r-kickoff-video/ and cannot find a link to a video. Lots of links to non-video things, but none that I could see to a video. There is a link on that page at: How Google and Facebook are using R by Michael E. Driscoll | February 19, 2009 http://dataspora.com/blog/predictive-analytics-using-r/ Following that link leads to a page, on which the first link, in: (March 26th Update: Video now available) Last night, I moderated our Bay Area R Users Group kick-off event with a panel discussion entitled The R and Science of Predictive Analytics, co-located with the Predictive Analytics World conference here in SF. leads you back to where you came from, and likewise the link at the bottom of the page: A video of the event is now available courtesy of Ron Fredericks and LectureMaker. Could you help by describing where on that web page it can be found? With thanks, Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 30-Mar-09 Time: 23:55:07 -- XFMail -- __ 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. __ 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.
Re: [R] Requesting help with lattice again
For the first question, add a groups argument. E.g. barchart(HSI ~ Scenario | Region, Wbirdsm, groups = HydroState) Also note that using Wbirdsm$HSI makes your call less readable, so I added the data argument. For your second question, setting the key does not set the color theme. You want to set the colors using par.setting. E.g. dotplot(HSI ~ Scenario | Region, Wbirdsm, groups = HydroState, par.settings = list(superpose.symbol = list(col = c(red, green, blue))), auto.key = list(space = right)) Then use auto.key instead of key. HTH, --sundar On Wed, Mar 25, 2009 at 7:02 AM, steve_fried...@nps.gov wrote: Hello, this is a request for assistance that I submitted earlier, this time with the dataset. My mistake for taking up bandwidth. I've also rephrased the question to address an additional concern. I'm working on a windows XP machine with R 2.8.1 1). I'd like a barchart (or other lattice type display) HSI ~ of the three factors (Region, Scenario and HydroState). However barchart complains library(reshape) library(lattice) barchart(Wbirdsm$HSI ~ WBirdsm$Scenario | WBridsm$Region) # This works, but only marginally. I'd like specify the interaction between the variables ( Scenario and HydroState) What is the best way to combine these factors so they can be treated as a single variable and rewrite the barchart call? barchart(Wbirdsm$HSI ~ WBirdsm$interaction between Scenario and HydroState) ~ Wbirdsm$Region) The second question refers back to my original posting. 2) # using dotplot instead of barchart I use the following code. key.variable - list(space = right, text = list(levels(wbirdm$variable)), points = list(pch = 1:3, col=c(red,green, blue))) dotplot(wbirdm$value ~ wbirdm$variable | wbirdm$Region, col=c(1:9), pch= rep(c(1:3), key = key.variable, groups=wbirdm$variable, ylab= Mean HSI)) # However the key legend doesn't appear in the plot with this sequence and the labels are too along the panels. Is there a way to address this? thank you very much for the assistance. Steve This is part of the larger data base, after I passed it thru melt. Region Species Scenario HydroState HSI 1 Eastern Panhandle WBLong NSM Ave 0.165945170 2 Eastern Panhandle WBLong NSM Dry 0.056244263 3 Eastern Panhandle WBLong NSM Wet 0.290692607 4 Eastern Panhandle WBLong ECB Ave 0.165945170 5 Eastern Panhandle WBLong ECB Dry 0.056244263 6 Eastern Panhandle WBLong ECB Wet 0.290692607 7 Eastern Panhandle WBLong CERP Ave 0.165945170 8 Eastern Panhandle WBLong CERP Dry 0.056244263 9 Eastern Panhandle WBLong CERP Wet 0.290692607 10 Long Pine Key / South Taylor Slough WBLong NSM Ave 0.151159734 11 Long Pine Key / South Taylor Slough WBLong NSM Dry 0.067348863 12 Long Pine Key / South Taylor Slough WBLong NSM Wet 0.20738 13 Long Pine Key / South Taylor Slough WBLong ECB Ave 0.151159734 14 Long Pine Key / South Taylor Slough WBLong ECB Dry 0.067348863 15 Long Pine Key / South Taylor Slough WBLong ECB Wet 0.20738 16 Long Pine Key / South Taylor Slough WBLong CERP Ave 0.151159734 17 Long Pine Key / South Taylor Slough WBLong CERP Dry 0.067348863 18 Long Pine Key / South Taylor Slough WBLong CERP Wet 0.20738 19 Northern Taylor Slough WBLong NSM Ave 0.115503291 20 Northern Taylor Slough WBLong NSM Dry 0.005617136 21 Northern Taylor Slough WBLong NSM Wet 0.252428530 22 Northern Taylor Slough WBLong ECB Ave 0.115503291 23 Northern Taylor Slough WBLong ECB Dry 0.005617136 24 Northern Taylor Slough WBLong ECB Wet 0.252428530 25 Northern Taylor Slough WBLong CERP Ave 0.115503291 26 Northern Taylor Slough WBLong CERP Dry 0.005617136 27 Northern Taylor Slough WBLong CERP Wet 0.252428530 28 East Slough WBLong NSM Ave 0.313215457 29 East Slough WBLong NSM Dry 0.046917053 30 East Slough WBLong NSM Wet 0.447002596 31 East Slough WBLong ECB Ave 0.313215457 32 East Slough WBLong ECB Dry 0.046917053 33 East Slough WBLong ECB Wet 0.447002596 34 East Slough WBLong CERP
Re: [R] XYplot simple question
Convert Plot to factor:
xyplot(AbvBioAnnProd ~ Year | Plot, type = c(b, r), pch = 16)
Also note that using the type argument with multiple values prevents
the necessity of a custom panel function.
HTH,
--sundar
On Mon, Mar 16, 2009 at 5:54 AM, AllenL allen.laroc...@gmail.com wrote:
Hello R friends,
Simple question today: I am desiring to do an xyplot with the below code,
which graphs time series across different experimental Plots-
xyplot(AbvBioAnnProd~Year|Plot) ### Plots each monoculture biomass vs
time
xyplot(AbvBioAnnProd~Year|Plot,panel=function(x,y){
panel.xyplot(x,y,type=b,pch=16)
panel.abline(lm(y~x))
})
What I want to add are unique labels for each panel, where instead of all
labeled plot with the slide-bar visual (although that is okay) I want
the Plot number to appear (i.e. the value of Plot for that panel).
This should be easy, right?
-Al
--
View this message in context:
http://www.nabble.com/XYplot-simple-question-tp22537350p22537350.html
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__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] XYplot simple question
Sorry, I should have
xyplot(AbvBioAnnProd ~ Year | factor(Plot), type = c(b, r), pch = 16)
On Mon, Mar 16, 2009 at 6:17 AM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Convert Plot to factor:
xyplot(AbvBioAnnProd ~ Year | Plot, type = c(b, r), pch = 16)
Also note that using the type argument with multiple values prevents
the necessity of a custom panel function.
HTH,
--sundar
On Mon, Mar 16, 2009 at 5:54 AM, AllenL allen.laroc...@gmail.com wrote:
Hello R friends,
Simple question today: I am desiring to do an xyplot with the below code,
which graphs time series across different experimental Plots-
xyplot(AbvBioAnnProd~Year|Plot) ### Plots each monoculture biomass vs
time
xyplot(AbvBioAnnProd~Year|Plot,panel=function(x,y){
panel.xyplot(x,y,type=b,pch=16)
panel.abline(lm(y~x))
})
What I want to add are unique labels for each panel, where instead of all
labeled plot with the slide-bar visual (although that is okay) I want
the Plot number to appear (i.e. the value of Plot for that panel).
This should be easy, right?
-Al
--
View this message in context:
http://www.nabble.com/XYplot-simple-question-tp22537350p22537350.html
Sent from the R help mailing list archive at Nabble.com.
__
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.
__
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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.
Re: [R] Get user system name
Assuming USER is defined on your system then Sys.getenv(USER) ought to work. --sundar On Mon, Mar 16, 2009 at 3:04 PM, Etienne Bellemare Racine etienn...@gmail.com wrote: I would like to get the name of the user form the system. Is it possible ? Something like system.user() returning something like [1] etber12 Thanks, Etienne [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] Get user system name
If that's the case, it's best to post the output from version to this thread. Type version from the R command prompt and paste the results here. On Mon, Mar 16, 2009 at 3:14 PM, Etienne Bellemare Racine etienn...@gmail.com wrote: It seems like it is not defined. I get an empty string. Is there a workaround or another solution ? -- Etienne Sundar Dorai-Raj a écrit : Assuming USER is defined on your system then Sys.getenv(USER) ought to work. --sundar On Mon, Mar 16, 2009 at 3:04 PM, Etienne Bellemare Racine etienn...@gmail.com wrote: I would like to get the name of the user form the system. Is it possible ? Something like system.user() returning something like [1] etber12 Thanks, Etienne [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] Matrix Construction; Subdiagonal
Does this help? A - matrix(0, 6, 6) vec - 1:5 A[row(A) == col(A) + 1] - vec --sundar On Wed, Mar 11, 2009 at 4:42 PM, Stu Field s...@colostate.edu wrote: I'm trying to enter a vector into the subdiagonal of a matrix but cannot find a command in R which corresponds to the MatLab version of diag(vec, k), where vec = the vector of interest, and k = the diagonal (k=0 for the diagonal; k=-1 for the subdiagonal; k=1 for superdiagonal, etc.) Is there an equivalent command in R? I'm looking for something like this: vec = seq(1, 5, 1) # vector of interest A = xyz(vec,-1) # creates a 6x6 matrix with vec on the subdiagonal where xyz is some function similar to diag, but with differing arguments. I can't believe there is not a simple way to do this... Thanks for your help, ~~ Stu Field, PhD Postdoctoral Fellow Department of Biology Colorado State University 1878 Campus Delivery Fort Collins, CO 80523-1878 Office: E208 Anatomy/Zoology Phone: (970) 491-5744 ~~ [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] Matrix Construction; Subdiagonal
You can always write your own function:
myDiag - function(x, vec, k) {
x[row(x) == col(x) - k] - vec
x
}
myDiag(A, vec, -1)
Of course, you should probably do some input checking too.
