[R] Low level algorithm conrol in Fisher's exact test
Hi folks, Forgive me if this question is a trivial issue. I was doing a series of Fishers' exact test using the fisher.test function in stats package. Since the counts I have were quite large (c(64, 3070, 2868, 4961135)), R suggested me to use *other algorithms* for the test which can be specified through the 'control' argument of the fisher.test function as I understood. But where can I find other algorithms that I can use? I hoped I could find relevant information in the manual but could not. Can anybody help me out there? Many thanks in advance. Kihwang __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Command line and R
Angelo Secchi wrote: On Wed, 09 Nov 2005 12:25:37 - (GMT) (Ted Harding) [EMAIL PROTECTED] wrote: On 09-Nov-05 Roger Bivand wrote: On Wed, 9 Nov 2005, Angelo Secchi wrote: Hi, I wrote a small R script (delta.R) using commandArgs(). The script works from the shell in usual way R --no-save arg1 delta2.R Suppose arg1 is the output of another shell command (e.g. gawk, sed ...). Is there a way to tell R to read arg1 from the output of the previous command? Any other workaround? Use shell variables, possibly also Sys.getenv() within R as well as or instead of commandArgs(). If it's a fairly simple shell comand (and even if it isn't, though it could get tricky for complicated ones) you can use the backquote trick (called, in well-spoken circles, command substitution): R --no-save `shellcmd` delta2.R As in all shell command lines, wherever you have a command (including arguments etc.) between backquotes, as exemplified by `shellcmd` above, the output of the command (as sent to stdout) replaces `shellcmd` in the command-line. This could be a lot of stuff (depending on what shellcmd is), or just one value, or whatever. ... and this behaviour is OS (or at least command shell specific) for anyone trying this on Windows and wondering why it doesn't work. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] help with legacy R code
David Zhao wrote:
Hi there,
Could somebody help me disect this legacy R script I inherited at work, I
have two questions:
1. I've tried to upgrade our R version from 1.6.2 (yeah, I know), to R 2.0,
but some of the lines in this script are not compatible with R 2.0, could
someone help me figure out where the problem is?
2. the jpeg generated (attached) seems to be off on some of the data, is
there a better way of doing this.
1a. R 2.0 must be a software I am not familar with, since for the R I
know such a version has never been released.
1b. We are unable to reproduce the stuff given below. Not even an error
message is given.
1c. Do you expect anybody has the time to make your own homework, in
particular on an unreproducible example? There are very convenient
debugging tools made available for you in R.
2. We do not see where your jpeg produced with your data is off.
Uwe Ligges
PS: Let me quote
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
Thanks very much in advance!
David
library(MASS)
jpeg(filename = diswrong.jpg, width = 800, height = 600, pointsize = 12,
quality = 75, bg = white)
myfunc - function(x, mean, sd, nfalse, ntotal, shape, rate) {
(nfalse*dgamma(x,shape,rate)+(ntotal-nfalse)*dnorm(x,mean,sd))/ntotal
}
wrong - scan(wrongrawdata.txt, list(x=0))
wrongfit - fitdistr(wrong$x, gamma)
wrongmean - mean(wrong$x)
wrongshape - wrongfit[[1]][1]
wrongrate - wrongfit[[1]][2]
good - scan(rawdata.txt, list(x=0))
xmin = 0
newx = good$x
xmean = mean(newx)
xmax = max(newx)+0.15
goodhist - hist(newx, br=seq(from=0,to=xmax,by=0.15), probability=T,
col=lightyellow)
initmean - (min(newx)+max(newx))/2
totalx - length(newx)
wrongmeanshift - wrongmean + 0.2
wrongper - pgamma(wrongmeanshift, wrongshape, wrongrate)
nfalseundermean -
which(abs(newx-wrongmeanshift)==min(abs(newx-wrongmeanshift)))
initnfalse - nfalseundermean / wrongper
fitmean - -1
fitsd - 0
fitnfalse - initnfalse
fitshape - wrongshape
fitrate - wrongrate
curve((fitnfalse*dgamma(x,fitshape,fitrate))/totalx, add=T, col=red,
lwd=2)
breaksllength - length(goodhist$breaks)
endi = breaksllength - 1
binprob = c(1)
for (i in 1:endi) {
expnegative - fitnfalse * (pgamma(goodhist$breaks[i+1],wrongshape,
wrongrate)-pgamma(goodhist$breaks[i],wrongshape, wrongrate))
if (goodhist$counts[i] == 0)
binprob[i] = 0
else
binprob[i] = (goodhist$counts[i] - expnegative) / goodhist$counts[i]
}
result = data.frame(newx)
prob = c(1)
for (i in 1:totalx) {
bini = which ((goodhist$breaks newx[i]) (goodhist$breaks newx[i]-0.15
))
if ((binprob[bini] 0.8) | (newx[i] wrongmean))
prob[i] = -1
else
prob[i] = binprob[bini]*100
}
result = data.frame(result, prob)
write.table(result, file=probwrong.txt, sep= , row.name=F, col.name=F)
fitpars = c(fitmean, fitsd, fitnfalse, fitshape, fitrate, totalx)
result = data.frame(fitpars)
write.table(result,file=parwrong.txt, sep= , row.name=F, col.name=F)
dev.off()
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Re: [R] Low level algorithm conrol in Fisher's exact test
Kihwang Lee wrote: Hi folks, Forgive me if this question is a trivial issue. I was doing a series of Fishers' exact test using the fisher.test function in stats package. Since the counts I have were quite large (c(64, 3070, 2868, 4961135)), R suggested me to use *other algorithms* for the test which can be specified through the 'control' argument of the fisher.test function as I understood. But where can I find other algorithms that I can use? I hoped I could find relevant information in the manual but could not. Can anybody help me out there? What about a chisq.test? And honestly, I know the answer before calculating anything Uwe Ligges Many thanks in advance. Kihwang __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] About: Error in FUN(X[[1]], ...) : symbol print-name too long
gsmatos1 wrote:
Hi,
I´m trying to use the Win2BUGS package from R and I have a similar problem
R2WinBUGS???
that reurns with the message:
Error in FUN(X[[1]], ...) : symbol print-name too long
But, there is no stray ` character in the file ( Sugestions given by: Duncan
Temple Lang duncan
Date: Mon, 26 Sep 2005 07:31:08 -0700 )
The progam in R is:
library(R2WinBUGS)
library(rbugs)
Hmm, mixing these two packages might not be a good idea...
dat -
list(x=c(49,48,50,44,54,56,48,48,51,51,50,53,51,50,51,54,50,53,50,49,51,47,53,50,49,55,53,48,54,46),
y=c(50,49,57,52,47,52,58,45,55,54,51,54,56,53,52,47,51,54,50,47,46,44,54,55,52,57,52,48,48,51))
dat - format4Bugs(dat, digits = 0)
What happens if you omit the line above?
Anyway, I can take closer look, but not within the next 24 hours ...
Uwe Ligges
parm - c(lbda)
bugs(dat, inits=list(NULL), parm, d2.bug,
n.chains = 1, n.iter = 5000, n.burnin = floor(n.iter/2),
n.thin = max(1, floor(n.chains * (n.iter - n.burnin)/1000)),
bin = (n.iter - n.burnin) / n.thin,
debug = TRUE, DIC = TRUE, digits = 5, codaPkg = FALSE,
bugs.directory = C:/WinBUGS14/,
working.directory = NULL, clearWD = FALSE)
The objective of the program is to compare means of two independent
samples
that results
in Beherens-Fisher posterior and in the model.file of WinBUGS d2.bug
there is the following codes:
model
{
for( i in 1 : 30 ) {
x[i] ~ dnorm(mu1,sig1)
}
for( i in 1 : 30 ) {
y[i] ~ dnorm(mu2,sig2)
}
mu1 ~ dnorm(50,1.0E-6)
sig1 ~ dgamma(0.001,0.001)
mu2 ~ dnorm(50,1.0E-6)
sig2 ~ dgamma(0.001,0.001)
lbda - mu1 - mu2
}
I´m a new user of WinBUGS and if someone detect error in the model codes
too, I´m grateful.
Thanks for help!
Gilberto Matos.
model
{
for( i in 1 : 30 ) {
x[i] ~ dnorm(mu1,sig1)
}
for( i in 1 : 30 ) {
y[i] ~ dnorm(mu2,sig2)
}
mu1 ~ dnorm(50,1.0E-6)
sig1 ~ dgamma(0.001,0.001)
mu2 ~ dnorm(50,1.0E-6)
sig2 ~ dgamma(0.001,0.001)
lbda - mu1 - mu2
}
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Re: [R] Problems with Shapiro Wilk's test of normality.
Never mind,
I solved it myself. It was an NA problem.
/Fredrik
On 11/9/05, Fredrik Karlsson [EMAIL PROTECTED] wrote:
Hi,
I am trying to create a table with information from Shapiro Wilk's
test of normality.
However, it fails due to lack of sample size, it says, but the way I
see it, this is not a problem.
(See the table of sample sizes (almost) at the bottom).
Applying a different function using a similar ftable call is not a
problem (See the bottom table).
This is R 2.1.0 on Linux (Gentoo).
/Fredrik
shapiro.p.value - function(x){
+ if(length(! is.na(x)) 3 length(! is.na(x)) 5000 ){
+ p - shapiro.test(x)$p.value
+ return(p)
+}else{
+ return(NA)
+ }
+ }
distribution.table.fun - function(x,na.rm=T,digits=1){
+
+ if(length(! is.na(x)) 3 length(! is.na(x)) 5000){
+# shapTest - shapiro.test(x)
+# W - shapTest$statistic
+ W - W
+ }
+
+
+
+ shap - shapiro.p.value(x)
+ stars - ''
+ premark - ''
+ postmark - ''
+ if(length(x) 10){
+ premark - '\\textit{'
+ postmark - '}'
+ }
+
+ #skapa stjärnor
+ if(! is.na(shap)){
+ if( shap = 0.001 ){
+ stars - '***'
+ }else{
+ if( shap = 0.01 ){
+ stars - '**'
+ }else{
+ if( shap = 0.05 ){
+ stars - '*'
+ }
+
+ }
+
+ }
+
+ outstr - paste(premark,'W=',W,',p=',shap,postmark,stars,sep=)
+ }
+ else{
+ outstr -
+ }
+
+
+ return(outstr)
+
+ }
ftable(tapply(aspvotwork$ampratio,list(Place=aspvotwork$Place,Age=aspvotwork$agemF,voicetype=aspvotwork$Type),FUN=length
))
voicetype Voiced Voiceless unaspirated Voiceless aspirated
Place Age
Velar 18 - 24 4441 34
24 - 30 7081 71
30 - 36 5966 64
36 - 42 2527 22
42 - 48 2223 23
48 - 54 12 9 7
Dental 18 - 24 4861 54
24 - 30 82 101 89
30 - 36 5782 72
36 - 42 1931 34
42 - 48 2533 31
48 - 54 1012 14
Labial 18 - 24 74 141 84
24 - 30 142 264 162
30 - 36 124 213 148
36 - 42 5091 50
42 - 48 4982 64
48 - 54 1726 16
ftable(tapply(aspvotwork$ampratio,list(Place=aspvotwork$Place,Age=aspvotwork$agemF,voicetype=aspvotwork$Type),FUN=distribution.table.fun,digits=4))
Error in shapiro.test(x) : sample size must be between 3 and 5000
ftable(tapply(aspvotwork$ampratio,list(Place=aspvotwork$Place,Age=aspvotwork$agemF,voicetype=aspvotwork$Type),FUN=mean,digits=4,na.rm=TRUE
))
voicetypeVoiced Voiceless unaspirated Voiceless aspirated
Place Age
Velar 18 - 24 0.4816810 0.4461307 0.4513994
24 - 30 0.5289028 0.4778686 0.4888445
30 - 36 0.5452949 0.5208633 0.4756369
36 - 42 0.5631310 0.4697789 0.4709779
42 - 48 0.4968318 0.4174068 0.4088855
48 - 54 0.3057712 0.4483639 0.4561953
Dental 18 - 24 0.4058078 0.4596251 0.4091731
24 - 30 0.4609731 0.4502778 0.4483340
30 - 36 0.5095430 0.4726149 0.4315419
36 - 42 0.4935719 0.4687774 0.4528758
42 - 48 0.4344465 0.4220429 0.4362018
48 - 54 0.3697664 0.4338549 0.4897856
Labial 18 - 24 0.4327926 0.4879985 0.4503917
24 - 30 0.5309634 0.4839031 0.5927699
30 - 36 0.4094516 0.757 0.3964693
36 - 42 0.5010130 0.480 0.4540598
42 - 48 0.4949510 0.4329442 0.3935921
48 - 54 0.5217893 0.5124186 0.5011346
--
My Gentoo + PVR-350 + IVTV + MythTV
Re: [R] Command line and R
On Thu, 10 Nov 2005, paul sorenson wrote: Angelo Secchi wrote: On Wed, 09 Nov 2005 12:25:37 - (GMT) (Ted Harding) [EMAIL PROTECTED] wrote: On 09-Nov-05 Roger Bivand wrote: On Wed, 9 Nov 2005, Angelo Secchi wrote: Hi, I wrote a small R script (delta.R) using commandArgs(). The script works from the shell in usual way R --no-save arg1 delta2.R Suppose arg1 is the output of another shell command (e.g. gawk, sed ...). Is there a way to tell R to read arg1 from the output of the previous command? Any other workaround? Use shell variables, possibly also Sys.getenv() within R as well as or instead of commandArgs(). If it's a fairly simple shell comand (and even if it isn't, though it could get tricky for complicated ones) you can use the backquote trick (called, in well-spoken circles, command substitution): R --no-save `shellcmd` delta2.R As in all shell command lines, wherever you have a command (including arguments etc.) between backquotes, as exemplified by `shellcmd` above, the output of the command (as sent to stdout) replaces `shellcmd` in the command-line. This could be a lot of stuff (depending on what shellcmd is), or just one value, or whatever. ... and this behaviour is OS (or at least command shell specific) for anyone trying this on Windows and wondering why it doesn't work. But it does work on Windows if you have a reasonable shell. Cmd.exe is and (especially) command.com are not shells in the usually accepted sense. Better to use Rterm than incur the additional overhead of R, though. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Low level algorithm conrol in Fisher's exact test
On Thu, 10 Nov 2005, Kihwang Lee wrote: Hi folks, Forgive me if this question is a trivial issue. I was doing a series of Fishers' exact test using the fisher.test function in stats package. Since the counts I have were quite large (c(64, 3070, 2868, 4961135)), R suggested me to use *other algorithms* for the test which can be specified through the 'control' argument of the fisher.test function as I understood. Not that I can reproduce. You cannot change the algorithm that way. But where can I find other algorithms that I can use? I hoped I could find relevant information in the manual but could not. Can anybody help me out there? What *exactly* did you see? I get fisher.test(matrix(c(64, 3070, 2868, 4961135), 2)) FEXACT error 40. Out of workspace. fisher.test(matrix(c(64, 3070, 2868, 4961135), 2), workspace=20e6) FEXACT error 501. The hash table key cannot be computed because the largest key is larger than the largest representable int. The algorithm cannot proceed. Reduce the workspace size or use another algorithm. Where does it say anything about using control= ? AFAIK R does not have a means of doing Fisher's test on such a table, and it really does not make much statistical sense to do so. With such numbers, the null hypothesis is almost always rejected (try the chisq test), even for negligible dependence. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Choosing the data type to improve accuracy in SVM, PPR and randomForest
Dear all, This question is not a pure R question but I believe it is quite related. I am trying to find some literature (without success) about the most appropriate type for the data I am using. For example: day of the week is it better represented as a factor or as a number? I am trying to answer this for SVM, PPR and randomForest. Thanks for any help Joao Moreira __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Low level algorithm conrol in Fisher's exact test
Uwe Ligges [EMAIL PROTECTED] writes: Kihwang Lee wrote: Hi folks, Forgive me if this question is a trivial issue. I was doing a series of Fishers' exact test using the fisher.test function in stats package. Since the counts I have were quite large (c(64, 3070, 2868, 4961135)), R suggested me to use *other algorithms* for the test which can be specified through the 'control' argument of the fisher.test function as I understood. But where can I find other algorithms that I can use? I hoped I could find relevant information in the manual but could not. Can anybody help me out there? What about a chisq.test? And honestly, I know the answer before calculating anything Actually, chisq.test complains that the expected values are too low... I.e. you expected less than 5 and got 64! So the chisquare approximation might not be perfect, but p 2e-16 should be close enough for jazz. There's a buglet in the internal FEXACT code that causes it to allocate a workspace that is way too big for cases like this. If you really want to know what the p value is, phyper() is less sensitive: phyper(63,2932,4964205,3134,lower=FALSE) [1] 4.512776e-74 (and in cases where one group is much larger than the other, you're not far off by assuming that the probability in that group is known, leading to a binomial test: binom.test(64,3134,p=2868/4961135)$p.value [1] 2.368985e-74 ) The control= argument is not too well documented, but according to my reading of the code, it is only used to set the mult argument to .C(fexact, ...) and has no effect on the current issue. Actually, the fexact C code is only used if or=1 (the default), so another way out is fisher.test(M,or=1+1e-15)$p.value [1] 4.512776e-74 fisher.test(M,or=1-1e-15)$p.value [1] 4.512776e-74 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Low level algorithm conrol in Fisher's exact test
Prof Brian Ripley [EMAIL PROTECTED] writes: AFAIK R does not have a means of doing Fisher's test on such a table, and it really does not make much statistical sense to do so. With such numbers, the null hypothesis is almost always rejected (try the chisq test), even for negligible dependence. I have to disagree a little here. If the count in the smaller group had been smaller we would have been well inside the scope of exact testing, e.g. fisher.test(matrix(c(4, 3070, 2868, 4961135), 2),or=1+1e-15) Fisher's Exact Test for Count Data data: matrix(c(4, 3070, 2868, 4961135), 2) p-value = 0.105 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.6132965 5.7830438 sample estimates: odds ratio 2.253824 (And the workspace issue still applies, hence the or= fiddle) A professional statistician would know enough to switch to the binomial (or Poisson) approximation, but others might need help. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Low level algorithm conrol in Fisher's exact test
Prof Brian Ripley wrote: Where does it say anything about using control= ? fisher.test(x, y = NULL, workspace = 20, hybrid = FALSE, control = list(), or = 1, alternative = two.sided, conf.int = TRUE, conf.level = 0.95) ... control a list with named components for low level algorithm control. I could not make any sense out of this, but it seems to indicate that this argument either selects or modifies the algorithm used to compute the output. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
On 11/9/2005 10:01 PM, Adaikalavan Ramasamy wrote: I think an alternative is to use a p-value from F distribution. Even tough it is not a statistics, it is much easier to explain and popular than 1/F. Better yet to report the confidence intervals. Just curious about your usage: why do you say a p-value is not a statistic? Duncan Murdoch Regards, Adai On Wed, 2005-11-09 at 17:09 -0600, Mike Miller wrote: On Wed, 9 Nov 2005, Gao Fay wrote: Hi there, Suppose mu is constant, and error is normally distributed with mean 0 and fixed variance s. I need to find a statistics that: Y_i = mu + beta1* I1_i beta2*I2_i + beta3*I1_i*I2_i + +error, where I_i is 1 Y_i is from group A, and 0 if Y_i is from group B. It is large when beta1=beta2=0 It is small when beta1 and/or beta2 is not equal to 0 How can I find it by R? Thank you very much for your time. That's a funny question. Usually we want a statistic that is small when beta1=beta2=0 and large otherwise. Why not compute the usual F statistic for the null beta1=beta2=0 and then use 1/F as your statistic? Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Help to multinomial analyses
Dear Sirs, Could you please be so kind as to send us some information on residuals in multinomial logistic models? Is it possible to use R software? We thank you in advance. Sincerely yours Luciana Alves,MSc Beatriz Leimann, MD -- Luciana Correia Alves Doutoranda em Saúde Pública ENSP - Fiocruz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] silly question on covariable declaration
Hi all, sory f it is too simple I am new on using R and have not been able to find the answer of my problem in the help archives. I am performing a relatively simple aov aov( var ~ block + block/plot/treatment + season:bloc) it's a bit longer but this is how it looks like. my var is lipid content, starch content,...in wood samples. Now I want to add the size of the tree from where the sample was taken as a covariable, because I think it could have something to do. How can I declare it ? Thanks in advance Bernat [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] match and %in%
R-help, I have two data frames with a commom column (but with different size) What I want is to get a column (say df1$mycolumn ) according to the matches of common columns in both data frames. I have tried this but it is not working: transform(fb, breidd = ifelse (match (as.character(df1$puntar), as.character(df2$puntar) ) , df2$breidd, no ) ) ) transform(fb, breidd = ifelse(as.character(df1$puntar) %in% as.character(df2$puntar) , df2$breidd, no ) ) ) Thank you in advance __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] match and %in%
On 11/10/05 6:33 AM, Luis Ridao Cruz [EMAIL PROTECTED] wrote: R-help, I have two data frames with a commom column (but with different size) What I want is to get a column (say df1$mycolumn ) according to the matches of common columns in both data frames. I have tried this but it is not working: transform(fb, breidd = ifelse (match (as.character(df1$puntar), as.character(df2$puntar) ) , df2$breidd, no ) ) ) transform(fb, breidd = ifelse(as.character(df1$puntar) %in% as.character(df2$puntar) , df2$breidd, no ) ) ) Try looking at ?merge or ?union to see if either will do what you like. Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] help with legacy R code
1) There is no 'R 2.0'. What version did you mean?
2) We cannot reproduce your script (no data files), and JPEGs are not
allowed on R-help: see http://www.r-project.org/mail.html. (PNGs are,
though).
3) You have given us no indication of what the problems are nor in which
lines.
Please give us more usable information.
On Wed, 9 Nov 2005, David Zhao wrote:
Could somebody help me disect this legacy R script I inherited at work, I
have two questions:
1. I've tried to upgrade our R version from 1.6.2 (yeah, I know), to R 2.0,
but some of the lines in this script are not compatible with R 2.0, could
someone help me figure out where the problem is?
2. the jpeg generated (attached) seems to be off on some of the data, is
there a better way of doing this.
library(MASS)
jpeg(filename = diswrong.jpg, width = 800, height = 600, pointsize = 12,
quality = 75, bg = white)
myfunc - function(x, mean, sd, nfalse, ntotal, shape, rate) {
(nfalse*dgamma(x,shape,rate)+(ntotal-nfalse)*dnorm(x,mean,sd))/ntotal
}
wrong - scan(wrongrawdata.txt, list(x=0))
...
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] How to find statistics like that.