--sundar
On Wed, Mar 11, 2009 at 4:57 PM, Stu Field s...@colostate.edu wrote:
Sure, that'll work fine, thanks.
But I guess I was looking for something more similar to MatLab, I'm really
surprised R doesn't have a preset command for this (?)
Thanks again,
Stu
On 11 • Mar • 2009, at 5:49 PM, Sundar Dorai-Raj wrote:
Does this help?
A - matrix(0, 6, 6)
vec - 1:5
A[row(A) == col(A) + 1] - vec
--sundar
On Wed, Mar 11, 2009 at 4:42 PM, Stu Field s...@colostate.edu wrote:
I'm trying to enter a vector into the subdiagonal of a matrix but
cannot find a command in R which corresponds to the MatLab version of
diag(vec, k), where vec = the vector of interest, and k = the diagonal
(k=0 for the diagonal; k=-1 for the subdiagonal; k=1 for
superdiagonal, etc.)
Is there an equivalent command in R?
I'm looking for something like this:
vec = seq(1, 5, 1) # vector of interest
A = xyz(vec,-1) # creates a 6x6 matrix with vec on the
subdiagonal
where xyz is some function similar to diag, but with differing
arguments.
I can't believe there is not a simple way to do this...
Thanks for your help,
~~
Stu Field, PhD
Postdoctoral Fellow
Department of Biology
Colorado State University
1878 Campus Delivery
Fort Collins, CO 80523-1878
Office: E208 Anatomy/Zoology
Phone: (970) 491-5744
~~
[[alternative HTML version deleted]]
__
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.
~~
Stu Field, PhD
Postdoctoral Fellow
Department of Biology
Colorado State University
1878 Campus Delivery
Fort Collins, CO 80523-1878
Office: E208 Anatomy/Zoology
Phone: (970) 491-5744
~~
__
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Re: [R] Lattice: Customizing point-sizes with groups
Try this: xyplot(y ~ x, temp, groups = groups, par.settings = list( superpose.symbol = list( cex = c(1, 3), pch = 19, col = c(blue, red See: str(trellis.par.get()) for other settings you might want to change. Also, you should drop the ; from all your scripts. HTH, --sundar On Mon, Mar 9, 2009 at 6:49 PM, Paul Boutros paul.bout...@utoronto.ca wrote: Hello, I am creating a scatter-plot in lattice, and I would like to customize the size of each point so that some points are larger and others smaller. Here's a toy example: library(lattice); temp - data.frame( x = 1:10, y = 1:10, cex = rep( c(1,3), 5), groups = c( rep(A, 5), rep(B, 5) ) ); xyplot(y ~ x, temp, cex = temp$cex, pch = 19); This works just fine if I create a straight xy-plot, without groups. However when I introduce groupings the cex argument specifies the point-size for the entire group. For example: xyplot(y ~ x, temp, cex = temp$cex, pch = 19, group = groups); Is it possible to combine per-spot sizing with groups in some way? One work-around is to manually specify all graphical parameters, but I thought there might be a better way than this: temp$col - rep(blue, 10); temp$col[temp$groups == B] - red; xyplot(y ~ x, temp, cex = temp$cex, pch = 19, col = temp$col); Any suggestions/advice is much appreciated! Paul __ 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. __ 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.
Re: [R] Lattice: Customizing point-sizes with groups
Sorry, I missed your point the first time. Why not create a group for each subset then? xyplot(y ~ x, temp, groups = interaction(cex, groups), par.settings = list( superpose.symbol = list( cex = c(1, 2, 3, 4), pch = 19, col = c(blue, red, green, purple On Tue, Mar 10, 2009 at 8:11 AM, Paul C. Boutros paul.bout...@utoronto.ca wrote: Hi Sundar, Thanks for your help! Unfortunately your code seems to give the same result. Compare this: temp - data.frame( x = 1:10, y = 1:10, cex = rep( c(1,3), 5), col = c( rep(blue, 5), rep(red, 5) ), groups = c( rep(A, 5), rep(B, 5) ) ); xyplot(y ~ x, temp, groups = groups, par.settings = list( superpose.symbol = list( cex = c(1, 3), pch = 19, col = c(blue, red And this: xyplot(y ~ x, temp, cex = temp$cex, col = temp$col, pch = 19); Once I introduce groups, I lose the ability to customize individual data-points and seem only to be able to customize entire groups. Paul -Original Message- From: Sundar Dorai-Raj [mailto:sdorai...@gmail.com] Sent: Tuesday, March 10, 2009 5:49 AM To: paul.bout...@utoronto.ca Cc: r-help@r-project.org Subject: Re: [R] Lattice: Customizing point-sizes with groups Try this: xyplot(y ~ x, temp, groups = groups, par.settings = list( superpose.symbol = list( cex = c(1, 3), pch = 19, col = c(blue, red See: str(trellis.par.get()) for other settings you might want to change. Also, you should drop the ; from all your scripts. HTH, --sundar On Mon, Mar 9, 2009 at 6:49 PM, Paul Boutros paul.bout...@utoronto.ca wrote: Hello, I am creating a scatter-plot in lattice, and I would like to customize the size of each point so that some points are larger and others smaller. Here's a toy example: library(lattice); temp - data.frame( x = 1:10, y = 1:10, cex = rep( c(1,3), 5), groups = c( rep(A, 5), rep(B, 5) ) ); xyplot(y ~ x, temp, cex = temp$cex, pch = 19); This works just fine if I create a straight xy-plot, without groups. However when I introduce groupings the cex argument specifies the point-size for the entire group. For example: xyplot(y ~ x, temp, cex = temp$cex, pch = 19, group = groups); Is it possible to combine per-spot sizing with groups in some way? One work-around is to manually specify all graphical parameters, but I thought there might be a better way than this: temp$col - rep(blue, 10); temp$col[temp$groups == B] - red; xyplot(y ~ x, temp, cex = temp$cex, pch = 19, col = temp$col); Any suggestions/advice is much appreciated! Paul __ 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. __ 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.
Re: [R] 2 Simple Lattice Plot Questions
Convert Year to a factor and both problems will be solved. --sundar On Tue, Mar 10, 2009 at 3:48 PM, jimdare jamesdar...@gmail.com wrote: Hi, I have created the plot below and have a few questions about changes. 1) How do I change the Year title of each plot so it reads from the top 2006,2007,2008,2009. 2) How do I get rid of those vertical grey bars in the title bar of each plot? I apologise for my ignorance... one of those days :( James http://www.nabble.com/file/p22445242/PLOT.jpg -- View this message in context: http://www.nabble.com/2-Simple-Lattice-Plot-Questions-tp22445242p22445242.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] 2 Simple Lattice Plot Questions
I don't believe Elena's suggestion will work. However, the following will: xyplot(..., scales = list(y = list(at = seq(5, 25, 5 though you may need to extend the limits a little as well: xyplot(..., ylim = lattice:::extend.limits(c(0, 30))) and add the scales argument from the first example to place explicit tick marks rather than let xyplot do so. HTH, --sundar On Tue, Mar 10, 2009 at 7:04 PM, Elena Wilson ewil...@dbmcons.com.au wrote: Have you tried specifying the levels of y's you want to display, e.g. ylim=c(0,5,10,15,20,30)? -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of jimdare Sent: Wednesday, 11 March 2009 12:50 PM To: r-help@r-project.org Subject: Re: [R] 2 Simple Lattice Plot Questions Thanks very much. One other thing... see how the Y axis begins before 0. At first glance it appears that the 0's are actually worth something. When I use ylim=c(0,30) I get the graph I want, but the tick marks show only 5,10,15,20,25. I want them to show 0,5,10,15,20,30. Does anyone know how to change this setting? Cheers, jimdare wrote: Hi, I have created the plot below and have a few questions about changes. 1) How do I change the Year title of each plot so it reads from the top 2006,2007,2008,2009. 2) How do I get rid of those vertical grey bars in the title bar of each plot? I apologise for my ignorance... one of those days :( James http://www.nabble.com/file/p22445242/PLOT.jpg -- View this message in context: http://www.nabble.com/2-Simple-Lattice-Plot-Questions-tp22445242p22447415.html Sent from the R help mailing list archive at Nabble.com. __ 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. -- Message protected by MailGuard: e-mail anti-virus, anti-spam and content filtering. http://www.mailguard.com.au/mg https://login.mailguard.com.au/report/1x1TEaABIw/4soNUVnBh04s1coMEHC4LA/7.998 __ 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. __ 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.
Re: [R] Date conversion
Hi, There are possibly several ways to do this. My approach would be: dates - strptime(as.character(DATE), %d%b%Y) year - dates$year + 1900 week - floor(dates$yday/365 * 52) HTH, --sundar On Thu, Mar 5, 2009 at 8:58 AM, Pele drdi...@yahoo.com wrote: Hi R users, I have a factor variable called date as shown below: Can anyone share the best / most efficient way to extract year and week (e.g. year = 2006, week = 52 for first record, etc..)? My data set has 1 million records. DATE 11DEC2006 11SEP2006 01APR2007 02DEC2007 Thanks in advance for any help! -- View this message in context: http://www.nabble.com/Date-conversion-tp22355788p22355788.html Sent from the R help mailing list archive at Nabble.com. __ 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. __ 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.
Re: [R] lattice: remove box around a wireframe
(Sorry for the repeat. Forgot to copy R-help) Try, test = data.frame(expand.grid(c(1:10), c(1:10))) z = test[,1] + test[,2] test = cbind(test, z) names(test) = c(x, y, z) require(lattice) wireframe(z ~ x*y, data = test, par.settings = list(axis.line = list(col = transparent)), par.box = c(col = transparent) ) --sundar On Wed, Mar 4, 2009 at 8:17 AM, Thomas Roth (geb. Kaliwe) hamstersqu...@web.de wrote: #Hi, # #somebody knows how to remove the outer box around a wireframe and reduce the height # # test = data.frame(expand.grid(c(1:10), c(1:10))) z = test[,1] + test[,2] test = cbind(test, z) names(test) = c(x, y, z) require(lattice) wireframe(z ~ x*y, data = test, par.box = c(col = transparent) ) #not this one but the remaining outer box. Thanks in advance Thomas Roth __ 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. __ 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.