If my usage is wrong please correct me. Thank you. Here are my reason : 1. p-value is a (cumulative) probability and always ranges from 0 to 1. A test statistic depending on its definition can wider range of possible values. 2. A test statistics is one that is calculated from the data without the need of assuming a null distribution. Whereas to calculate p-values, you need to assume a null distribution or estimate it empirically using permutation techniques. 3. The directionality of a test statistics may be ignored. For example a t-statistics of -5 and 5 are equally interesting in a two-sided testing. But the smaller the p-value, more evidence against the null hypothesis. Regards, Adai On Thu, 2005-11-10 at 06:05 -0500, Duncan Murdoch wrote: On 11/9/2005 10:01 PM, Adaikalavan Ramasamy wrote: I think an alternative is to use a p-value from F distribution. Even tough it is not a statistics, it is much easier to explain and popular than 1/F. Better yet to report the confidence intervals. Just curious about your usage: why do you say a p-value is not a statistic? Duncan Murdoch Regards, Adai On Wed, 2005-11-09 at 17:09 -0600, Mike Miller wrote: On Wed, 9 Nov 2005, Gao Fay wrote: Hi there, Suppose mu is constant, and error is normally distributed with mean 0 and fixed variance s. I need to find a statistics that: Y_i = mu + beta1* I1_i beta2*I2_i + beta3*I1_i*I2_i + +error, where I_i is 1 Y_i is from group A, and 0 if Y_i is from group B. It is large when beta1=beta2=0 It is small when beta1 and/or beta2 is not equal to 0 How can I find it by R? Thank you very much for your time. That's a funny question. Usually we want a statistic that is small when beta1=beta2=0 and large otherwise. Why not compute the usual F statistic for the null beta1=beta2=0 and then use 1/F as your statistic? Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
The definition of a statistic that I learned in grad school is that it's a function of a random sample from a population. Any p-value would fit that definition. Andy From: Adaikalavan Ramasamy If my usage is wrong please correct me. Thank you. Here are my reason : 1. p-value is a (cumulative) probability and always ranges from 0 to 1. A test statistic depending on its definition can wider range of possible values. 2. A test statistics is one that is calculated from the data without the need of assuming a null distribution. Whereas to calculate p-values, you need to assume a null distribution or estimate it empirically using permutation techniques. 3. The directionality of a test statistics may be ignored. For example a t-statistics of -5 and 5 are equally interesting in a two-sided testing. But the smaller the p-value, more evidence against the null hypothesis. Regards, Adai On Thu, 2005-11-10 at 06:05 -0500, Duncan Murdoch wrote: On 11/9/2005 10:01 PM, Adaikalavan Ramasamy wrote: I think an alternative is to use a p-value from F distribution. Even tough it is not a statistics, it is much easier to explain and popular than 1/F. Better yet to report the confidence intervals. Just curious about your usage: why do you say a p-value is not a statistic? Duncan Murdoch Regards, Adai On Wed, 2005-11-09 at 17:09 -0600, Mike Miller wrote: On Wed, 9 Nov 2005, Gao Fay wrote: Hi there, Suppose mu is constant, and error is normally distributed with mean 0 and fixed variance s. I need to find a statistics that: Y_i = mu + beta1* I1_i beta2*I2_i + beta3*I1_i*I2_i + +error, where I_i is 1 Y_i is from group A, and 0 if Y_i is from group B. It is large when beta1=beta2=0 It is small when beta1 and/or beta2 is not equal to 0 How can I find it by R? Thank you very much for your time. That's a funny question. Usually we want a statistic that is small when beta1=beta2=0 and large otherwise. Why not compute the usual F statistic for the null beta1=beta2=0 and then use 1/F as your statistic? Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Help to multinomial analyses
hi, hi all, Dear Sirs, Could you please be so kind as to send us some information on residuals in multinomial logistic models? here are some references to multinomial models: Agresti A.(1996) An Introduction to Categorical Data Analysis Agresti A. (2002) Categorical Data Analysis, 2nd Edition McCullagh P. and Nelder J. (1989). Generalized Linear Models. Chapman and Hall, London. http://data.princeton.edu/wws509/notes/c6.pdf http://www.statslab.cam.ac.uk/~pat/Splusdiscrete2.pdf (see chapter 11 for Multinomial response) etc ... Is it possible to use R software? You can consult these links : http://www.stat.ufl.edu/~aa/cda/software.html http://www.statslab.cam.ac.uk/~pat/Splusdiscrete2.pdf (see chapter 11 for Multinomial response) cheers, P.BADY __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
-Original Message- From: [EMAIL PROTECTED] [SMTP:[EMAIL PROTECTED] On Behalf Of Adaikalavan Ramasamy Sent: Thursday, November 10, 2005 10:31 AM To: Duncan Murdoch Cc: r-help@stat.math.ethz.ch Subject: Re: [R] How to find statistics like that. If my usage is wrong please correct me. Thank you. Here are my reason : 1. p-value is a (cumulative) probability and always ranges from 0 to 1. A test statistic depending on its definition can wider range of possible values. 2. A test statistics is one that is calculated from the data without the need of assuming a null distribution. Whereas to calculate p-values, you need to assume a null distribution or estimate it empirically using permutation techniques. 3. The directionality of a test statistics may be ignored. For example a t-statistics of -5 and 5 are equally interesting in a two-sided testing. But the smaller the p-value, more evidence against the null hypothesis. Regards, Adai Hi: A statistic is any real-valued or vector-valued function whose domain includes the sample space of a random sample. The p-value is a real-valued function and its domain includes the sample space of a random sample. The p-value has a sampling distribution. The code below, found with Google (sampling distribution of the p-value R command) shows the sampling distribution of the p-value for a t-test of a mean when the null hypothesis is true. Ruben n-18 mu-40 pop.var-100 n.draw-200 alpha-0.05 draws-matrix(rnorm(n.draw * n, mu, sqrt(pop.var)), n) get.p.value-function(x) t.test(x, mu = mu)$p.value pvalues-apply(draws, 2, get.p.value) hist(pvalues) sum(pvalues = alpha) [1] 6 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] paste argument of a function as a file name
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the argument to
the function as a file name
something like this:
MY.function- function(df)
{
...
...
write.table(df,argument.xls)
}
MY.function(argument)
Thank you
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Re: [R] Help to multinomial analyses
Luciana Alves asked Dear Sirs, Could you please be so kind as to send us some information on residuals in multinomial logistic models? Peter L. Flom, PhD Assistant Director, Statistics and Data Analysis Core Center for Drug Use and HIV Research National Development and Research Institutes 71 W. 23rd St http://cduhr.ndri.org www.peterflom.com New York, NY 10010 (212) 845-4485 (voice) (917) 438-0894 (fax) [EMAIL PROTECTED] 11/10/2005 8:03:56 AM gave some very good references here are some references to multinomial models: Agresti A.(1996) An Introduction to Categorical Data Analysis Agresti A. (2002) Categorical Data Analysis, 2nd Edition McCullagh P. and Nelder J. (1989). Generalized Linear Models. Chapman and Hall, London. http://data.princeton.edu/wws509/notes/c6.pdf http://www.statslab.cam.ac.uk/~pat/Splusdiscrete2.pdf (see chapter 11 for Multinomial response) etc ... Is it possible to use R software? You can consult these links : http://www.stat.ufl.edu/~aa/cda/software.html http://www.statslab.cam.ac.uk/~pat/Splusdiscrete2.pdf (see chapter 11 for Multinomial response) I would suggest, in addition, Hosmer Lemeshow Applied Logistic Regression esp. p 280-288, and references therein. In particular, they cite Lesaffre Albert, Multiple-group regression diagnostics, Applied Statistics, 38, 425-440 but note that the techniques recommended there are not implemented in 'available software'. I would be interested to know if these techniques have been implemented since H and L. Regards Peter cheers, P.BADY __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Remove levels
Daer All, I have a factor variable, X with 5 levels. When I type tables(X) it gives me: table(X) 1 2 34 5 10 50 0 0 How to drop the levels with zeros such that when I will type: table(X) it will give me: table(X) 1 2 10 5 Thank a lot, Bernard - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] paste argument of a function as a file name
On 11/10/05 8:28 AM, Luis Ridao Cruz [EMAIL PROTECTED] wrote:
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the argument to
the function as a file name
help('paste')
Sean
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[R] write.table read.table with Dates
I've found several similar issues with write.table/read.table with Dates on this list, but trying to follow this advice I still get an error. First, I read in data from several files, constructing several date/time columns using ISOdatetime str(Tall$Begin) 'POSIXct', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02 00:00:00 ... length(Tall$Begin) [1] 40114 class(Tall$Begin) [1] POSIXt POSIXct This looks good (time is not always 00:00:00 ...) This data came from several files, now I want to store the result I have in data.frame Tall and be able to retrieve this quickly some other time. This is what I do: write.table(Tall, file=somefile.csv, sep=,, qmethod=double, row.names=FALSE) Later, I do this to read the file again: fieldnames=c(Begin,test-a,test-b,Eind) T=read.table(file = somefile.csv, col.names = fieldnames, header = TRUE, sep = ,, quote=\, fill=FALSE) I understand T$Begin now is a factor. I tried to simply convert it again using (as I read on this mailinglist ...): Q = strptime(as.character(T$Begin),format=%Y-%m-%d %H:%M:%S) Q is looking good, though its length I don't understand .. is it a list or something? It seems there are 40114 values in there somewhere... class(Q) [1] POSIXt POSIXlt length(Q) [1] 9 str(Q) 'POSIXlt', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02 00:00:00 ... T$Begin = Q ### yields this error Error in $-.data.frame(`*tmp*`, Begin, value = list(sec = c(0, 0, : replacement has 9 rows, data has 40114 Could somebody explain me how to convert the date column? Or perhaps there is an easier way? Thanks in advance for your time. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] paste argument of a function as a file name
Le 10.11.2005 14:28, Luis Ridao Cruz a écrit :
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the argument to
the function as a file name
something like this:
MY.function- function(df)
{
...
...
write.table(df,argument.xls)
}
MY.function(argument)
Thank you
Hi,
Maybe sprintf or paste.
MY.function- function(df, arg=argument)
{
...
...
write.table(df,paste(arg,.xls,sep=))
# or :
# write.table(df,sprintf(%s.xls,arg))
}
MY.function(argument)
--
visit the R Graph Gallery : http://addictedtor.free.fr/graphiques
+---+
| Romain FRANCOIS - http://francoisromain.free.fr |
| Doctorant INRIA Futurs / EDF |
+---+
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Re: [R] paste argument of a function as a file name
Luis Ridao Cruz wrote:
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the argument to
the function as a file name
something like this:
More like this:
foo = function(df){
fn=paste(deparse(substitute(df)),'.xls',sep='')
write.table(df,file=fn)
}
Then:
x=1:10
foo(x)
produces a file: x.xls
deparse(substitute(df)) is used in plot() to label the Y-axis with the
name of the object passed to plot(), which is similar to what you want
to do here.
Baz
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Re: [R] paste argument of a function as a file name
my.write - function( obj, name ){
filename - file=paste( name, .txt, sep=)
write.table( obj, file=filename, sep=\t, quote=F)
}
my.write( df, output )
Regards, Adai
On Thu, 2005-11-10 at 13:28 +, Luis Ridao Cruz wrote:
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the argument to
the function as a file name
something like this:
MY.function- function(df)
{
...
...
write.table(df,argument.xls)
}
MY.function(argument)
Thank you
__
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Re: [R] paste argument of a function as a file name
Why not use something like
MY.function - function(x){
filn - deparse(substitute(x))
filename - paste(filn,xls,sep=.)
...
...
write.table(x,file=filename)
}
Med venlig hilsen
Frede Aakmann Tøgersen
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Luis Ridao Cruz
Sendt: 10. november 2005 14:28
Til: r-help@stat.math.ethz.ch
Emne: [R] paste argument of a function as a file name
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the
argument to the function as a file name
something like this:
MY.function- function(df)
{
...
...
write.table(df,argument.xls)
}
MY.function(argument)
Thank you
__
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PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
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Re: [R] Remove levels
Marc Bernard wrote: Daer All, I have a factor variable, X with 5 levels. When I type tables(X) it gives me: table(X) 1 2 34 5 10 50 0 0 How to drop the levels with zeros such that when I will type: table(X) it will give me: table(X) 1 2 10 5 table(X[,drop=TRUE]) Uwe Ligges Thank a lot, Bernard - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
On 11/10/2005 7:31 AM, Adaikalavan Ramasamy wrote: If my usage is wrong please correct me. Thank you. Here are my reason : 1. p-value is a (cumulative) probability and always ranges from 0 to 1. A test statistic depending on its definition can wider range of possible values. 2. A test statistics is one that is calculated from the data without the need of assuming a null distribution. Whereas to calculate p-values, you need to assume a null distribution or estimate it empirically using permutation techniques. 3. The directionality of a test statistics may be ignored. For example a t-statistics of -5 and 5 are equally interesting in a two-sided testing. But the smaller the p-value, more evidence against the null hypothesis. Regards, Adai Thanks for your explanation. I think your interpretation is one that is sometimes taught, but I think it's more useful to think of a p-value as just another statistic, whose null distribution (in the ideal case, but not always in practice) is a uniform distribution on (0,1), and whose distribution when the alternative is true (again, ideally) tends to be more concentrated near 0. This takes a lot of the mysticism out of them. Duncan Murdoch On Thu, 2005-11-10 at 06:05 -0500, Duncan Murdoch wrote: On 11/9/2005 10:01 PM, Adaikalavan Ramasamy wrote: I think an alternative is to use a p-value from F distribution. Even tough it is not a statistics, it is much easier to explain and popular than 1/F. Better yet to report the confidence intervals. Just curious about your usage: why do you say a p-value is not a statistic? Duncan Murdoch Regards, Adai On Wed, 2005-11-09 at 17:09 -0600, Mike Miller wrote: On Wed, 9 Nov 2005, Gao Fay wrote: Hi there, Suppose mu is constant, and error is normally distributed with mean 0 and fixed variance s. I need to find a statistics that: Y_i = mu + beta1* I1_i beta2*I2_i + beta3*I1_i*I2_i + +error, where I_i is 1 Y_i is from group A, and 0 if Y_i is from group B. It is large when beta1=beta2=0 It is small when beta1 and/or beta2 is not equal to 0 How can I find it by R? Thank you very much for your time. That's a funny question. Usually we want a statistic that is small when beta1=beta2=0 and large otherwise. Why not compute the usual F statistic for the null beta1=beta2=0 and then use 1/F as your statistic? Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] paste argument of a function as a file name
Luis Ridao Cruz wrote:
R-help,
I have a function which is exporting the output to a file via
write.table(df, file = file name.xls )
What I want is to paste the file name (above) by taking the argument to
the function as a file name
something like this:
MY.function- function(df)
{
...
...
write.table(df,argument.xls)
}
MY.function(argument)
Has been asked hundreds of times on this list. Please check the archives
as the posting guide asks to do ...