Re: [R] levelplot help needed
To reorder the y-labels, simply reorder the factor levels:
df - data.frame(x_label = factor(x_label),
y_label = factor(y_label, rev(y_label)),
values = as.vector(my.data))
Not sure about putting the strips at the bottom. A quick scan of
?xyplot and ?strip.default suggests that this is not possible, but I'm
sure Deepayan will correct me if I'm wrong (he often does).
--sundar
On Fri, Feb 27, 2009 at 5:51 AM, Antje niederlein-rs...@yahoo.de wrote:
Hi there,
I'm looking for someone who can give me some hints how to make a nice
levelplot. As an example, I have the following code:
# create some example data
# --
xl - 4
yl - 10
my.data - sapply(1:xl, FUN = function(x) { rnorm( yl, mean = x) })
x_label - rep(c(X Label 1, X Label 2, X Label 3, X Label 4), each =
yl)
y_label - rep(paste(Y Label , 1:yl, sep=), xl)
df - data.frame(x_label = factor(x_label),y_label = factor(y_label), values
= as.vector(my.data))
df1 - data.frame(df, group = rep(Group 1, xl*yl))
df2 - data.frame(df, group = rep(Group 2, xl*yl))
df3 - data.frame(df, group = rep(Group 3, xl*yl))
mdf - rbind(df1,df2,df3)
# plot
# --
graph - levelplot(mdf$values ~ mdf$x_label * mdf$y_label | mdf$group,
aspect = xy, layout = c(3,1),
scales = list(x = list(labels =
substr(levels(factor(mdf$x_label)),0,5), rot = 45)))
print(graph)
# --
(I need to put this strange x-labels, because in my real data the values of
the x-labels are too long and I just want to display the first 10 characters
as label)
My questions:
* I'd like to start with Y Label 1 in the upper row (that's a more general
issue, how can I have influence on the order of x,y, and groups?)
* I'd like to put the groups at the bottom
Can anybody give me some help?
Antje
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Re: [R] lattice contourplot line types
The only way I can figure out to do this is to use two calls to
panel.contourplot:
library(lattice)
x - seq(-2, 2, length = 20)
y - seq(-2, 2, length = 20)
grid - expand.grid(x=x, y=y)
grid$z - dnorm(grid$x) * dnorm(grid$y)
contourplot(z ~ x * y, grid,
panel = function(at, lty, col, ...) {
at.o - at[seq(1, length(at), 2)]
at.e - at[seq(2, length(at), 2)]
panel.contourplot(at = at.o, lty = 1, col = blue, ...)
panel.contourplot(at = at.e, lty = 4, col = red, ...)
},
at = pretty(grid$z, 10))
HTH,
--sundar
On Sun, Feb 22, 2009 at 12:45 PM, Andrew Beckerman
a.becker...@sheffield.ac.uk wrote:
Dear all -
I would like to adjust the line type of specific contours in contourplot
from the lattice package, but it seems like lty does not take a list in the
call.
Here is my call to contourplot:
contourplot(preds~size+trt|Size.Name,
data=pred.dat,layout=c(2,4),
at=c(0.025,0.5,0.975),
par.strip.text=list(cex=1.2),
scales=list(cex=0.5),
xlab=list(Size,cex=1.2),
ylab=list(Treatment,cex=1.2),
panel=function(x,y,z,...){
panel.contourplot(x,y,z,lwd=1,...)
panel.grid(h=-1,v=-1,col=grey,...)})
I would like to specify lty=c(2,1,2) corresponding to the
at=c(0.025,0.5,0.975), and have tried this in both the core part of the
call, and in panel.countourplot. However, it only recognises the first
type.
If there is no straightforward answer, I can provide the data.
Best wishes,
Andrew
R version 2.8.1 (2008-12-22)
i386-apple-darwin8.11.1
locale:
en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] MASS_7.2-45 lattice_0.17-17
loaded via a namespace (and not attached):
[1] grid_2.8.1
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Re: [R] Uninstall question
This is on the Mac FAQ: http://cran.cnr.berkeley.edu/bin/macosx/RMacOSX-FAQ.html#How-can-R-for-Mac-OS-X-be-uninstalled_003f HTH, --sundar On Tue, Feb 17, 2009 at 7:17 AM, ANJAN PURKAYASTHA anjan.purkayas...@gmail.com wrote: I need to uninstall R 2.7.1 from my Mac. What is the best way to uninstall it? Simply delete the R icon in the Applications folder? Or is it more involved? TIA, Anjan -- = anjan purkayastha, phd bioinformatics analyst whitehead institute for biomedical research nine cambridge center cambridge, ma 02142 purkayas [at] wi [dot] mit [dot] edu 703.740.6939 [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] Adding abline in Lattice graph
Try:
coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
panel= function(...) {
panel.smooth(...)
panel.abline(h = 126, col = red)
panel.abline(v = 6.5, col = blue)
},
xlab=ABC, ylab=FBG)
Also note that you removed your with call and give coplot a data argument.
HTH,
--sundar
On Thu, Feb 12, 2009 at 3:41 PM, Veerappa Chetty chett...@gmail.com wrote:
Hi,I would like add a horizontal line at 126 (col=red) and a vertical line
at 6.5 ( col= blue) in each panel .How should I use the panel.abline
function in the following code I am using:
--
library(lattice)
with(reg.dat.5,coplot(lbxglu~lbxgh|eth,panel=panel.smooth,xlab=ABC,
ylab=FBG))
Thanks a lot.
Professor of Family Medicine
Boston University
Tel: 617-414-6221, Fax:617-414-3345
emails: chett...@gmail.com,vche...@bu.edu
[[alternative HTML version deleted]]
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Re: [R] Adding abline in Lattice graph
Sorry, that was my lack of understanding on how coplot works. Try the following:
coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
panel= function(...) {
panel.smooth(...)
abline(h = 126, col = red)
abline(v = 6.5, col = blue)
},
xlab=ABC, ylab=FBG)
On Thu, Feb 12, 2009 at 3:49 PM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Try:
coplot(lbxglu~lbxgh|eth, data = reg.dat.5,
panel= function(...) {
panel.smooth(...)
panel.abline(h = 126, col = red)
panel.abline(v = 6.5, col = blue)
},
xlab=ABC, ylab=FBG)
Also note that you removed your with call and give coplot a data argument.
HTH,
--sundar
On Thu, Feb 12, 2009 at 3:41 PM, Veerappa Chetty chett...@gmail.com wrote:
Hi,I would like add a horizontal line at 126 (col=red) and a vertical line
at 6.5 ( col= blue) in each panel .How should I use the panel.abline
function in the following code I am using:
--
library(lattice)
with(reg.dat.5,coplot(lbxglu~lbxgh|eth,panel=panel.smooth,xlab=ABC,
ylab=FBG))
Thanks a lot.
Professor of Family Medicine
Boston University
Tel: 617-414-6221, Fax:617-414-3345
emails: chett...@gmail.com,vche...@bu.edu
[[alternative HTML version deleted]]
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Re: [R] changing settings on a barchart (lattice)
Add the adj argument to panel.text to left (adj = 0) or right(adj =
1) justify the text. Add the font argument to change the font. See
?text.
--sundar
On Wed, Feb 11, 2009 at 1:25 PM, Dimitri Liakhovitski ld7...@gmail.com wrote:
Thanks a lot, Sundar. I experimented somewhat and here is the code
that works well - it allows me to modify most of the stuff I want to
modify:
p-as.vector(c(0.1, 0.2, 0.3, 0.4))
names(p)-c(A,BB,,)
barchart(~sort(p), main=list(Chart Title,cex=1),xlab=list(X axis
title,cex=1),xlim=c(0,0.42),
layout = c(1,1),
stack = TRUE,
auto.key = list(points = FALSE, rectangles = TRUE, space =
top),
panel = function(y,x,...){
panel.grid(h = 0, v = -1, col = gray60, lty
=dotted)
panel.barchart(x,y,col=brown)
panel.text(x,y,label = round(x,2),cex=1)
}
)
One last question: How can I modify the way the value labels (those
that are at the end of the bars) appear? Can I make them bold? Make
them appear a bit to the right or to the left of where they currently
are?
Thanks a lot!
Dimitri
On Wed, Feb 11, 2009 at 12:46 PM, Sundar Dorai-Raj sdorai...@gmail.com
wrote:
Pass a list to xlab and main for the font sizes:
barchart(..., xlab = list(x-axis, cex = 2), main = list(title, cex = 2))
For value labels and a grid you'll need a custom panel function:
barchart(..., panel = function(x, y, ...) {
panel.barchart(x, y, ...)
panel.text(x, y, format(y), cex = 1.2)
panel.grid(h = -1, v = -1)
})
This is untested, but I think it should get you started.
--sundar
On Wed, Feb 11, 2009 at 9:10 AM, Dimitri Liakhovitski ld7...@gmail.com
wrote:
Hello!
I apologize - I never used lattice before, so my question is probably
very basic - but I just can't find the answer in the archive nor in
the documentation:
I have a named numeric vector p of 6 numbers (of the type 6 numbers
with people's names to whom those numbers belong). I want a simple bar
chart.
I am doing:
library(lattice)
trellis.par.set(fontsize=list(text=12)) # Changes only axes font
barchart(~sort(p),main=Text for main,xlab=Text for X axis,
col=PrimaryColors[3])
It works just fine.
Question: Where how can I change such things as font size for X axis
label (below the numbers), font size for Title, value labels (to label
bars with actual numbers), add grids, etc.?
Thanks a lot!
--
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com
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--
Dimitri Liakhovitski
MarketTools, Inc.
dimitri.liakhovit...@markettools.com
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Re: [R] Efficent way to create an nxn upper triangular matrix of one's
Try
x - diag(n)
x[upper.tri(x)] - 1
On Wed, Feb 11, 2009 at 1:22 PM, Dale Steele dale.w.ste...@gmail.com wrote:
The code below create an nxn upper triangular matrix of one's. I'm
stuck on finding a more efficient vectorized way - Thanks. --Dale
n - 9
data - matrix(data=NA, nrow=n, ncol=n)
data
for (i in 1:n) {
data[,i] - c(rep(1,i), rep(0,n-i))
}
data
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Re: [R] sapply
I'm not sure what you really want, so perhaps a simple example would help (i.e. what a sample of the input looks like and what the output you need looks like). My guess would be sapply(df, diff) but again, I'm not sure. --sundar On Sun, Feb 8, 2009 at 4:24 PM, glenn g1enn.robe...@btinternet.com wrote: Newbie question sorry (have tried the help pages I promise) I have a dataframe (date,stockprice) say and looking how I might get the return of: dataframe (difference in days, change in stock price) using sapply - I require a very simple function and don't really want to go down the zoo and quant mod route Regards glenn [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] subsets problem
you can try lapply(lapply(uniques, function(x) subset(df, date == x)), myfun) or possibly more accurate (subset may be finicky due to scoping): lapply(lapply(uniques, function(x) df[df$date == x, ]), myfun) or use ?split lapply(split(df, df$date), myfun) HTH, --sundar On Sun, Feb 8, 2009 at 5:00 PM, glenn g1enn.robe...@btinternet.com wrote: Help with this much appreciated I have a large dataframe that I would like to subset where the constraint Test1 - subset(df, date == uniques[[1]]), where uniques is a list of dates that must be matched to create Test1. I would like to perform an operation on Test1 that results in a single column of data. So far so good. How do loop through all values in the uniques list (say there is 50), perform an operationon Test1Test50, and then bolt all the lists together in a single list please ? Regards Glenn [[alternative HTML version deleted]] __ 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. __ 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.
Re: [R] Foreign function call
You're missing that R_TSConv is an R object. You can use stats:::R_TSConv to see the value. Not sure how this helps you though. On Wed, Feb 4, 2009 at 10:08 AM, rkevinbur...@charter.net wrote: Let me get more specific. I think it this can be answered then I can translate the information to other calls. In the arima 'R' code there is a reference to .Call(R_TSconv, a, b) If from the console I type: .Call(R_TSConv, c(1,-1), c(1,-1)) I get: Error: object R_TSConv not found If I do getNativeSymbolInfo(R_TSConv) I get: Error in FUN(R_TSConv[[1L]], ...) : no such symbol R_TSConv What am I missing? Thank you. Kevin __ 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. __ 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.
Re: [R] color and fontfamily in lattice
Try this:
dados - data.frame(varsep = factor(rep(1:2,10)),
i = runif(20))
library(lattice)
font.settings - list(
font = 2,
cex = 2,
fontfamily = serif)
my.theme - list(
box.umbrella = list(col = red),
box.rectangle = list(col = purple),
box.dot = list(col = blue, pch = 15),
par.xlab.text = font.settings,
par.ylab.text = font.settings,
axis.text = font.settings)
bwplot(varsep ~ i, dados,
xlab = names(dados)[1],
ylab = names(dados)[2],
panel = function(...) {
panel.grid(v = -1, h = 0)
panel.bwplot(...)
},
par.settings = my.theme)
Type trellis.par.get() at the R command line to see other parameters
you can change. To change the settings on all plots, you can remove
the par.settings from the call to bwplot and simply use:
trellis.par.set(theme = my.theme)
HTH,
--sundar
2009/2/3 Leandro Marino lean...@cesgranrio.org.br:
Hi,
I am having some problems using bwplot(lattice) in my data. I want change
some parameters:
1) Fontfamily to serif
2) The size of the font
3) Put it in a bold face
4) Change de color of the lines
How can I do that?! Now, I am using this to plot my boxplot.
dados - data.frame(varsep=as.factor(rep(1:2,10)),i=runif(20))
bwplot(dados[,'varsep']~dados[,'i'],xlab=names(dados)[2],ylab=names(dados)[1],panel
=function(...){panel.grid(v = -1, h =
0);panel.bwplot(...)},font=2,fontfamily='serif')
Thanks for any help on advance and sorry about my English.
Atenciosamente,
Leandro Lins Marino
Centro de Avaliação
Fundação CESGRANRIO
Rua Santa Alexandrina, 1011 - 2º andar
Rio de Janeiro, RJ - CEP: 20261-903
R (21) 2103-9600 R.:236
0 (21) 8777-7907
( lean...@cesgranrio.org.br
Aquele que suporta o peso da sociedade
é precisamente aquele que obtém
as menores vantagens. (SMITH, Adam)
Antes de imprimir pense em sua responsabilidade e compromisso com o MEIO
AMBIENTE
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Re: [R] xyplot with lowess curves
You'll need a custom panel function. It would also help if you
provided a reproducible example:
xyplot (
SnowLineElevation ~ Year | Model,
data = data,
panel = function(x, y, col, ...) {
col - ifelse(panel.number() == 1, red, green)
panel.xyplot(x, y, col = blue, ...)
panel.loess(x, y, col = col)
},
ylim = c(0,1800),
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
Alternatively, you can use the group argument in conjunction with the panels:
xyplot(SnowLineElevation ~ Year | Model, data, groups = Model, type =
c(p, smooth))
if you want the points and the lines to be the same color.
--sundar
On Mon, Feb 2, 2009 at 10:20 AM, Hutchinson,David [PYR]
david.hutchin...@ec.gc.ca wrote:
I am trying to change the attributes of the lowess lines fit to an
xyplot command, but have been unsuccessful in my search of the online
help. Right now, both the points and lowess line come out in the same
color (blue). I am unsure how I can change the properties of the lowess
line separately.
xyplot (
SnowLineElevation ~ Year | Model,
data = data,
ylim = c(0,1800),
type = c('p','smooth'),
col = 'blue',
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
Any help would be much appreciated,
Dave
[[alternative HTML version deleted]]
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Re: [R] xyplot with lowess curves
Does this do what you want? The panel argument has the custom pane
function I referred to before.
Col - c(red, green, blue, purple)
xyplot (
SnowLineElevation ~ Year | Model,
data = d,
panel = function(x, y, col, ...) {
Col - Col[panel.number()]
panel.xyplot(x, y, col = blue, ...)
panel.loess(x, y, col = Col)
},
ylim = c(0,100),
type = c('p','smooth'),
col = 'blue',
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
On Mon, Feb 2, 2009 at 12:18 PM, Hutchinson,David [PYR]
david.hutchin...@ec.gc.ca wrote:
I haven't had much luck with a custom panel function; mainly because I
don't truly understand how to embedd the functionality into the xyplot
command.
Here's a reproducible example if you can help out.
Thanks,
Dave
library (lattice)
d - NULL
models - c('A','B','C','D')
n = 100
for (i in seq(along = models)){
d - rbind(
d, data.frame (
Model = models[i],
Year = seq(1960, length.out=n, by = 1),
SnowLineElevation = runif(n, 0, 100)
)
)
}
xyplot (
SnowLineElevation ~ Year | Model,
data = d,
ylim = c(0,100),
type = c('p','smooth'),
col = 'blue',
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
-Original Message-
From: Sundar Dorai-Raj [mailto:sdorai...@gmail.com]
Sent: Monday, February 02, 2009 11:43 AM
To: Hutchinson,David [PYR]
Cc: r-help@r-project.org
Subject: Re: [R] xyplot with lowess curves
You'll need a custom panel function. It would also help if you provided
a reproducible example:
xyplot (
SnowLineElevation ~ Year | Model,
data = data,
panel = function(x, y, col, ...) {
col - ifelse(panel.number() == 1, red, green)
panel.xyplot(x, y, col = blue, ...)
panel.loess(x, y, col = col)
},
ylim = c(0,1800),
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
Alternatively, you can use the group argument in conjunction with the
panels:
xyplot(SnowLineElevation ~ Year | Model, data, groups = Model, type =
c(p, smooth))
if you want the points and the lines to be the same color.
--sundar
On Mon, Feb 2, 2009 at 10:20 AM, Hutchinson,David [PYR]
david.hutchin...@ec.gc.ca wrote:
I am trying to change the attributes of the lowess lines fit to an
xyplot command, but have been unsuccessful in my search of the online
help. Right now, both the points and lowess line come out in the same
color (blue). I am unsure how I can change the properties of the
lowess line separately.
xyplot (
SnowLineElevation ~ Year | Model,
data = data,
ylim = c(0,1800),
type = c('p','smooth'),
col = 'blue',
pch = 21,
xlab = 'Year',
ylab = 'Snowline Elevation [m]'
)
Any help would be much appreciated,
Dave
[[alternative HTML version deleted]]
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Re: [R] Assigning colnames in loop
It's always best to do this with list operations (e.g. lapply) rather
than a loop:
DF1 - split(DF, DF$male)
DF2 - lapply(DF1, function(x) {
x2 - t(as.matrix(x[3:5], dimnames = list(levels(x$age), NULL)))
as.data.frame(x2)
})
Then DF2[[0]] and DF2[[1]] are the data.frames you want.
HTH,
--sundar
On Mon, Feb 2, 2009 at 2:30 PM, Peter Jepsen p...@dce.au.dk wrote:
Dear R-listers,
I am trying to assign colnames to a data frame within a loop, but I keep
getting a target of assignment expands to non-language object-error. I
need to split up a large dataset into about 20 smaller ones, and I would
like to assign colnames within the loop, so I won't have to type the
same thing 20 times over.
I have concocted this really goofy example which constructs two
datasets:
-
male - rep(0:1, each=5)
age - factor(c(10:14,10:14))
DF - data.frame(male, age, res1=rnorm(10), res2=rnorm(10),
res3=rnorm(10))
for(n in 0:1) {
assign(paste(test,n, sep=.), as.data.frame(t(subset(DF,
male==n, select=c(res1, res2, res3)
colnames(get(paste(test,n, sep=.))) -
paste(age,levels(age), m, n, sep=) # This line gives an error.
assign(colnames(paste(test,n, sep=.))) -
paste(age,levels(age), m, n, sep=) # This line gives the same
error.
}
---
The following command assigns the right colnames to the 'test.0' data
frame, but I want this line inside the loop so I won't have to type it
20 times over.
colnames(test.0) - paste(age,levels(age), m, 0, sep=)
Thank you in advance for any assistance.