Hint: paste()
Uwe Ligges
Thank you
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Re: [R] Remove levels
Marc Bernard [EMAIL PROTECTED] writes: Daer All, I have a factor variable, X with 5 levels. When I type tables(X) it gives me: table(X) 1 2 34 5 10 50 0 0 How to drop the levels with zeros such that when I will type: table(X) it will give me: table(X) 1 2 10 5 table(factor(X)) or table(X[drop=TRUE]) should do it. The latter runs the former, but the intention might be clearer. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
On Thu, 10 Nov 2005, Ruben Roa wrote: A statistic is any real-valued or vector-valued function whose domain includes the sample space of a random sample. The p-value is a real-valued function and its domain includes the sample space of a random sample. The p-value has a sampling distribution. The code below, found with Google (sampling distribution of the p-value R command) shows the sampling distribution of the p-value for a t-test of a mean when the null hypothesis is true. Ruben n-18 mu-40 pop.var-100 n.draw-200 alpha-0.05 draws-matrix(rnorm(n.draw * n, mu, sqrt(pop.var)), n) get.p.value-function(x) t.test(x, mu = mu)$p.value pvalues-apply(draws, 2, get.p.value) hist(pvalues) sum(pvalues = alpha) [1] 6 The sampling distribution of a p-value when the null hypothesis is true can be given more simply by this R code: runif() That holds for any valid test, not just a t test, that produces p-values distributed continuously on [0,1]. Discrete distributions can't quite do that without special tweaking. Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] write.table read.table with Dates
On 11/10/05, JeeBee [EMAIL PROTECTED] wrote: I've found several similar issues with write.table/read.table with Dates on this list, but trying to follow this advice I still get an error. First, I read in data from several files, constructing several date/time columns using ISOdatetime str(Tall$Begin) 'POSIXct', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02 00:00:00 ... length(Tall$Begin) [1] 40114 class(Tall$Begin) [1] POSIXt POSIXct This looks good (time is not always 00:00:00 ...) This data came from several files, now I want to store the result I have in data.frame Tall and be able to retrieve this quickly some other time. This is what I do: write.table(Tall, file=somefile.csv, sep=,, qmethod=double, row.names=FALSE) Later, I do this to read the file again: fieldnames=c(Begin,test-a,test-b,Eind) T=read.table(file = somefile.csv, col.names = fieldnames, header = TRUE, sep = ,, quote=\, fill=FALSE) I understand T$Begin now is a factor. I tried to simply convert it again using (as I read on this mailinglist ...): Q = strptime(as.character(T$Begin),format=%Y-%m-%d %H:%M:%S) Q is looking good, though its length I don't understand .. is it a list or something? It seems there are 40114 values in there somewhere... class(Q) [1] POSIXt POSIXlt length(Q) [1] 9 str(Q) 'POSIXlt', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02 00:00:00 ... T$Begin = Q ### yields this error Error in $-.data.frame(`*tmp*`, Begin, value = list(sec = c(0, 0, : replacement has 9 rows, data has 40114 Could somebody explain me how to convert the date column? Or perhaps there is an easier way? You are converting it to POSIXlt (which represents date/times as a 9 element structure) but its likely you really wanted to convert it to POSIXct. as.POSIXct(T$Begin) Also, you might need to use the tz= argument depending on what result you want. See the Help Desk article in RNews 4/1 for more info. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
-Original Message- From: Mike Miller [SMTP:[EMAIL PROTECTED] Sent: Thursday, November 10, 2005 12:32 PM To: Ruben Roa Cc: [EMAIL PROTECTED]; Duncan Murdoch; r-help@stat.math.ethz.ch Subject: Re: [R] How to find statistics like that. On Thu, 10 Nov 2005, Ruben Roa wrote: A statistic is any real-valued or vector-valued function whose domain includes the sample space of a random sample. The p-value is a real-valued function and its domain includes the sample space of a random sample. The p-value has a sampling distribution. The code below, found with Google (sampling distribution of the p-value R command) shows the sampling distribution of the p-value for a t-test of a mean when the null hypothesis is true. Ruben n-18 mu-40 pop.var-100 n.draw-200 alpha-0.05 draws-matrix(rnorm(n.draw * n, mu, sqrt(pop.var)), n) get.p.value-function(x) t.test(x, mu = mu)$p.value pvalues-apply(draws, 2, get.p.value) hist(pvalues) sum(pvalues = alpha) [1] 6 The sampling distribution of a p-value when the null hypothesis is true can be given more simply by this R code: runif() That holds for any valid test, not just a t test, that produces p-values distributed continuously on [0,1]. Discrete distributions can't quite do that without special tweaking. Mike Theorem 2.1.4 in Casella and Berger (1990, p. 52). Ruben __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Interpretation of output from glm
Dear Pedro, The basic point, which relates to the principle of marginality in formulating linear models, applies whether the predictors are factors, covariates, or both. I think that this is a common topic in books on linear models; I certainly discuss it in my Applied Regression, Linear Models, and Related Methods. Regards, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Pedro de Barros Sent: Wednesday, November 09, 2005 10:45 AM To: r-help@stat.math.ethz.ch Subject: Re: [R] Interpretation of output from glm Importance: High Dear John, Thanks for the quick reply. I did indeed have these ideas, but somehow floating, and all I could find about this mentioned categorical predictors. Can you suggest a good book where I could try to learn more about this? Thanks again, Pedro At 01:49 09/11/2005, you wrote: Dear Pedro, -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Pedro de Barros Sent: Tuesday, November 08, 2005 9:47 AM To: r-help@stat.math.ethz.ch Subject: [R] Interpretation of output from glm Importance: High I am fitting a logistic model to binary data. The response variable is a factor (0 or 1) and all predictors are continuous variables. The main predictor is LT (I expect a logistic relation between LT and the probability of being mature) and the other are variables I expect to modify this relation. I want to test if all predictors contribute significantly for the fit or not I fit the full model, and get these results summary(HMMaturation.glmfit.Full) Call: glm(formula = Mature ~ LT + CondF + Biom + LT:CondF + LT:Biom, family = binomial(link = logit), data = HMIndSamples) Deviance Residuals: Min 1Q Median 3Q Max -3.0983 -0.7620 0.2540 0.7202 2.0292 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) -8.789e-01 3.694e-01 -2.379 0.01735 * LT 5.372e-02 1.798e-02 2.987 0.00281 ** CondF -6.763e-02 9.296e-03 -7.275 3.46e-13 *** Biom-1.375e-02 2.005e-03 -6.856 7.07e-12 *** LT:CondF 2.434e-03 3.813e-04 6.383 1.74e-10 *** LT:Biom 7.833e-04 9.614e-05 8.148 3.71e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 10272.4 on 8224 degrees of freedom Residual deviance: 7185.8 on 8219 degrees of freedom AIC: 7197.8 Number of Fisher Scoring iterations: 8 However, when I run anova on the fit, I get anova(HMMaturation.glmfit.Full, test='Chisq') Analysis of Deviance Table Model: binomial, link: logit Response: Mature Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(|Chi|) NULL822410272.4 LT 1 2873.8 8223 7398.7 0.0 CondF 1 0.1 8222 7398.5 0.7 Biom1 0.2 8221 7398.3 0.7 LT:CondF1142.1 8220 7256.3 9.413e-33 LT:Biom 1 70.4 8219 7185.8 4.763e-17 Warning message: fitted probabilities numerically 0 or 1 occurred in: method(x = x[, varseq = i, drop = FALSE], y = object$y, weights = object$prior.weights, I am having a little difficulty interpreting these results. The result from the fit tells me that all predictors are significant, while the anova indicates that besides LT (the main variable), only the interaction of the other terms is significant, but the main effects are not. I believe that in the first output (on the glm object), the significance of all terms is calculated considering each of them alone in the model (i.e. removing all other terms), while the anova output is (as it says) considering the sequential addition of the terms. So, there are 2 questions: a) Can I tell that the interactions are significant, but not the main effects? In a model with this structure, the main effects represent slopes over the origin (i.e., where the other variables in the product terms are 0), and aren't meaningfully interpreted as main effects. (Is there even any data near the origin?) b) Is it legitimate to consider a model where the interactions are considered, but not the main effects CondF and Biom? Generally, no: That is, such a model is interpretable, but it places strange constraints on the regression surface -- that the CondF and Biom slopes are 0 over the origin. None of this is specific to logistic
[R] Fonts, Plus
Dear R Wizards: sorry, I need more help. hopefully, it will help others in the future. I am using R 2.2.0 Patched (2005-11-07 r36217). [a] # copy from the postscriptFont documentation CMitalic - postscriptFont(ComputerModern, c(CM_regular_10.afm, CM_boldx_10.afm, cmti10.afm, cmbxti10.afm, CM_symbol_10.afm)) postscriptFonts(CMitalic=CMitalic) # trying this one out. I copied the syntax that worked for lucida pdf(file=test.pdf, fonts=CMitalic, version=1.4); par(family=CMitalic); plot( c(0,1),c(0,1) ); myeq - bquote((w[I]==.(1/7))); text( 0.5, 0.3, myeq ); text( 0.5, 0.7, this is computer modern); dev.off(); Now, pdffonts test.pdf (from the xpdf distribution) gives me $ pdffonts test.pdf name type emb sub uni object ID --- --- --- - Error (4149): Dictionary key must be a name object Error (4152): Dictionary key must be a name object ZapfDingbats Type 1 no no no 5 0 HelveticaType 1 no no no 10 0 Helvetica-Bold Type 1 no no no 11 0 Helvetica-ObliqueType 1 no no no 12 0 Helvetica-BoldObliqueType 1 no no no 13 0 Symbol Type 1 no no no 14 0 CMR10Type 1 no no no 15 0 CMBX10 Type 1 no no no 16 0 CMTI10 Type 1 no no no 17 0 CMBXTI10 Type 1 no no no 18 0 Error (4149): Dictionary key must be a name object Error (4152): Dictionary key must be a name object CMSY10 Type 1 no no no 19 0 so, something is still wrong. [b] I am looking at the docs for postscriptFonts. ?postscriptFonts. May I suggest that we add two or three more lines to show usage? something like plot(c(0,1),c(0,1)); text(0.2, 0.5, hello, font=2); dev.off(). More generally, a documented sample example file that shows usage of many/multiple postscript fonts and families within one graph would be a great help. This is of course all just my own ignorance. In general, I am not yet sure about the whole font syntax. I wonder what a font=something statement in the plot statement itself does. I believe the par(family=) changes the font used for the figure [e.g., axis labels], although I am wondering why I am giving a string [CMItalic] rather than a variable [CMItalic]. the ?text documentation does not have an example of font selection, especially if I want to mix multiple fonts and from different font families. [c] how do I tell a CMD BATCH not to execute the site file? R --no-init-file works only interactively. ( Would it not make sense to allow this options also for CMD BATCH?) I probably have an incorrect installation, because the suggestion from R --help fails for me $ R CMD command --help /usr/local/lib64/R/bin/Rcmd: line 45: exec: command: not found However, R CMD BATCH my.R works just fine. Further suggestion: let's have an abbreviation for --no-init-file, too; e.g., -I. [d] regarding my earlier suggestion of a variable that contains the currently executing file, I know I can put an argv0 - filename into each file, but it would be nice if this happened automatically and was available everywhere. just a suggestion... please don't see the above as a complaint. R is great, and the effort you guys put in is terrific. It's just that I am struggling with the syntax here, and this one is not easy to figure out. Regards, /ivo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] write.table read.table with Dates
I see that strptime returns a list of year, mon, mday, hour, min, sec, etc. The following works for me (for each column that is a date/time field in my imported file) cat(Converting date/time fields...\n) Q = strptime(as.character(data$myfield), format=%Y-%m-%d%H:%M:%S) data$myfield = ISOdatetime(year = Q$year + 1900, month = Q$mon + 1, day = Q$mday, hour =Q$hour, min = Q$min, sec = Q$sec, tz = ) ISOdatetime does return a vector, which is, I guess, what I want. It is quite slow like this though, and I don't think it's the best way. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] write.table read.table with Dates
On Thu, 10 Nov 2005, JeeBee wrote: I've found several similar issues with write.table/read.table with Dates on this list, but trying to follow this advice I still get an error. First, I read in data from several files, constructing several date/time columns using ISOdatetime str(Tall$Begin) 'POSIXct', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02 00:00:00 ... length(Tall$Begin) [1] 40114 class(Tall$Begin) [1] POSIXt POSIXct This looks good (time is not always 00:00:00 ...) This data came from several files, now I want to store the result I have in data.frame Tall and be able to retrieve this quickly some other time. This is what I do: write.table(Tall, file=somefile.csv, sep=,, qmethod=double, row.names=FALSE) Later, I do this to read the file again: fieldnames=c(Begin,test-a,test-b,Eind) T=read.table(file = somefile.csv, col.names = fieldnames, header = TRUE, sep = ,, quote=\, fill=FALSE) You can avoid all this trouble by using colClasses as documented on the help page. I understand T$Begin now is a factor. I tried to simply convert it again using (as I read on this mailinglist ...): Q = strptime(as.character(T$Begin),format=%Y-%m-%d %H:%M:%S) Or just as.POSIXct(as.character(T$Begin)) Q is looking good, though its length I don't understand .. is it a list or something? It seems there are 40114 values in there somewhere... It is a list of length 9. Try names(Q) class(Q) [1] POSIXt POSIXlt length(Q) [1] 9 str(Q) 'POSIXlt', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02 00:00:00 ... T$Begin = Q ### yields this error Error in $-.data.frame(`*tmp*`, Begin, value = list(sec = c(0, 0, : replacement has 9 rows, data has 40114 Could somebody explain me how to convert the date column? Or perhaps there is an easier way? You started with POSIXct, and you need to convert back to POSIXct with as.POSIXct(Q). Reading ?DateTimeClasses should explain to you what you are missing. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R-help: conversion of long decimal numbers into hexadecimal
Hi there, could somebody help me to convert a decimal number into a hexadecimal number? I know that there is the function sprintf, but the numbers I want to convert consist of 20 or more numbers. Spintf is not able to convert these big numbers. Thanks for any help. Antje [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] question about the dataset fgl
Dear sir or ma'am, I have a question about the dataset fgl. The dataset seems to be in the VR package, so I tried to download it from CRAN. However, after downloading, when I tried to load the package, it was not in my package list. I am wondering what is wrong. Any advice on how to access the fgl dataset would be appreciated. Thanks. Seungho Huh [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] estimating significance P-value between 2 matricesv
Hi all, Being a new user of R I am having some troubles, I hope to get some solutions.., I have tried looking everywhere but so far not been so successful. i have two matrices mat1=[1:26], mat2[1:47] there are some missing values in mat1[5:6,15:26] I want to calculate Pearson correlation r between mat1 * mat2 and also estimate the significance P-value between mat1 * mat2 By using cor(mat1[2:26],mat2[2:47], use=pairwise.complete.obs) I was able to calcualte the Pearson correlation r between mat1 * mat2 I am not able to estimate the significance P-value using cor.test My Qs are? How to estimate significance P-value between variables of 2 matrices? OR is there any other approach to do this.. any idea or suggestions ?? thank you for your time, balaji [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] different functions on different vector subsets
Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 I found a solution in term of: sapply(a,function(x) if (is(s.na)) NA else if (x0) (-1)/2^x else 2^x) but still I would like to understand why the solution above did not work. I think is more ellegant. my R version is: sessionInfo() R version 2.2.0, 2005-10-06, i386-pc-mingw32 attached base packages: [1] methods stats graphics grDevices utils datasets [7] base Thanks, Ron [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] error in rowSums:'x' must be numeric
Dear All, It's Eszter again from Hungary. I could not solve my problem form yesterday, so I still have to ask your help. I have a binary dataset of vegetation samples and species as a comma separated file. I would like to calculate the Jaccard distance of the dataset. I have the following error message: Error in rowSums(x, prod(dn), p, na.rm) : 'x' must be numeric In addition: Warning message: results may be meaningless because input data have negative entries in: vegdist(t2, method = jaccard, binary = FALSE, diag = FALSE, Do you have any idea what can be the problem? I have only 0 and 1 in the dataset. Thank you very much! All the best: Eszter ___ KGFB 2006 - Garantáltan a legjobb ár! Nyerje meg az új Swiftet + garantált 10,000,- Ft értékű ajándék. WWW.NETRISK.HU __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to find statistics like that.