Peter.
sessionInfo()
R version 2.8.1 (2008-12-22)
i386-pc-mingw32
locale:
LC_COLLATE=Danish_Denmark.1252;LC_CTYPE=Danish_Denmark.1252;LC_MONETARY=
Danish_Denmark.1252;LC_NUMERIC=C;LC_TIME=Danish_Denmark.1252
attached base packages:
[1] tools stats graphics grDevices utils datasets methods
base
other attached packages:
[1] epiR_0.9-14 maptools_0.7-18 sp_0.9-29 foreign_0.8-30
chron_2.3-28
loaded via a namespace (and not attached):
[1] grid_2.8.1 lattice_0.17-20
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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.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] capturing stderr/stdout
Hi, Prof. Ripley,
Thanks for the reply. Mostly I want to capture output as it is written
to the stream. For example, I quite often do the following to view the
progress of a log file from a computationally intensive script.
1. Open a console and type:
Rscript script.R script.log
which directs both stdout and stderr to script.log
2. Open another console and type:
tail -f script.log
This way I get both the script's log file and its current progress.
I guess my question is: Is there a way to accomplish the tail -f
command in R?
Thanks,
--sundar
Prof Brian Ripley said the following on 11/20/2008 11:43 PM:
I am not sure what the issue is here. Do you want to capture both stderr
and stdout (use 21 in the command with an sh-like shell), or is the
problem that you don't get immediate output?
The latter is a Perl issue: you need to circumvent output buffering.
See e.g
http://perl.plover.com/FAQs/Buffering.html
Sundar Dorai-Raj wrote:
Hi,
I have an application in perl that prints some output to either stderr
or stdout.
Here's an example:
# tmp.pl
print STDERR starting iterator\n;
for(my $i = 0; $i 100; $i++) {
print $i . \n;
}
# tmp.R
con - pipe(perl tmp.pl)
r - readLines(con, n = -1)
close(con)
However, the second line stalls until the perl for-loop finishes. What
I would like is to process each line as it comes. E.g. something like:
while(TRUE) {
r - readLines(con, n = 1) # read one line
if(r == 1) print(r)
if(length(r) == 0) break
}
Of course, this won't work since I'm not calling readLines
appropriately. Answers must work on Windows but may include cygwin
utilities if necessary. Any advice would be appreciated. Version info
at the end if it matters.
Thanks, --sundar
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 8.0
year 2008
month 10
day20
svn rev46754
language R
version.string R version 2.8.0 (2008-10-20)
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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.
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R-help@r-project.org mailing list
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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] capturing stderr/stdout
Hi,
I have an application in perl that prints some output to either stderr
or stdout.
Here's an example:
# tmp.pl
print STDERR starting iterator\n;
for(my $i = 0; $i 100; $i++) {
print $i . \n;
}
# tmp.R
con - pipe(perl tmp.pl)
r - readLines(con, n = -1)
close(con)
However, the second line stalls until the perl for-loop finishes. What I
would like is to process each line as it comes. E.g. something like:
while(TRUE) {
r - readLines(con, n = 1) # read one line
if(r == 1) print(r)
if(length(r) == 0) break
}
Of course, this won't work since I'm not calling readLines
appropriately. Answers must work on Windows but may include cygwin
utilities if necessary. Any advice would be appreciated. Version info at
the end if it matters.
Thanks, --sundar
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 8.0
year 2008
month 10
day20
svn rev46754
language R
version.string R version 2.8.0 (2008-10-20)
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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.
Re: [R] determining plot location in lattice
Hi, Greg,
Thanks again for your reply. I looked at TkIdentify which lead me to
plt, which with RSiteSearch(grid plt) lead me to gridBase::gridPLT.
Here's my solution:
stopifnot(require(tkrplot),
require(lattice),
require(gridBase))
makePlot - function() {
print(xyplot(1 ~ 1))
if(click) {
trellis.focus(highlight = FALSE)
plt - gridPLT()
out - ((x plt[1] || x plt[2]) ||
(y plt[3] || y plt[4]))
cat(if(out) outside else inside, \n)
}
}
clickPlot - function(x, y) {
width - as.numeric(tclvalue(tkwinfo(reqwidth, img)))
height - as.numeric(tclvalue(tkwinfo(reqheight, img)))
x - as.numeric(x)/width
y - 1 - as.numeric(y)/height
click - TRUE
tkrreplot(img)
click - FALSE
}
x - y - -1
click - FALSE
tt - tktoplevel()
img - tkrplot(tt, makePlot)
tkbind(img, 1, clickPlot)
tkpack(img)
tkwm.resizable(tt, 0, 0)
Greg Snow said the following on 11/7/2008 8:23 PM:
As far as R is concerned, the plot in tkrplot is being written to a file and
therefore is not interactive. So functions like grid.locator and trellis.focus
are not going to work.
You need to use the Tk commands to determine where the mouse is and where a
click occurred, then this needs to be converted to the coordinates of the
trellis plot. There are some examples of the first part in the TeachingDemos
package, eg TkBrush, TkIdentify, TkApprox, and a couple of others (also the
play.sudoku function in the Sudoku package). These all work with base graphics
rather than trellis/grid so will not help with the 2nd part. Possibly
something in the grid package can tell you the coordinates of the various
viewports (but you need to save this information before the end of the plotting
command because the information will probably be lost when the plot finishes).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
project.org] On Behalf Of Sundar Dorai-Raj
Sent: Friday, November 07, 2008 12:43 PM
To: r-help@r-project.org
Subject: [R] determining plot location in lattice
Hi,
I'm dealing with a lattice plot inserted into a tk widget and would
like
to know when a user has clicked on the plot area of a plot (i.e. inside
the axes). For example,
library(tkrplot)
library(lattice)
tt - tktoplevel()
makePlot - function() print(xyplot(1 ~ 1))
printCoords - function(x, y) print(c(x, y))
img - tkrplot(tt, makePlot)
tkbind(img, 1, printCoords)
tkpack(img)
I would like to know when a user clicks inside the axes, but don't know
how to determine where the plot region is. Essentially, this comes down
to answering the following question: Assuming the entire plot is on a
unit square, what are the coordinates of the plotting area?
This question is may seem unrelated to tkrplot, but outside of this
context I may not get answers that work. I have been playing with both
lattice::trellis.focus and grid::grid.locator to no avail.
Thanks,
--sundar
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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] determining plot location in lattice
Hi, I'm dealing with a lattice plot inserted into a tk widget and would like to know when a user has clicked on the plot area of a plot (i.e. inside the axes). For example, library(tkrplot) library(lattice) tt - tktoplevel() makePlot - function() print(xyplot(1 ~ 1)) printCoords - function(x, y) print(c(x, y)) img - tkrplot(tt, makePlot) tkbind(img, 1, printCoords) tkpack(img) I would like to know when a user clicks inside the axes, but don't know how to determine where the plot region is. Essentially, this comes down to answering the following question: Assuming the entire plot is on a unit square, what are the coordinates of the plotting area? This question is may seem unrelated to tkrplot, but outside of this context I may not get answers that work. I have been playing with both lattice::trellis.focus and grid::grid.locator to no avail. Thanks, --sundar __ 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.
Re: [R] tklistbox selection
I solved this problem by adding exportselection = 0 to the call to tklistbox. I.e. tb1 - tklistbox(tt, listvariable = tcl1, exportselection = 0, selectmode = multiple) Thanks, --sundar Sundar Dorai-Raj said the following on 10/29/2008 5:56 PM: Hi, I'm posting yet another question about tcltk since I'm still struggling with the package. I'm trying to create a tklistbox and a ttkcombobox on the same parent and am having a problem. Here's an example: library(tcltk) tt - tktoplevel() tcl1 - tclVar() tcl2 - tclVar() tclObj(tcl1) - letters[1:5] tclObj(tcl2) - LETTERS[1] tb1 - tklistbox(tt, listvariable = tcl1, selectmode = multiple) tb2 - ttkcombobox(tt, values = LETTERS[1:2], textvariable = tcl2) tkpack(tb1, tb2) First, I select some values in the list box. But when I select a value from the combo box, the selection from the list box is no longer highlighted. Is this a bug or am I missing something in the documentation? My goal is to allow highlighted text simultaneously in both widgets. Thanks, --sundar __ 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. __ 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.
[R] tklistbox selection
Hi, I'm posting yet another question about tcltk since I'm still struggling with the package. I'm trying to create a tklistbox and a ttkcombobox on the same parent and am having a problem. Here's an example: library(tcltk) tt - tktoplevel() tcl1 - tclVar() tcl2 - tclVar() tclObj(tcl1) - letters[1:5] tclObj(tcl2) - LETTERS[1] tb1 - tklistbox(tt, listvariable = tcl1, selectmode = multiple) tb2 - ttkcombobox(tt, values = LETTERS[1:2], textvariable = tcl2) tkpack(tb1, tb2) First, I select some values in the list box. But when I select a value from the combo box, the selection from the list box is no longer highlighted. Is this a bug or am I missing something in the documentation? My goal is to allow highlighted text simultaneously in both widgets. Thanks, --sundar __ 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.
[R] ttkcombobox
Hi, all, (sessionInfo at the end) I've been struggling with the tcltk package and can't seem to get the ttkcombobox to work. Here's an example: library(tcltk) p - tktoplevel() l - tclVar() ## I don't know if I'm even calling it correctly cb - ttkcombobox(p, values = letters[1:4], textvariable = l) tkpack(cb) 1. How do I know when the value of the combobox has been changed from, say, a to b? 2. How can I get the value of the box? E.g. if I switch the box to b, how can I access this value? 3. How can I set the default value to something of my choice? I've tried the set argument as described in the Tcl/Tk 8.5 Manual but I received an error: cb - ttkcombobox(p, values = letters[1:4], textvariable = l, set = b) Error in structure(.External(dotTclObjv, objv, PACKAGE = tcltk), class = tclObj) : [tcl] unknown option -set. Thanks, --sundar sessionInfo() R version 2.7.2 (2008-08-25) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods [8] base __ 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.