On 11/10/2005 9:32 AM, Mike Miller wrote: On Thu, 10 Nov 2005, Ruben Roa wrote: A statistic is any real-valued or vector-valued function whose domain includes the sample space of a random sample. The p-value is a real-valued function and its domain includes the sample space of a random sample. The p-value has a sampling distribution. The code below, found with Google (sampling distribution of the p-value R command) shows the sampling distribution of the p-value for a t-test of a mean when the null hypothesis is true. Ruben n-18 mu-40 pop.var-100 n.draw-200 alpha-0.05 draws-matrix(rnorm(n.draw * n, mu, sqrt(pop.var)), n) get.p.value-function(x) t.test(x, mu = mu)$p.value pvalues-apply(draws, 2, get.p.value) hist(pvalues) sum(pvalues = alpha) [1] 6 The sampling distribution of a p-value when the null hypothesis is true can be given more simply by this R code: runif() That holds for any valid test, not just a t test, that produces p-values distributed continuously on [0,1]. Discrete distributions can't quite do that without special tweaking. Nor can most composite null hypotheses, e.g. H0: mu = 0 versus H1: mu 0 A t-test may be an appropriate test, but its p-value is not uniformly distributed when mu is -1, even though the null is true. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] ltext - adding text to each panel from a matrix
Hi all (really probably just Deepayan):
In the plot below I want to add text on either side of each violin plot that
indicates the number of observations that are either positive or negative.
I'm trying to do this with ltext() and I've also monkeyed about with
panel.text(). The code below is generally what I want but my calls to
ltext() are wrong and I'm not sure how to fix them. Right now they replicate
the first column of the matrices obs.pos and obs.neg for each panel. How do
I tell ltext to advance to the next column when the next panel is plotted? I
don't see how subscripts can do it, but I bet it's something along that
line...
Thanks, Andy
rm(list = ls())
set.seed(354)
# make a bimodal dataset with three groups and three treatments
foo - c(rnorm(150,-1,0.5), rnorm(150,1,0.25))
treatment - factor(rep(seq(1,3),100), labels = c(Treatment 1, Treatment
2, Treatment 3))
group - factor(rep(seq(1,3),100), labels = c(Group A, Group B, Group
C))
group - sample(group)
# corrupt Group A, Treatment 2 for fun.
foo[group==Group A treatment==Treatment 2][1:8] - rnorm(8,1,1)
dat - data.frame(foo,treatment,group)
# set the limits for the plot, which also tells where to put the text
my.xlim - c(-6, 6)
# make a matrix that counts the number of obs greater or less than zero
obs.pos - tapply(dat[dat[,1] 0,1], dat[dat[,1] 0,-1], length)
obs.neg - tapply(dat[dat[,1] = 0,1], dat[dat[,1] = 0,-1], length)
#write some coordinate data
x.obs.pos - rep(my.xlim[2],dim(obs.pos)[2])
y.obs.pos - 1:dim(obs.pos)[2]
x.obs.neg - rep(my.xlim[1],dim(obs.neg)[2])
y.obs.neg - 1:dim(obs.neg)[2]
bwplot(treatment~foo|group, data = dat,
panel=function(...) {
panel.violin(..., col = transparent, varwidth = F)
panel.abline(v=0, lty = dotted)
ltext(x.obs.pos, y.obs.pos, obs.pos, pos = 2)
ltext(x.obs.neg, y.obs.neg, obs.neg, pos = 4)
},
par.strip.text = list(cex = 0.8), xlim = my.xlim)
obs.pos
obs.neg
# note that the numbers in the plot only match the matrices for Group A,
# which is the first panel. Alas.
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[R] R-help: conversion of long decimal numbers into hexadecimal numbers
Hi there, could somebody help me to convert a decimal number into a hexadecimal number? I know that there is the function sprintf, but the numbers I want to convert consist of 20 or more numbers. Spintf is not able to convert these big numbers. Thanks for any help. Antje Döring [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] question about the dataset fgl
`VR' is a bundle consisting of `MASS', `class', `nnet' and `spatial', as the description says. The fgl data is in the MASS package, so you need to load that one. In any case, data() would have told you that after the bundle is installed. Andy From: Huh, Seungho Dear sir or ma'am, I have a question about the dataset fgl. The dataset seems to be in the VR package, so I tried to download it from CRAN. However, after downloading, when I tried to load the package, it was not in my package list. I am wondering what is wrong. Any advice on how to access the fgl dataset would be appreciated. Thanks. Seungho Huh [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Help regarding mas5 normalization
Hello everybody, I am trying to use mas5 to normalize some array data and using mas5 and mas5calls. But I received these warning message. If anybody can explain the problem I would really appreciate that. Thanks in advance. background correction: mas PM/MM correction : mas expression values: mas background correcting...Warning message: 'loadURL' is deprecated. Use 'load(url())' instead. See help(Deprecated) Warning message: 'loadURL' is deprecated. Use 'load(url())' instead. See help(Deprecated) Warning message: 'loadURL' is deprecated. Use 'load(url())' instead. See help(Deprecated) There were 14 warnings (use warnings() to see them) Note: http://www.bioconductor.org/repository/devel/package/Win32 does not seem to have a valid repository, skipping Note: You did not specify a download type. Using a default value of: Source This will be fine for almost all users Error in FUN(X[[1]], ...) : no slot of name Uses for this object of class localPkg [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] order statistics / sample quantiles
Are there any R functions or packages that can compute distributions,
expectations, or quantiles of order statistics (or sample quantiles or
extreme values) for a given distribution such as a normal distribution?
Both exact and asymptotic calculations are of interest. I am already
aware of the 'quantile' function of 'stats'.
David
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Research Scientist
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515-334-4739 Tel
515-334-4473 Fax
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Re: [R] different functions on different vector subsets
The error messages mean what they say. I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. ## I assume you mean 1/(2^a). If not, modify the following appropriately. 2^(a*sign(a)) ## will do As for your error message for: a[a0]-(-1)/2^a[a0] a0 has an NA at the first index and so R doesn't know what index you want to assign the value to. Ergo the error message. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ron Ophir Sent: Thursday, November 10, 2005 7:26 AM To: r-help@stat.math.ethz.ch Subject: [R] different functions on different vector subsets Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 I found a solution in term of: sapply(a,function(x) if (is(s.na)) NA else if (x0) (-1)/2^x else 2^x) but still I would like to understand why the solution above did not work. I think is more ellegant. my R version is: sessionInfo() R version 2.2.0, 2005-10-06, i386-pc-mingw32 attached base packages: [1] methods stats graphics grDevices utils datasets [7] base Thanks, Ron [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
On Thu, 10 Nov 2005, Ron Ophir wrote: Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 The reason NAs are not allowed in subscripted assignments is based on numeric rather than logical subscripts. For numeric subscripts the problem is ambiguity about what the NA index should do (we know there is ambiguity because two parts of the R code did different things). For logical subscripts you could argue that the ambiguity isn't present and that if the index was NA the element should just be set to NA. This change might be worth making. I found a solution in term of: sapply(a,function(x) if (is(s.na)) NA else if (x0) (-1)/2^x else 2^x) but still I would like to understand why the solution above did not work. I think is more ellegant. A better general solution is a-ifelse(a0, -1/2^a, 2^a) An alternative for this problem that is faster when a is very large is a-sign(a)*2^abs(a) -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
Oops. Sorry. Should be: sign(a)*2^a where I assume you meant the inverse value should be -1/2^|a| = - 2^a for a0 -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: bgunter Sent: Thursday, November 10, 2005 9:00 AM To: Ron Ophir; r-help@stat.math.ethz.ch Subject: RE: [R] different functions on different vector subsets The error messages mean what they say. I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. ## I assume you mean 1/(2^a). If not, modify the following appropriately. 2^(a*sign(a)) ## will do As for your error message for: a[a0]-(-1)/2^a[a0] a0 has an NA at the first index and so R doesn't know what index you want to assign the value to. Ergo the error message. -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ron Ophir Sent: Thursday, November 10, 2005 7:26 AM To: r-help@stat.math.ethz.ch Subject: [R] different functions on different vector subsets Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 I found a solution in term of: sapply(a,function(x) if (is(s.na)) NA else if (x0) (-1)/2^x else 2^x) but still I would like to understand why the solution above did not work. I think is more ellegant. my R version is: sessionInfo() R version 2.2.0, 2005-10-06, i386-pc-mingw32 attached base packages: [1] methods stats graphics grDevices utils datasets [7] base Thanks, Ron [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Concurrent structures
@BOOK{Mandel1995,
title = {Analysis of two-way layouts},
publisher = {Chapman \ Hall},
year = {1995},
address = {New York, NY, USA},
author = {John Mandel},
}
describes diagnostics for identifying concurrent structures (i.e., y_
{ij} = A + B_{i} * C_{j} + e_{ij} ). Does anyone know of an
implementation in R?