Re: [R] ttkcombobox
Hi, Greg,
Yes, that helps immensely!
Thanks,
--sundar
Greg Snow said the following on 10/27/2008 7:14 PM:
Here is one example of using ttkcombobox, hopefully it helps:
library(tcltk)
have.ttk - function () { # from Prof. Ripley
as.character(tcl(info, tclversion)) = 8.5
}
testfunc - function() {
tt - tktoplevel()
if( !have.ttk() ) stop( Need version 8.5 or greater of tcl )
tkpack(tklabel(tt, text='Choose a package'), side='top')
tmp - tclVar()
tclvalue(tmp) - 'TeachingDemos'
tkpack(ttkcombobox( tt, values=c('blockrand',
'ObsSens','TeachingDemos','sudoku'), textvariable=tmp ),
side='left' )
tkpack(tkbutton(tt,text='print',command=function()
print(tclvalue(tmp
tkpack(tkbutton(tt,text='exit', command=function()
tkdestroy(tt)),side='right')
}
testfunc()
If you want something to happen when you change the combobox (rather than just
getting the value when you press a button or something else), then you will
probably have to use tkbind to bind an event to it, or use tkafter to check it
on a regular basis. I usually do as above and just have a button or something
else run the code when I want and it just gets the current value.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
project.org] On Behalf Of Sundar Dorai-Raj
Sent: Monday, October 27, 2008 3:02 PM
To: r-help@r-project.org
Subject: [R] ttkcombobox
Hi, all,
(sessionInfo at the end)
I've been struggling with the tcltk package and can't seem to get the
ttkcombobox to work. Here's an example:
library(tcltk)
p - tktoplevel()
l - tclVar()
## I don't know if I'm even calling it correctly
cb - ttkcombobox(p, values = letters[1:4], textvariable = l)
tkpack(cb)
1. How do I know when the value of the combobox has been changed from,
say, a to b?
2. How can I get the value of the box? E.g. if I switch the box to b,
how can I access this value?
3. How can I set the default value to something of my choice? I've
tried
the set argument as described in the Tcl/Tk 8.5 Manual but I
received an error:
cb - ttkcombobox(p, values = letters[1:4], textvariable = l, set =
b)
Error in structure(.External(dotTclObjv, objv, PACKAGE = tcltk),
class = tclObj) :
[tcl] unknown option -set.
Thanks,
--sundar
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] tcltk stats graphics grDevices utils datasets methods
[8] base
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] using bquote to construct function
Thanks, Gabor! This workaround fixes my immediate needs.
--sundar
Gabor Grothendieck said the following on 10/1/2008 10:42 PM:
That may be a bug in R but I think there is another problem on top of that
as I don't think bquote descends into function bodies:
z - 2
bquote(function(x) {x^.(z)})
function(x) {x^.(z)}
bquote(function(x, y) { x^.(z) + y})
function(x, y) { x^.(z) + y}
R.version.string # Vista
[1] R version 2.7.2 (2008-08-25)
Try it this way:
z - 2
f - function(x, y) {}
body(f) - bquote({ x^.(z) + y })
eval(f)(2, 3)
On Thu, Oct 2, 2008 at 1:14 AM, Sundar Dorai-Raj
[EMAIL PROTECTED] wrote:
Hi, R-help,
(sessionInfo at the end)
I'm trying to construct a function using bquote and running into a strange
error message. As an example, what I would like to do is this:
z - 2
eval(bquote(function(x, y) { x^.(z) + y }))(2, 3)
However, I get the following:
Error in eval(expr, envir, enclos) :
invalid formal argument list for function
However, if I change the command to following, it works:
z - 2
eval(bquote(function(x) { x^.(z) }))(2)
# [1] 4
In other words, I remove the second argument. Is there a workaround for this
without using eval(parse(text = ))?
Thanks,
--sundar
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
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and provide commented, minimal, self-contained, reproducible code.
[R] using bquote to construct function
Hi, R-help,
(sessionInfo at the end)
I'm trying to construct a function using bquote and running into a
strange error message. As an example, what I would like to do is this:
z - 2
eval(bquote(function(x, y) { x^.(z) + y }))(2, 3)
However, I get the following:
Error in eval(expr, envir, enclos) :
invalid formal argument list for function
However, if I change the command to following, it works:
z - 2
eval(bquote(function(x) { x^.(z) }))(2)
# [1] 4
In other words, I remove the second argument. Is there a workaround for
this without using eval(parse(text = ))?
Thanks,
--sundar
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] use expression() in a loop
Nanye Long said the following on 8/18/2008 3:00 PM:
Hi all,
I want to do plot() in a loop to make 10 graphs, so I have some code like
for (i in 1:10) {
plot(... ... , xlab = expression(g[i]) )
}
I expect g_1, g_2, and so on appear on x labels, but it simply prints
g_i for each graph. Does anybody know how to get around this problem?
Thanks.
NL
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Try:
par(mfrow = c(2, 5))
for (i in 1:10) {
plot(1, 1, xlab = bquote(g[.(i)]))
}
HTH,
--sundar
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Re: [R] upgrade to 2.5
Iasonas Lamprianou said the following on 5/2/2007 8:25 AM: Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without having to install all the packages again? Thanks Jason You may find the following link relevant. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359.html HTH, --sundar __ [EMAIL PROTECTED] 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. __ 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.
Re: [R] Accessing items in a list of lists
[EMAIL PROTECTED] said the following on 5/14/2008
12:40 PM:
Using R 2.6.2, say I have the following list of lists, comb:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1 = data1, data2 = data2, data3 = data3)
So that all names for the lowest level list are common. How can I most
efficiently access all of the sublist items a indexed by the outer
list names? For example, I can loop through comb[[i]], unlisting as I
go, and then look up the field a, as below, but there has got to be a
cleaner way.
finaldata - double(0)
for(i in 1:length(names(comb))) {
test - unlist(comb[[i]])
finaldata - c(finaldata, test[which(names(test) == a)])
}
data.frame(names(comb), finaldata)
Gives what I want:
names.comb. finaldata
1 data1 1
2 data2 4
3 data3 3
Any help you can give would be greatly appreciated. Thanks.
Try
data.frame(names.comb = names(comb),
finaldata = sapply(comb, [[, a))
HTH,
--sundar
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Re: [R] as.numeric with tclvalue redux
Erin Hodgess said the following on 3/24/2008 10:39 AM:
Hi again R People:
This works fine:
library(tcltk)
a - tclVar(4.5)
as.numeric(tclvalue(a))
[1] 4.5
#But if you have:
b - tclVar(pi)
as.numeric(tclvalue(b))
[1] NA
Warning message:
NAs introduced by coercion
Is anyone aware of a way around this, please?
thanks,
Erin
Does this help?
eval.tclvalue - function(x, ...) {
x - type.convert(tclvalue(x), as.is = TRUE)
if(is.character(x) exists(x, ...)) {
get(x)
} else {
x
}
}
a - tclVar(4.5)
b - tclVar(pi)
c - tclVar(abcd)
eval.tclvalue(a)
eval.tclvalue(b)
eval.tclvalue(c)
HTH,
--sundar
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Re: [R] Lme does not work without a random effect (UNCLASSIFIED)
Park, Kyong H Mr ECBC said the following on 3/14/2008 12:25 PM: Classification: UNCLASSIFIED Caveats: NONE Dear R users, I'm interested in finding a random effect of the Block in the data shown below, but 'lme' does not work without the random effect. I'm not sure how to group the data without continuous value which is shown in the error message at the bottom line. If I use 'aov' with Error(Block), is there a test method comparing between with and without the Block random effect. I'm using R 2.4.1. Appreciate your help. Kyong LCU ST1 SURF Block 1 6.71 AN 1 2 6.97 AY 1 3 6.77 BN 1 4 6.90 BY 1 5 6.63 CN 1 6 6.94 CY 1 7 6.79 DN 1 8 6.93 DY 1 9 6.23 AN 2 10 6.83 AY 2 11 6.61 BN 2 12 6.86 BY 2 13 6.51 CN 2 14 6.90 CY 2 15 5.90 DN 2 16 6.97 DY 2 A result with the random effect: Anal1-lme(LCU~ST1*SURF,random=~1|Block,data=data1) summary(Anal1) Linear mixed-effects model fit by REML Data: data1 AIC BIClogLik 25.38958 26.18399 -2.694789 Random effects: Formula: ~1 | Block (Intercept) Residual StdDev: 0.1421141 0.218483 Fixed effects: LCU ~ ST1 * SURF Value Std.Error DF t-value p-value (Intercept) 6.470 0.1842977 7 35.10625 0. ST1B 0.220 0.2184830 7 1.00694 0.3475 ST1C 0.100 0.2184830 7 0.45770 0.6610 ST1D-0.125 0.2184830 7 -0.57213 0.5851 SURFY0.430 0.2184830 7 1.96812 0.0897 ST1B:SURFY -0.240 0.3089816 7 -0.77675 0.4627 ST1C:SURFY -0.080 0.3089816 7 -0.25892 0.8031 ST1D:SURFY 0.175 0.3089816 7 0.56638 0.5888 Without the random effect: Anal2-lme(LCU~ST1*SURF,data=data1) Error in getGroups.data.frame(dataMix, groups) : Invalid formula for groups Classification: UNCLASSIFIED Caveats: NONE Use lm to fit the model without random effect and use anova to compare: z - read.table(con - textConnection( LCU ST1 SURF Block 1 6.71 AN 1 2 6.97 AY 1 3 6.77 BN 1 4 6.90 BY 1 5 6.63 CN 1 6 6.94 CY 1 7 6.79 DN 1 8 6.93 DY 1 9 6.23 AN 2 10 6.83 AY 2 11 6.61 BN 2 12 6.86 BY 2 13 6.51 CN 2 14 6.90 CY 2 15 5.90 DN 2 16 6.97 DY 2), header = TRUE) close(con) library(nlme) fit - lme(LCU~ST1*SURF,random=~1|Block,data=z) fit0 - lm(LCU~ST1*SURF,data=z) anova(fit, fit0) HTH, --sundar __ 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.