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/
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Re: [R] R-help: conversion of long decimal numbers into hexadeci
On 10-Nov-05 Antje Döring wrote: Hi there, could somebody help me to convert a decimal number into a hexadecimal number? I know that there is the function sprintf, but the numbers I want to convert consist of 20 or more numbers. Spintf is not able to convert these big numbers. If I understand aright, you have decimal integers with 20 or more digits (and you want to get these as hexadecimal). You are probably out of luck for a direct approach, since 10^20 2^64 (indeed 2^66), so you will have overflowed a 64-bit integer. However, I'm not sure what the limitations on integer types are in R on all platforms. If, however, all you need is to do these conversions, and you do not really need to use R (how off-topic can I get ... ?), then (at any rate on Linux/Unix systems where the program is installed by default) you can use the aribitrary-precision calculator 'bc'. Session: $ 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'. obase=16 1234567898765432123456789 1056E0F555A18EBDA7D15 123456789876543212345678987654321 6163E6712EBBAA4E3D62B41F4B1 12345678987654321234567898765432123456789876543212345678987654321 1E02BC221DC9369C8981C6F859501BD313D339F09180862B41F4B1 quit Und so weiter ... and of course you can go in the opposite direction by ibase=16 (to set hex as the input base) and obase=10 (to set decimal as the output base). 'bc' is a classic Unix tool, and features as an illoustration of complex programming in C, with lex and yacc and all, in The Unix Programming Environment (as I recall) by Kernighan and Ritchie. I don't need it often, but when you need it it's very handy (e.g. now). Hoping this helps, Ted. PS: $ bc -l 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'. scale=1000 pi=4*a(1) pi 3.141592653589793238462643383279502884197169399375105820974944592307\ 81640628620899862803482534211706798214808651328230664709384460955058\ 22317253594081284811174502841027019385211055596446229489549303819644\ 28810975665933446128475648233786783165271201909145648566923460348610\ .. 08302642522308253344685035261931188171010003137838752886587533208381\ 42061717766914730359825349042875546873115956286388235378759375195778\ 18577805321712268066130019278766111959092164201988 (last digit wrong because of truncation) E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 10-Nov-05 Time: 17:28:05 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
On Thu, 10 Nov 2005, Ron Ophir wrote: Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 I found a solution in term of: sapply(a,function(x) if (is(s.na)) NA else if (x0) (-1)/2^x else 2^x) but still I would like to understand why the solution above did not work. I think is more ellegant. What do you think the NA value in a 0 [1]NA TRUE TRUE TRUE FALSE FALSE FALSE means? Should you replace a[1] or not? You are saying you don't know, so what is R to do? It tells you to make up your mind. Try ind - !is.na(a) a 0 a[ind] - 2^a[ind] ind - !is.na(a) a 0 a[ind] - (-1)/2^a[ind] or use ifelse as in ifelse(a 0, 2^a, -1/2^a) which is a lot more elegant. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] ltext - adding text to each panel from a matrix
Haven't checked it too carefully, but how about:
bwplot(treatment~foo|group, data = dat,
panel=function(x,y,...) {
panel.violin(x,y, ..., col = transparent, varwidth = F)
gt0 - table( x 0, y)
panel.abline(v=0, lty = dotted)
grid.text(as.character(gt0[1,]), unit(1, 'lines'), unit(1:3,
'native'), just='left')
grid.text(as.character(gt0[2,]), unit(1, 'npc') - unit(1,
'lines'), unit(1:3, 'native'), just='left')
},
par.strip.text = list(cex = 0.8), xlim = my.xlim)
--Matt
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of Andy Bunn
Sent: Thursday, November 10, 2005 8:19 AM
To: R-Help
Subject: [R] ltext - adding text to each panel from a matrix
Hi all (really probably just Deepayan):
In the plot below I want to add text on either side of each
violin plot that
indicates the number of observations that are either positive
or negative.
I'm trying to do this with ltext() and I've also monkeyed about with
panel.text(). The code below is generally what I want but my calls to
ltext() are wrong and I'm not sure how to fix them. Right now
they replicate
the first column of the matrices obs.pos and obs.neg for each
panel. How do
I tell ltext to advance to the next column when the next
panel is plotted? I
don't see how subscripts can do it, but I bet it's something
along that
line...
Thanks, Andy
rm(list = ls())
set.seed(354)
# make a bimodal dataset with three groups and three treatments
foo - c(rnorm(150,-1,0.5), rnorm(150,1,0.25))
treatment - factor(rep(seq(1,3),100), labels = c(Treatment
1, Treatment
2, Treatment 3))
group - factor(rep(seq(1,3),100), labels = c(Group A,
Group B, Group
C))
group - sample(group)
# corrupt Group A, Treatment 2 for fun.
foo[group==Group A treatment==Treatment 2][1:8] - rnorm(8,1,1)
dat - data.frame(foo,treatment,group)
# set the limits for the plot, which also tells where to put the text
my.xlim - c(-6, 6)
# make a matrix that counts the number of obs greater or less
than zero
obs.pos - tapply(dat[dat[,1] 0,1], dat[dat[,1] 0,-1], length)
obs.neg - tapply(dat[dat[,1] = 0,1], dat[dat[,1] = 0,-1], length)
#write some coordinate data
x.obs.pos - rep(my.xlim[2],dim(obs.pos)[2])
y.obs.pos - 1:dim(obs.pos)[2]
x.obs.neg - rep(my.xlim[1],dim(obs.neg)[2])
y.obs.neg - 1:dim(obs.neg)[2]
bwplot(treatment~foo|group, data = dat,
panel=function(...) {
panel.violin(..., col = transparent, varwidth = F)
panel.abline(v=0, lty = dotted)
ltext(x.obs.pos, y.obs.pos, obs.pos, pos = 2)
ltext(x.obs.neg, y.obs.neg, obs.neg, pos = 4)
},
par.strip.text = list(cex = 0.8), xlim = my.xlim)
obs.pos
obs.neg
# note that the numbers in the plot only match the matrices
for Group A,
# which is the first panel. Alas.
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[R] specifying a key for a trellis display
Folks: The key argument of trellis commands (e.g. xyplot()) allows one to place a key at the top of a trellis display using key=list(space='top',...) I would like to increase the space between the bottom of the key and the trellis plots beyond the default. Is there a simple way to do this? At present, I add an empty row (e.g. text = '', point colored in background color) to the bottom of each key column. This seems a bit of a kludge. Is there a slicker way to do it, i.e. a parameter that I have missed? Cheers, Bert -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] ltext - adding text to each panel from a matrix
On 11/10/05, Andy Bunn [EMAIL PROTECTED] wrote:
Hi all (really probably just Deepayan):
In the plot below I want to add text on either side of each violin plot
that
indicates the number of observations that are either positive or negative.
I'm trying to do this with ltext() and I've also monkeyed about with
panel.text().
They are the same (at least for now).
The code below is generally what I want but my calls to
ltext() are wrong and I'm not sure how to fix them. Right now they
replicate
the first column of the matrices obs.pos and obs.neg for each panel. How do
I tell ltext to advance to the next column when the next panel is plotted?
I
don't see how subscripts can do it, but I bet it's something along that
line...
Maybe, but there's a more direct solution (somewhat artificial perhaps):
bwplot(treatment~foo|group, data = dat,
panel=function(..., packet.number) {
panel.violin(..., col = transparent, varwidth = F)
panel.abline(v=0, lty = dotted)
ltext(x.obs.pos, y.obs.pos, obs.pos[, packet.number], pos = 2)
ltext(x.obs.neg, y.obs.neg, obs.neg[, packet.number], pos = 4)
},
par.strip.text = list(cex = 0.8), xlim = my.xlim)
This is documented under 'panel' in ?xyplot.
Deepayan
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Re: [R] write.table read.table with Dates
In addition to the solutions already provided, note that if *all* you
want to do is save your dataframe in a file, and later recreate it
from that file, you can use dump().
dump('Tall',file='Tall.r')
rm(Tall) ## just to demonstrate that the next command will recreate Tall
source('Tall.r')
-Don
At 2:21 PM +0100 11/10/05, JeeBee wrote:
I've found several similar issues with write.table/read.table
with Dates on this list, but trying to follow this advice I still
get an error.
First, I read in data from several files, constructing several date/time
columns using ISOdatetime
str(Tall$Begin)
'POSIXct', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02
00:00:00 ...
length(Tall$Begin)
[1] 40114
class(Tall$Begin)
[1] POSIXt POSIXct
This looks good (time is not always 00:00:00 ...)
This data came from several files, now I want to store the result I have
in data.frame Tall and be able to retrieve this quickly some other time.
This is what I do:
write.table(Tall, file=somefile.csv, sep=,, qmethod=double,
row.names=FALSE)
Later, I do this to read the file again:
fieldnames=c(Begin,test-a,test-b,Eind)
T=read.table(file = somefile.csv, col.names = fieldnames,
header = TRUE, sep = ,, quote=\, fill=FALSE)
I understand T$Begin now is a factor. I tried to simply convert it
again using (as I read on this mailinglist ...):
Q = strptime(as.character(T$Begin),format=%Y-%m-%d %H:%M:%S)
Q is looking good, though its length I don't understand .. is it a list or
something? It seems there are 40114 values in there somewhere...
class(Q)
[1] POSIXt POSIXlt
length(Q)
[1] 9
str(Q)
'POSIXlt', format: chr [1:40114] 2005-10-02 00:00:00 2005-10-02
00:00:00 ...
T$Begin = Q ### yields this error
Error in $-.data.frame(`*tmp*`, Begin, value = list(sec = c(0, 0, :
replacement has 9 rows, data has 40114
Could somebody explain me how to convert the date column?
Or perhaps there is an easier way?
Thanks in advance for your time.
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--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
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Re: [R] order statistics / sample quantiles
On Thu, 10 Nov 2005, Bickel, David wrote: Are there any R functions or packages that can compute distributions, expectations, or quantiles of order statistics (or sample quantiles or extreme values) for a given distribution such as a normal distribution? Both exact and asymptotic calculations are of interest. I am already aware of the 'quantile' function of 'stats'. The density function of the order statistics is given as an example in the FAQ. This can then be integrated with integrate() to give expectations. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] specifying a key for a trellis display
On 11/10/05, Berton Gunter [EMAIL PROTECTED] wrote: Folks: The key argument of trellis commands (e.g. xyplot()) allows one to place a key at the top of a trellis display using key=list(space='top',...) I would like to increase the space between the bottom of the key and the trellis plots beyond the default. Is there a simple way to do this? At present, I add an empty row (e.g. text = '', point colored in background color) to the bottom of each key column. This seems a bit of a kludge. Is there a slicker way to do it, i.e. a parameter that I have missed? Yes, trellis.par.get(layout.heights). e.g. xyplot(1 ~ 1, key = list(text = list(letters[1:3]), points = list(col = 1:3)), par.settings = list(layout.heights = list(key.axis.padding = 5))) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] specifying a key for a trellis display
Many thanks, Deepayan. As I suspected... I'll fool around with these key arguments to see what they do. -- Bert -Original Message- From: Deepayan Sarkar [mailto:[EMAIL PROTECTED] Sent: Thursday, November 10, 2005 11:31 AM To: Berton Gunter Cc: r-help@stat.math.ethz.ch Subject: Re: specifying a key for a trellis display On 11/10/05, Berton Gunter [EMAIL PROTECTED] wrote: Folks: The key argument of trellis commands (e.g. xyplot()) allows one to place a key at the top of a trellis display using key=list(space='top',...) I would like to increase the space between the bottom of the key and the trellis plots beyond the default. Is there a simple way to do this? At present, I add an empty row (e.g. text = '', point colored in background color) to the bottom of each key column. This seems a bit of a kludge. Is there a slicker way to do it, i.e. a parameter that I have missed? Yes, trellis.par.get(layout.heights). e.g. xyplot(1 ~ 1, key = list(text = list(letters[1:3]), points = list(col = 1:3)), par.settings = list(layout.heights = list(key.axis.padding = 5))) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
Thanks Thomas, ...For logical subscripts you could argue that the ambiguity isn't present and that if the index was NA the element should just be set to NA. This change might be worth making. I see you got my point. NA should return NA no matter what the comparison is. But any way thanks Brian, Jim, and Berton, I have leaned a lot. It was a good practice. Ron Ron Ophir, Ph.D. Bioinformatician, Biological Services Weizmann Institute of Science POB 26 Rehovot 76100 Israel e-mail: [EMAIL PROTECTED] Phone: 972-8-9342614 Fax:972-8-9344113 Thomas Lumley [EMAIL PROTECTED] 11/10/05 7:04 PM On Thu, 10 Nov 2005, Ron Ophir wrote: Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 The reason NAs are not allowed in subscripted assignments is based on numeric rather than logical subscripts. For numeric subscripts the problem is ambiguity about what the NA index should do (we know there is ambiguity because two parts of the R code did different things). For logical subscripts you could argue that the ambiguity isn't present and that if the index was NA the element should just be set to NA. This change might be worth making. I found a solution in term of: sapply(a,function(x) if (is(s.na)) NA else if (x0) (-1)/2^x else 2^x) but still I would like to understand why the solution above did not work. I think is more ellegant. A better general solution is a-ifelse(a0, -1/2^a, 2^a) An alternative for this problem that is faster when a is very large is a-sign(a)*2^abs(a) -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
On Thu, 10 Nov 2005, Thomas Lumley wrote: On Thu, 10 Nov 2005, Ron Ophir wrote: Hi, I am trying to apply two different functions on on a vector as follow: a-c(NA,1,2,3,-3,-4,-6) if a0 I would like to raise it by the power of 2: 2^a and if the a0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a0]-2^[a0] a[a0]-(-1)/2^a[a0] I got the following error Error: NAs are not allowed in subscripted assignments any other manupulation that I did with is.na() but did not succeed. What is funny that the two sides of the assignment work and return the same vector size: 2^a[a0] [1] NA 2 4 8 a[a0] [1] NA 1 2 3 The reason NAs are not allowed in subscripted assignments is based on numeric rather than logical subscripts. For numeric subscripts the problem is ambiguity about what the NA index should do (we know there is ambiguity because two parts of the R code did different things). For logical subscripts you could argue that the ambiguity isn't present and that if the index was NA the element should just be set to NA. This change might be worth making. That presumes NA is a valid value, but in general it is not. (Not for raw, not for lists, not for data frames, ) I don't think we want such inconsistent behaviour. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] grid.remove() doesn't remove output
I've found that grid.remove() doesn't clear the output when the grob is the only one on the device (or viewport; I didn't test it). For example: library(grid) grid.newpage() grid.circle(name=cir, x=.5, y=.5, r=.3, gp=gpar(lwd=5)) grid.lines(c(.2, .8), c(.3, .7), name=lin) grid.remove(cir) # circle disappears grid.remove(lin) # object deleted, but line remains on output If I now draw another primitive, lin disappears. If I plotted only one thing, say cir, in the first place, grid.remove() won't clear the output. I'm using R 2.2. Any pointer is appreciated. Zepu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
On Thu, 10 Nov 2005, Ron Ophir wrote: Thanks Thomas, ...For logical subscripts you could argue that the ambiguity isn't present and that if the index was NA the element should just be set to NA. This change might be worth making. I see you got my point. NA should return NA no matter what the comparison is. I'm not sure that I did get your point. As Brian said, you aren't specifying whether or not to set the value. In your example it didn't matter because it would end up NA either way. I was saying that for eg a-c(1,2,3,4) b-c(NA,T,F,T) a[b]-7 we could relax the prohibition on NA indexing to give c(NA,7,7,7) as the result. In your case that would give what you wanted, but in other cases it might not. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] different functions on different vector subsets
On Thu, 10 Nov 2005, Thomas Lumley wrote: On Thu, 10 Nov 2005, Ron Ophir wrote: Thanks Thomas, ...For logical subscripts you could argue that the ambiguity isn't present and that if the index was NA the element should just be set to NA. This change might be worth making. I see you got my point. NA should return NA no matter what the comparison is. I'm not sure that I did get your point. As Brian said, you aren't specifying whether or not to set the value. In your example it didn't matter because it would end up NA either way. And as Brian later pointed out, this approach wouldn't work for things other than simple vectors. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] make check failed on linux-amd64 using PGI compilers
Dear R-help, I am trying to build R-2.2.0-patched (2005-11-07 r36217) on the head node of a Scyld cluster (dual Opteron 250s) using PGI compilers (v6.0). I used the flags suggested by Jennifer Lai on R-devel (taken from R-admin, except that I had to add -L/usr/X11R6/lib64 to LDFLAGS). The build went fine, but make check-all failed when running tests/Examples/graphics-Ex.R, at: plot(1:2, xaxs = i) # 'inner-axis' w/o extra space stopifnot(par(xaxp)[1:2] == 1:2 + par(usr)[1:2] == 1:2) Error: par(xaxp)]1:2] == 1:2 par(usr)[1:2] == 1:2 is not TRUE The above looks a bit strange to me, as running the R built as above, par(xaxp)[1:2] - 1:2 gives [1] -1.110223e-16 -2.220446e-16 Is my R build faulty? I'd very much appreciate any advise. Best, Andy __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] IF/Else
Hi,
I am trying to write a for loop with if else statements to calculate
biomass density estimates for different types of sampling gear.
My code is:
bmd=for (i in 1:length(Gear)){
if (Gear==20) {bioden=Biomass/141}
else {if (Gear==23) {bioden=Biomass/68}}
else {if (Gear==160) {bioden=Biomass/4120}}
else {if (Gear==170) {bioden=Biomass/2210}}
else {if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)}}
else {if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}}
}
The syntax that is returned is:
bmd=for (i in 1:length(Gear)){
+ if (Gear==20) {bioden=Biomass/141}
+ else {if (Gear==23) {bioden=Biomass/68}}
+ else {if (Gear==160) {bioden=Biomass/4120}}
Error: syntax error in:
else {if (Gear==23) {bioden=Biomass/68}}
else
else {if (Gear==170) {bioden=Biomass/2210}}
Error: syntax error in else
else {if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)}}
Error: syntax error in else
else {if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}}
Error: syntax error in else
}
Error: syntax error in }
It appears that the code works for the first two if/else statements and
then fails there after. Any suggestions?
Cameron Guenther
Associate Research Scientist
FWC/FWRI, Marine Fisheries Research
100 8th Avenue S.E.
St. Petersburg, FL 33701
(727)896-8626 Ext. 4305
[EMAIL PROTECTED]
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[R] How to export multiple files using write.table in the loop?
Hi,
I tried to split a big file into some small files seperately by R. I can only
do that writing duplicated codes. When I tried to write a loop, I only got one
appned or destroyed exported file. For example:
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-subset(data1,select=c(V1,V2,V3))
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-subset(data1,select=c(V1,V4,V5))
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
or
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-data.frame(data1[,1],data1[,2],data1[,3]
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-data.frame(data1[,1],data1[,4],data1[,5]
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
I tried to write a loop for it like :
i - 1
while (i 3)
{
qq-data.frame(test[,1],test[,2[i],test[,2[i]+1])
write.table(qq,file=C:\\Alice\\bb.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
i - i+1;
}
But it's not right. I tried hard to revise it but I just can't make it. If
someone can help me out here, your help is really greatly appreciated. Thank
you sooo much!
Best,
Alice
-
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[R] converting a character string to a subscripted numeric variable
Dear R-helpers: It seems that I have a mental block. (Some say that it sits atop my shoulders.) For reasons too tedious to retell I have an R object: input.line[7] [1] -13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76, I would like to convert this into a subscripted variable, Beta, something that should be straightforward if I had *almost* what I have. I'd like to say Beta - c(-13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76) but I can't because: input.line[7] is a character string, and it ends in a comma. This cannot be as difficult as I have found it to be. Can anyone help? Copious Thanks. Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] converting a character string to a subscripted numeric variable
Charles Annis, P.E. wrote: Dear R-helpers: It seems that I have a mental block. (Some say that it sits atop my shoulders.) For reasons too tedious to retell I have an R object: input.line[7] [1] -13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76, I would like to convert this into a subscripted variable, Beta, something that should be straightforward if I had *almost* what I have. I'd like to say Beta - c(-13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76) but I can't because: input.line[7] is a character string, and it ends in a comma. This cannot be as difficult as I have found it to be. Can anyone help? Copious Thanks. How about: x - -13.24, -11.24, x - as.numeric(strsplit(x, ,)[[1]]) x[!is.na(x)] --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] converting a character string to a subscripted numeric variable
Eternal gratitude to Sunbar and Matt and Patrick! The easy solution is Beta - as.numeric(strsplit(input.line [7], ,)[[1]]) Beta - Beta[!is.na(Beta)] Beta I have a slew of files to interrogate and need to know from some of the input, what to look for in the remainder of the input. Thanks to all! Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: Sundar Dorai-Raj [mailto:[EMAIL PROTECTED] Sent: Thursday, November 10, 2005 5:00 PM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [R] converting a character string to a subscripted numeric variable Charles Annis, P.E. wrote: Dear R-helpers: It seems that I have a mental block. (Some say that it sits atop my shoulders.) For reasons too tedious to retell I have an R object: input.line[7] [1] -13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76, I would like to convert this into a subscripted variable, Beta, something that should be straightforward if I had *almost* what I have. I'd like to say Beta - c(-13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76) but I can't because: input.line[7] is a character string, and it ends in a comma. This cannot be as difficult as I have found it to be. Can anyone help? Copious Thanks. How about: x - -13.24, -11.24, x - as.numeric(strsplit(x, ,)[[1]]) x[!is.na(x)] --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] make check failed on linux-amd64 using PGI compilers
Liaw, Andy [EMAIL PROTECTED] writes: Dear R-help, I am trying to build R-2.2.0-patched (2005-11-07 r36217) on the head node of a Scyld cluster (dual Opteron 250s) using PGI compilers (v6.0). I used the flags suggested by Jennifer Lai on R-devel (taken from R-admin, except that I had to add -L/usr/X11R6/lib64 to LDFLAGS). The build went fine, but make check-all failed when running tests/Examples/graphics-Ex.R, at: plot(1:2, xaxs = i) # 'inner-axis' w/o extra space stopifnot(par(xaxp)[1:2] == 1:2 + par(usr)[1:2] == 1:2) Error: par(xaxp)]1:2] == 1:2 par(usr)[1:2] == 1:2 is not TRUE The above looks a bit strange to me, as running the R built as above, par(xaxp)[1:2] - 1:2 gives [1] -1.110223e-16 -2.220446e-16 Is my R build faulty? I'd very much appreciate any advise. Relative errors of 1 ULP would not ruin my sleep. It's not obvious to me which is the floating point operation responsible for the effect, though, and it is a bit suspicious given that it should just be the x-range of the data. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] converting a character string to a subscripted numeric variable
Charles Annis, P.E. [EMAIL PROTECTED] writes: Eternal gratitude to Sunbar and Matt and Patrick! The easy solution is Beta - as.numeric(strsplit(input.line [7], ,)[[1]]) Beta - Beta[!is.na(Beta)] Beta Also x - input.line [7] eval(parse(text=paste(c(, x, or x - sub(, *$,,x) scan(textConnection(x), sep=,) I have a slew of files to interrogate and need to know from some of the input, what to look for in the remainder of the input. Thanks to all! Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: Sundar Dorai-Raj [mailto:[EMAIL PROTECTED] Sent: Thursday, November 10, 2005 5:00 PM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [R] converting a character string to a subscripted numeric variable Charles Annis, P.E. wrote: Dear R-helpers: It seems that I have a mental block. (Some say that it sits atop my shoulders.) For reasons too tedious to retell I have an R object: input.line[7] [1] -13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76, I would like to convert this into a subscripted variable, Beta, something that should be straightforward if I had *almost* what I have. I'd like to say Beta - c(-13.24, -11.24, -9.24, -7.24, -5.24, -3.24, -1.24, 0.76, 2.76, 4.76, 6.76, 8.76, 10.76, 12.76, 14.76, 16.76, 18.76, 20.76, 22.76, 24.76, 26.76, 28.76, 30.76, 32.76, 34.76, 36.76, 38.76, 40.76, 42.76, 44.76) but I can't because: input.line[7] is a character string, and it ends in a comma. This cannot be as difficult as I have found it to be. Can anyone help? Copious Thanks. How about: x - -13.24, -11.24, x - as.numeric(strsplit(x, ,)[[1]]) x[!is.na(x)] --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] IF/Else
On 10-Nov-05 Guenther, Cameron wrote:
Hi,
I am trying to write a for loop with if else statements to calculate
biomass density estimates for different types of sampling gear.
My code is:
bmd=for (i in 1:length(Gear)){
if (Gear==20) {bioden=Biomass/141}
else {if (Gear==23) {bioden=Biomass/68}}
else {if (Gear==160) {bioden=Biomass/4120}}
else {if (Gear==170) {bioden=Biomass/2210}}
else {if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)}}
else {if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}}
}
In the above, you have in effect written a string of elses
following a single if, since the {if(Gear==160){...}} isolates
its if from the preceding else, etc.. But there are other
things wrong.
The following should work:
bmd=for (i in 1:length(Gear)) {
if (Gear==20) {bioden=Biomass/141} else
if (Gear==23) {bioden=Biomass/68} else
if (Gear==160) {bioden=Biomass/4120} else
if (Gear==170) {bioden=Biomass/2210} else
if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)} else
if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}
}
where I have removed superfluous { on the left and } on the
right.
Putting the else at the end of the line, rather than on the
next, is needed since otherwise the line is a completed
statement befor the else is encountered and then the else
is a syntax error. With else at the end of the line, the
end of line is reached on an incomplete statement so R goes
on to the next line before making up its mind about it.
I.e. Good:
A-2
if(A==1){B-1} else
if(A==2){B-2} else
if(A==3){B-3}
while Bad:
A-2
if(A==1){B-1}
else if(A==2){B-2}
else if(A==3){B-3}
Also Bad:
A-2
if(A==1){B-1} else
{if(A==2){B-2}} else
{if(A==3){B-3}}
since each {if(A==2){B-2}} is a complete statement delimited
by {...} and the else is then again a syntax error, The final
statement is OK, however.
Mind you, if the cases in your code are the only possibilites
for Gear, then you don't need the final if at all -- just the
assignment, since if you get that far it's the only case left,
and you'll only reach it by falling through the preceding else.