Re: [R] Distances between two datasets of x and y co-ordinates
Andrew McFadden said the following on 3/12/2008 1:47 PM: Hi all I am trying to determine the distances between two datasets of x and y points. The number of points in dataset One is very small i.e. perhaps 5-10. The number of points in dataset Two is likely to be very large i.e. 20,000-30,000. My initial approach was to append the first dataset to the second and then carry out the calculation: dists - as.matrix(dist(gis data from 2 * datasets)) However, the memory of the computer is not sufficient. A lot of calculations carried out in this situation are unnecessary as I only want approx 5 * 20,000 calculations versus 20,000 *20,000. x - c(2660156,2663703,2658165,2659303,2661531,2660914) y - c(6476767,6475013,6475487,6479659,6477004,6476388) data2-cbind(x,y) x - c(266500,261) y - c(6478767,6485013) data1-cbind(x,y) Any suggestions on how to do this would be appreciated. Regards Andrew If you're trying to find only the closest point in data1 to data2, then use knn (or knn1) in the 'class' package: library(class) nn - knn1(data2, data1, 1:nrow(data2)) which gives you the rows in data1 closest to each row in data2. Then compute the distance: rowSums((data2[nn, ] - data1)^2)^0.5 HTH, --sundar __ 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.
Re: [R] LaTeX in R
Or my personal favorite if the length of mySigma is variable: mySigma - 2:3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10,sd = mySigma[2]),lty = 2) leg - as.expression(lapply(mySigma, function(x) bquote(sigma == .(x legend(x = topright, lty = c(1,2),legend = leg) Thanks, --sundar Uwe Ligges said the following on 3/7/2008 8:15 AM: You might want to read Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation in Plots. R News 2 (3), 32-34. with an example at the end that meets your requirements: (please note that I removed those ugly ; mySigma[1] - 2 mySigma[2] - 3 plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') lines(dnorm(1:10, sd = mySigma[2]), lty = 2) legend1 - substitute(sigma == myS, list(myS = mySigma[1])) legend2 - substitute(sigma == myS, list(myS = mySigma[2])) legend(x = topright, lty = c(1,2), legend = do.call(expression, list(legend1, legend2))) Uwe Ligges Mario Maiworm wrote: Finally, this should work for an array of sigmas. I just realized that the substitute()-command is not evaluated within a c()-environment :( mySigma[1] - 2; mySigma[2] - 3; plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') ; lines(dnorm(1:10,sd = mySigma[2]),lty = 2); legend(x = topright, lty = c(1,2),legend = c(substitute(sigma == myS, list(myS = mySigma[1])),substitute(sigma == myS, list(myS = mySigma[2] Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 16:30 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: AW: [R] LaTeX in R Mario Maiworm wrote: Thank you, uwe and jeremy. I was actually looking exactly for that! But something still doesn't work: I want to plot a symbol in a legend of a plot, lets say \sigma = 2. 2 should be the value of a variable. So, when I try mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=paste(expression(sigma), = ,mySigma),lty=1) , the sigma is not plotted as a symbol. This version: mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l') legend(x=topright,legend=expression(paste(sigma, = ,mySigma)),lty=1) gives me a 'real' sigma but the mySigma variable is not evaluated. Any ideas? Yes: mySigma - 2 plot(1:10, dnorm(1:10, sd = mySigma), type='l') legend(x = topright, lty = 1, legend = substitute(sigma == myS, list(myS = mySigma))) Uwe Ligges Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ -Ursprüngliche Nachricht- Von: Uwe Ligges [mailto:[EMAIL PROTECTED] Gesendet: Freitag, 7. März 2008 15:27 An: Mario Maiworm Cc: r-help@r-project.org Betreff: Re: [R] LaTeX in R Mario Maiworm wrote: Dear Rers, I understand that I can include R-code in LaTeX using Sweave. Is there a way to do it the other way round? Particularly, I need some TeX symbols in the legend of an R-plot. This can be done in matlab easily, so I am optimistic with R. Any suggestions for a command or package? See ?plotmath Uwe Ligges Best, Mario. __ Mario Maiworm Biological Psychology and Neuropsychology University of Hamburg Von-Melle-Park 11 D-20146 Hamburg Tel.: +49 40 42838 3515 Fax.: +49 40 42838 6591 http://bpn.uni-hamburg.de/Maiworm_e.html http://cinacs.org __ 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. __ 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. __ 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.
Re: [R] wire.frame tick labels from matrix
Marlin Keith Cox said the following on 2/15/2008 11:39 AM:
Dear R Users, close to the end of this I used wireframe to create a 3D plot
from a matrix. The x and y axis tick labels (1-6) for each were created
from the matrix being a 6X6 matrix. I need the axis tick labels to be the
row and column headings which you can see in the output (mat.x). I have
tried several work arounds, but they have not been successful.
Thanks in advance. keith
rm(list=ls())
sen-read.csv(brooktroutsort.csv)
#Resistance
R=c(660, 548, 676, 763, 768, 692, 657, 630, 748, 680, 786, 645, 710, 677,
692, 732, 737, 651, 396,
601, 640, 448, 464, 472, 434, 487, 495, 426, 429, 456)
#Detector length
Lend=c(37.0, 39.0, 39.0, 39.0, 40.0, 41.5, 44.0, 45.0, 46.0, 47.0,
47.0, 48.0,
48.5, 49.0, 51.0, 53.0, 53.0, 60.0, 89.0, 103.0, 108.5, 118.0, 118.0,
123.0,
126.0, 138.0, 139.0, 141.0, 141.0, 151.0)
#Errors to be multiplied by Restistance
x=c(0,.05,.10,.15,.20,.25)
#Errors to be multiplied by Detector length
y=c(0,.01,.02,.03,.04,.05)
#equation to predict water weight in grams
a=3.453*((Lend^2)/R)+1.994
X=(R%o%x+R)
Y=((Lend%o%y+Lend)^2)
num.x.col - length(X[1,])
num.y.col - length(Y[1,])
num.rows - length(X[,1])
Z - matrix(nrow=num.rows, ncol=num.x.col*num.y.col)
for( i in 1:num.rows) {
Z[i,] - as.vector( Y[i,] %*% t(X[i,])^-1 )
}
pred.est - 3.453*Z+1.994
z=(pred.est-a)/a
colnames(z)- rep(c(X1,X2,X3,X4,X5,X6),6)
meanz=colMeans(z)
mat.x - matrix(meanz, nrow=6, ncol=6, byrow=TRUE)
colnames(mat.x)- c(0,1,2,3,4,5)
rownames(mat.x)-c(0,5,10,15,20,25)
mat.x
library(lattice)
wireframe(mat.x,drape=TRUE,zlab=list(Proportion Error of Estimate,
rot=90), xlab=Resistance Error (%) ,ylab=Length Error
(%),scale=list(arrows=FALSE))
detach(z)
detach(sen)
Try:
mat.df - data.frame(z = as.vector(mat.x))
mat.df$x - rep(c(0,5,10,15,20,25), each = 6)
mat.df$y - rep(c(0,1,2,3,4,5), times = 6)
library(lattice)
wireframe(z ~ x * y, mat.df,
drape = TRUE,
zlab = list(Proportion Error of Estimate, rot=90),
xlab = Resistance Error (%) ,
ylab = Length Error (%),
scales = list(arrows = FALSE))
HTH,
--sundar
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Re: [R] fun.aggregate=mean in reshape
[Ricardo Rodriguez] Your XEN ICT Team said the following on 2/12/2008 12:23 AM: Hi all, We are facing a problem while introducing ourselves to Reshape package use. Melt seems to work fine, but cast fails when we use mean as fun.aggregate. As you see here, length and sum work fine, but mean throws this same error whatever dataset we use. cast(aqm, month ~ variable, length) month ozone solar.r wind temp 1 526 27 31 31 2 6 9 30 30 30 3 726 31 31 31 4 826 28 31 31 5 929 30 30 30 cast(aqm, month ~ variable, sum) month ozone solar.r wind temp 1 5 6144895 360.3 2032 2 6 2655705 308.0 2373 3 7 15376711 277.2 2601 4 8 15594812 272.6 2603 5 9 9125023 305.4 2307 cast(aqm, month ~ variable, mean) Error in get(as.character(FUN), mode = function, envir = envir) : variable fun of mode function was not found Our environment: version _ platform i386-apple-darwin8.10.1 arch i386 os darwin8.10.1 system i386, darwin8.10.1 status major 2 minor 6.2 year 2008 month 02 day08 svn rev44383 language R version.string R version 2.6.2 (2008-02-08) installed.packages() reshapereshape /Library/Frameworks/R.framework/Resources/library 0.8.0 NANA Please, could you help use to work out this issue? Thanks! Do you have an object called 'mean' that's masking the base::mean function? I can replicate your error using the following: library(reshape) names(airquality) - tolower(names(airquality)) aqm - melt(airquality, id=c(month, day), na.rm=TRUE) mean - 1 cast(aqm, month ~ variable, mean) Error in get(as.character(FUN), mode = function, envir = envir) : variable fun of mode function was not found cast(aqm, month ~ variable, base::mean) monthozone solar.r wind temp 1 5 23.61538 181.2963 11.622581 65.54839 2 6 29.4 190.1667 10.27 79.1 3 7 59.11538 216.4839 8.941935 83.90323 4 8 59.96154 171.8571 8.793548 83.96774 5 9 31.44828 167.4333 10.18 76.9 find(mean) [1] .GlobalEnv package:base HTH, --sundar __ 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.
Re: [R] lattice color problem with symbols: bug?
Stefan Grosse said the following on 1/11/2008 10:04 AM: Dear useR's, I have a problem with the lattice plotting of some symbols: library(lattice) test-data.frame(x=c(2,3,1,5),u=c(rep(1,2),rep(2,2)),g=c(rep(c(1,2),2))) xyplot(x~u,groups=g, data=test, par.settings=list( superpose.symbol=list(pch=c(22, 23),cex=c(1.7,1.6),col=black) ), key=list( text=list(c(t1,t2)), space = bottom,pch=c(23, 22), points=F, cex=1.0, col=black ), ) As you see the symbols which have been plotted into the plotting area appear to have some filling color while in the legend there is no filling color although the specification of the symbols is the same. If I use a normal plot command, the symbols are also not filled: plot(c(1,2,3),c(1,2,3),pch=c(22,23,24),cex=1.5) That problem must have occured during a recent lattice update since the color was not there when I was plotting a year ago with R2.5.x and some older lattice. It occurs on my linux as well as on my windows machine both with R 2.6.1 and latest lattice from CRAN. So here my questions: How do I get rid of the color? (or is it a bug?) If it is a feature, so how do I determine the color in both the symbols in the legend and in the plot itself? (Actually I was overlaying two plots e.g. one colored diamond and one empty diamond to achieve that effect but if there is a more efficient way to draw bordered symbols with customized color that would be preferable...) Stefan Hi, Stefan You need to use fill = 'transparent' in your par.settings call. If you're not using a fill color, you probably should use pch = c(0, 5) instead. This is explained under pch on the ?points help page. Here's a modified version of your example: library(lattice) test - data.frame(x = c(2, 3, 1, 5), u = rep(1:2, each = 2), g = paste(t, rep(1:2, 2), sep = )) xyplot(x ~ u, data = test, groups = g, par.settings = list( superpose.symbol = list( pch = c(22, 23), cex = c(1.7, 1.6), fill = c(red, blue), col = black)), auto.key = list(space = bottom)) HTH, --sundar __ 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.
Re: [R] side by side plots
[EMAIL PROTECTED] said the following on 1/7/2008 2:59 PM: Hello everyone, I have an overlay plot it's nice but you can't see all the data. I would like to know if there is a way to get a plot that gives a side by side plot so that each plot would be next to each other. The two plots have the same data are of different species. At the moment this is the code I'm using: exp-cbind(abs(round(rnorm(10),2)*10), seq(100, 200, by=10)) ref-cbind(abs(round(rnorm(10),2)*10), seq(100, 200, by=10)) plot(ref, ylab=Intensity, xlab=wavelength, type=h) points(exp, type=h, col=red) This is working in a script and I would like to have a single pdf/png file for the user with this plot, rather than asking the user to manually compare them. Any ideas on how I would do this? Thanks Paul __ 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. Hi, Paul, Simple enough in lattice: library(lattice) z - rbind(cbind(as.data.frame(exp), type = exp), cbind(as.data.frame(ref), type = ref)) z$type - factor(z$type, levels = c(ref, exp)) xyplot(V2 ~ V1 | type, data = z, type = h) HTH, --sundar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mfrow for levelplot?
marcg said the following on 1/5/2008 3:48 AM:
hello
could anyone tell my, why I do not suceed with mfrow?
par(mfrow=c(4,4))
for (i in 5:17){
levelplot(maxwater[,i]~maxwater$V1*maxwater$V2, col.regions=whiteblue(5),
xlab=, cuts=4)
}
Thanks
Marc
--
Because par settings have little to no effect on lattice. From ?Lattice:
Lattice plots are highly customizable via user-modifiable
settings. However, these are completely unrelated to base graphics
settings; in particular, changing 'par()' settings usually have no
effect on lattice plots.
To do what you're doing, you need to understand how Lattice works with
panels. Try:
## Since you didn't supply the data.frame 'maxwater'
## here's a fake dataset to demonstrate
library(lattice)
set.seed(1)
z - expand.grid(V1 = 1:10, V2 = 1:10)
z - cbind(z, matrix(rnorm(100 * 16), 100, 16))
names(z) - sprintf(V%d, 1:ncol(z))
## now create a formula
left - paste(names(z)[3:18], collapse = +)
right - paste(names(z)[1:2], collapse = *)
form - formula(sprintf(%s~%s, left, right))
## not sure how you defined 'whiteblue', but here's my version
whiteblue - colorRampPalette(c(white, blue))
## now call levelplot using a formula
levelplot(form, z, col.regions = whiteblue(5),
xlab = , cuts = 4, as.table = TRUE)
In the future, please read the posting guide about providing commented,
minimal, self-contained, reproducible code.
HTH,
--sundar
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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.
Re: [R] Accesing the value
get(x)
This is a FAQ:
http://cran.cnr.berkeley.edu/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
--sundar
Shubha Vishwanath Karanth said the following on 12/28/2007 8:38 AM:
Hi R,
x=A
A=5
I need to get the value of A using x only. How do I do this?
Thanks in advance, Shubha
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
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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.
__
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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.
Re: [R] How can R be used to solve algebra equations?
francogrex said the following on 11/29/2007 1:00 PM: suppose I have this equation: (x^2+y^2+3z^3)/(5*z^2*x^3)=0 and I want to find x in relation to the other variables which actually is: x=sqrt(-3*z^3-y^2) or x=-sqrt(-3*z^3-y^2) Can R give me this expression solution? I know there is uniroot, but here I want the expression not the value, because I do not have values for y and z. Thanks Try: library(Ryacas) yacas(Solve((x^2+y^2+3*z^3)/(5*z^2*x^3)==0, x)) expression(list(x == root(-(4 * (y^2 + 3 * z^3)), 2)/2, x == -root(-(4 * (y^2 + 3 * z^3)), 2)/2)) HTH, --sundar __ 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.
Re: [R] cumsum
threshold said the following on 11/3/2007 4:41 AM: Hi, my problem belongs to the basic ones. I want to get cumulated sum over the matrix columns by one command (if such exists). Ordinary R's cumsum(x) when x is: [,1] [,2] [1,]15 [2,]26 [3,]37 [4,]48 yields: 1 3 6 10 15 21 28 36 I want: [,1] [,2] [1,]15 [2,]3 11 [3,]6 18 [4,] 10 26 Is there any command to do so?? best, robert Try apply(x, 2, cumsum) HTH, --sundar __ 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.
Re: [R] plot for binomial glm
Radek John said the following on 10/28/2007 5:53 AM: Hello everybody! I am trying to plot glm with family=binomial and can`t work it out. My Data are: mort temp num numdead 132 7 0 232 8 0 332 8 0 437 15 3 537 15 1 637 17 3 742 11 8 842 28 14 942 15 12 10 47 10 10 11 47 12 12 12 47 13 13 13 52 18 18 14 52 19 19 15 52 22 22 I fitted glm glm.mort-glm(cbind(numdead, num - numdead) ~ temp, family=binomial) But now I don`t know, how to plot it. I need a plot with some points for variable numdead and some curve for the model. Thanks. Radek John How about: t1 - min(mort$temp) t2 - max(mort$temp) mort2 - data.frame(temp = seq(t1, t2, len = 100)) prd.mort - predict(glm.mort, mort2, type = r, se.fit = TRUE) plot(mort2$temp, prd.mort$fit, type = l, xlab = Tempature, ylab = Proportion Dead) lines(mort2$temp, prd.mort$fit - 1.96 * prd.mort$se.fit, lty = 2) lines(mort2$temp, prd.mort$fit + 1.96 * prd.mort$se.fit, lty = 2) points(mort$temp, mort$numdead/mort$num) See ?predict.glm. HTH, --sundar __ 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.
Re: [R] Wildcards
subura said the following on 10/15/2007 12:04 PM: Care to explain how i can use a wildcard expression to source all files ending with .R in a subdirectory ? I've tried something like this 'source(glob2rx(*.R))' without success. Thank you Try R.files - list.files(my.path, pattern = glob2rx(*.R), full = TRUE) invisible(sapply(R.files, source)) HTH, --sundar __ 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.
Re: [R] use 'lapply' to creat 2 new columns based on old ones in a data frame
runner said the following on 10/12/2007 4:46 PM:
There is a dataset 'm', which has 3 columns: 'index', 'old1' and 'old2';
I want to create 2 new columns: 'new1' and 'new2' on this condition:
if 'index'==i, then 'new1'='old1'+add[i].
'add' is a vector of numbers to be added to old columns, e.g. add=c(10,20,30
...)
Like this:
index old1old2new1new2
1 5 6 15 16
2 5 6 2526
3 5 6 35 36
3 50 60 80 90
Since the actual dataset is huge, I use 'lapply'. I am able to add 1 column:
do.call(rbind, lapply( 1:nrow(m),
function(i) {m$new1[i]=m[i,2]+add[m[i,1]];
return (m[i,])}
))
but don't know how to do for 2 columns at the same time, sth. like this
simply doesn't work:
do.call(rbind,lapply(1:nrow(m),
function(i){ m$new1[i]=m[i,2]+add[m[i,1]];
m$new2[i]=m[i,3]+add[m[i,1]];
return (m[i,])}
))
Could you please tell me how? or any other better approach?
No need for lapply.
x$new1 - x$old1 + add[x$index]
x$new2 - x$old2 + add[x$index]
To see how this works, try:
add - c(10, 20, 30)
index - c(1, 2, 1, 3, 1, 2, 3)
add[index]
but be careful if 'add' is length 3 and 'index' has a 4 in it you will get
index - c(index, 4)
add[index] ## produces an 'NA'
I hope I understood your question correctly. It's happy hour on the U.S.
east coast.
HTH,
--sundar
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[R] Checking for adequate disk space
Hi, all, (version info at end) I'm running a script which takes input files, does some analysis, and writes the output to csv files. Last night I ran the script (it took ~6.5 hours) thinking all would go well since it ran on a subset of the data without issue. However, when I returned this morning more than half the output files had no data. I checked the Rout file for errors and there were none. After spending about an hour debugging the script I learned the problem was not the script, but I ran out of disk space. But write.table, along with the rest of the script, still continued as if nothing was wrong. My question is: How can I programmatically determine if a user has adequate space to use write.* and then throw an error if they don't? I'm running R on Linux (RHEL4). For the short term, I will accept Linux-only solutions but would prefer an OS-free solution. My quick-and-dirty solution is to use: write.csv(x, file) if(file.info(file)[size] == 0) stop(you *may* have run out of disk space) However, this solution may not work as there may be other reasons the file size is 0 (e.g. x is NULL or 0-length?). x - character(0) write.table(x, file, col.names = FALSE, row.names = FALSE) file.info(file)[size] size file0 version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) __ 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.