Indeed, you could even dispense with all of the elses and
just use if statements, since only one of them will be
satisfied; but of course that is a bit inefficient in that
every statement will be tested even if one has already been
satisfied.
Hoping this helps,
Ted.
The syntax that is returned is:
bmd=for (i in 1:length(Gear)){
+ if (Gear==20) {bioden=Biomass/141}
+ else {if (Gear==23) {bioden=Biomass/68}}
+ else {if (Gear==160) {bioden=Biomass/4120}}
Error: syntax error in:
else {if (Gear==23) {bioden=Biomass/68}}
else
else {if (Gear==170) {bioden=Biomass/2210}}
Error: syntax error in else
else {if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)}}
Error: syntax error in else
else {if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}}
Error: syntax error in else
}
Error: syntax error in }
It appears that the code works for the first two if/else statements and
then fails there after. Any suggestions?
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 10-Nov-05 Time: 22:44:59
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Re: [R] make check failed on linux-amd64 using PGI compilers
From: [EMAIL PROTECTED] Liaw, Andy [EMAIL PROTECTED] writes: Dear R-help, I am trying to build R-2.2.0-patched (2005-11-07 r36217) on the head node of a Scyld cluster (dual Opteron 250s) using PGI compilers (v6.0). I used the flags suggested by Jennifer Lai on R-devel (taken from R-admin, except that I had to add -L/usr/X11R6/lib64 to LDFLAGS). The build went fine, but make check-all failed when running tests/Examples/graphics-Ex.R, at: plot(1:2, xaxs = i) # 'inner-axis' w/o extra space stopifnot(par(xaxp)[1:2] == 1:2 + par(usr)[1:2] == 1:2) Error: par(xaxp)]1:2] == 1:2 par(usr)[1:2] == 1:2 is not TRUE The above looks a bit strange to me, as running the R built as above, par(xaxp)[1:2] - 1:2 gives [1] -1.110223e-16 -2.220446e-16 Is my R build faulty? I'd very much appreciate any advise. Relative errors of 1 ULP would not ruin my sleep. It's not obvious to me which is the floating point operation responsible for the effect, though, and it is a bit suspicious given that it should just be the x-range of the data. It's the par() output that has storage mode double, as that stores the coordinates. 1:2 has storage mode integer. Those comparisons looked a bit suspicious to me. Isn't it recommended time and again on this list not to test for exact equality of numerics? all.equal() of the above does return TRUE. Andy -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Interpretation of output from glm
Dear John, Thanks for the pointers. I will read this. Pedro At 14:41 10/11/2005, you wrote: Dear Pedro, The basic point, which relates to the principle of marginality in formulating linear models, applies whether the predictors are factors, covariates, or both. I think that this is a common topic in books on linear models; I certainly discuss it in my Applied Regression, Linear Models, and Related Methods. Regards, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Pedro de Barros Sent: Wednesday, November 09, 2005 10:45 AM To: r-help@stat.math.ethz.ch Subject: Re: [R] Interpretation of output from glm Importance: High Dear John, Thanks for the quick reply. I did indeed have these ideas, but somehow floating, and all I could find about this mentioned categorical predictors. Can you suggest a good book where I could try to learn more about this? Thanks again, Pedro At 01:49 09/11/2005, you wrote: Dear Pedro, -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Pedro de Barros Sent: Tuesday, November 08, 2005 9:47 AM To: r-help@stat.math.ethz.ch Subject: [R] Interpretation of output from glm Importance: High I am fitting a logistic model to binary data. The response variable is a factor (0 or 1) and all predictors are continuous variables. The main predictor is LT (I expect a logistic relation between LT and the probability of being mature) and the other are variables I expect to modify this relation. I want to test if all predictors contribute significantly for the fit or not I fit the full model, and get these results summary(HMMaturation.glmfit.Full) Call: glm(formula = Mature ~ LT + CondF + Biom + LT:CondF + LT:Biom, family = binomial(link = logit), data = HMIndSamples) Deviance Residuals: Min 1Q Median 3Q Max -3.0983 -0.7620 0.2540 0.7202 2.0292 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) -8.789e-01 3.694e-01 -2.379 0.01735 * LT 5.372e-02 1.798e-02 2.987 0.00281 ** CondF -6.763e-02 9.296e-03 -7.275 3.46e-13 *** Biom-1.375e-02 2.005e-03 -6.856 7.07e-12 *** LT:CondF 2.434e-03 3.813e-04 6.383 1.74e-10 *** LT:Biom 7.833e-04 9.614e-05 8.148 3.71e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 10272.4 on 8224 degrees of freedom Residual deviance: 7185.8 on 8219 degrees of freedom AIC: 7197.8 Number of Fisher Scoring iterations: 8 However, when I run anova on the fit, I get anova(HMMaturation.glmfit.Full, test='Chisq') Analysis of Deviance Table Model: binomial, link: logit Response: Mature Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(|Chi|) NULL822410272.4 LT 1 2873.8 8223 7398.7 0.0 CondF 1 0.1 8222 7398.5 0.7 Biom1 0.2 8221 7398.3 0.7 LT:CondF1142.1 8220 7256.3 9.413e-33 LT:Biom 1 70.4 8219 7185.8 4.763e-17 Warning message: fitted probabilities numerically 0 or 1 occurred in: method(x = x[, varseq = i, drop = FALSE], y = object$y, weights = object$prior.weights, I am having a little difficulty interpreting these results. The result from the fit tells me that all predictors are significant, while the anova indicates that besides LT (the main variable), only the interaction of the other terms is significant, but the main effects are not. I believe that in the first output (on the glm object), the significance of all terms is calculated considering each of them alone in the model (i.e. removing all other terms), while the anova output is (as it says) considering the sequential addition of the terms. So, there are 2 questions: a) Can I tell that the interactions are significant, but not the main effects? In a model with this structure, the main effects represent slopes over the origin (i.e., where the other variables in the product terms are 0), and aren't meaningfully interpreted as main effects. (Is there even any data near the origin?) b) Is it legitimate to consider a model where the interactions are considered, but not the main effects CondF and Biom? Generally, no: That is, such a model
Re: [R] IF/Else
Try using switch:
bioden - Biomass /
switch(paste(Gear),
`20` = 141,
`23` = 68,
# ... fill in the ones I have omitted ...,
`301` = DIST_TOW*4*1853)
On 11/10/05, Guenther, Cameron [EMAIL PROTECTED] wrote:
Hi,
I am trying to write a for loop with if else statements to calculate
biomass density estimates for different types of sampling gear.
My code is:
bmd=for (i in 1:length(Gear)){
if (Gear==20) {bioden=Biomass/141}
else {if (Gear==23) {bioden=Biomass/68}}
else {if (Gear==160) {bioden=Biomass/4120}}
else {if (Gear==170) {bioden=Biomass/2210}}
else {if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)}}
else {if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}}
}
The syntax that is returned is:
bmd=for (i in 1:length(Gear)){
+ if (Gear==20) {bioden=Biomass/141}
+ else {if (Gear==23) {bioden=Biomass/68}}
+ else {if (Gear==160) {bioden=Biomass/4120}}
Error: syntax error in:
else {if (Gear==23) {bioden=Biomass/68}}
else
else {if (Gear==170) {bioden=Biomass/2210}}
Error: syntax error in else
else {if (Gear==300) {bioden=Biomass/(DIST_TOW*4*1853)}}
Error: syntax error in else
else {if (Gear==301) {bioden=Biomass/(DIST_TOW*4*1853)}}
Error: syntax error in else
}
Error: syntax error in }
It appears that the code works for the first two if/else statements and
then fails there after. Any suggestions?
Cameron Guenther
Associate Research Scientist
FWC/FWRI, Marine Fisheries Research
100 8th Avenue S.E.
St. Petersburg, FL 33701
(727)896-8626 Ext. 4305
[EMAIL PROTECTED]
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Re: [R] How to export multiple files using write.table in the loop?
Try this:
col.list - list(1:3, c(1,4:5))
for(cols in col.list) write.table(data1[,cols], ...whatever...)
On 11/10/05, alice.0309 [EMAIL PROTECTED] wrote:
Hi,
I tried to split a big file into some small files seperately by R. I can
only do that writing duplicated codes. When I tried to write a loop, I only
got one appned or destroyed exported file. For example:
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-subset(data1,select=c(V1,V2,V3))
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-subset(data1,select=c(V1,V4,V5))
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
or
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-data.frame(data1[,1],data1[,2],data1[,3]
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-data.frame(data1[,1],data1[,4],data1[,5]
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
I tried to write a loop for it like :
i - 1
while (i 3)
{
qq-data.frame(test[,1],test[,2[i],test[,2[i]+1])
write.table(qq,file=C:\\Alice\\bb.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
i - i+1;
}
But it's not right. I tried hard to revise it but I just can't make it. If
someone can help me out here, your help is really greatly appreciated. Thank
you sooo much!
Best,
Alice
-
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[R] following Appendix A results in plot.new has not been called yet
Hello. I was exploring the R software package and received the error message plot.new has not been called yet when following Appendix A A Sample Session of R-intro.pdf. I searched the message archives and found no similiar report to mine. I am using R on CentOS and I am using the latest from R-project, version 2.2.0. Here is my .Rhistory file recorded when following the appendix after the first removal of variables (e.g. rm(x,y) ): x - 1:20 w - 1 + sqrt(x)/2 dummy - data.frame(x=x, y=x + rnorm(x)*w) dummy fm - lm(y ~ x, data=dummy) summary(fm) fm1 - lm(y ~ x, data=dummy, weight=1/w^2) summary(fm1) attach(dummy) lrf - lowess(x,y) plot(x,y) lines(x, lrf$y) Results in: Error in plot.xy(xy.coords(x, y), type = type, col = col, lty = lty, ...) : plot.new has not been called yet If I continue along with the list of commands, I yield similiar errors. I am unsure if this is related to using CentOS, my installation, bad steps in the pdf file, or a bug in the software. I am leaning towards the steps in the pdf. Regards, Daniel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] following Appendix A results in plot.new has not been called yet
You left out y - rnorm(x) On 11/10/05, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Hello. I was exploring the R software package and received the error message plot.new has not been called yet when following Appendix A A Sample Session of R-intro.pdf. I searched the message archives and found no similiar report to mine. I am using R on CentOS and I am using the latest from R-project, version 2.2.0. Here is my .Rhistory file recorded when following the appendix after the first removal of variables (e.g. rm(x,y) ): x - 1:20 w - 1 + sqrt(x)/2 dummy - data.frame(x=x, y=x + rnorm(x)*w) dummy fm - lm(y ~ x, data=dummy) summary(fm) fm1 - lm(y ~ x, data=dummy, weight=1/w^2) summary(fm1) attach(dummy) lrf - lowess(x,y) plot(x,y) lines(x, lrf$y) Results in: Error in plot.xy(xy.coords(x, y), type = type, col = col, lty = lty, ...) : plot.new has not been called yet If I continue along with the list of commands, I yield similiar errors. I am unsure if this is related to using CentOS, my installation, bad steps in the pdf file, or a bug in the software. I am leaning towards the steps in the pdf. Regards, Daniel __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Help regarding mas5 normalization
Please do not post to both BioConductor and R. On Thu, 2005-11-10 at 09:51 -0700, Nayeem Quayum wrote: Hello everybody, I am trying to use mas5 to normalize some array data and using mas5 and mas5calls. But I received these warning message. If anybody can explain the problem I would really appreciate that. Thanks in advance. background correction: mas PM/MM correction : mas expression values: mas background correcting...Warning message: 'loadURL' is deprecated. Use 'load(url())' instead. See help(Deprecated) Warning message: 'loadURL' is deprecated. Use 'load(url())' instead. See help(Deprecated) Warning message: 'loadURL' is deprecated. Use 'load(url())' instead. See help(Deprecated) There were 14 warnings (use warnings() to see them) Note: http://www.bioconductor.org/repository/devel/package/Win32 does not seem to have a valid repository, skipping Note: You did not specify a download type. Using a default value of: Source This will be fine for almost all users Error in FUN(X[[1]], ...) : no slot of name Uses for this object of class localPkg [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to export multiple files using write.table in the loop?
Hi Gabor,
Thank you very very much for your help! I'm still a little confused. Let me
re-state my problems here:
There are 29 columns in my large dataset(with 5 rows)
I need to split out 14 different files like this:
Col1, Col2, Col3--split1.txt
Col1, Col4, Col5--split2.txt
.
.
Col1, Col28,Col29-split14.txt
How can I do that using some functions or loop statement?
Sincerely,
Alice
Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
col.list - list(1:3, c(1,4:5))
for(cols in col.list) write.table(data1[,cols], ...whatever...)
On 11/10/05, alice.0309 wrote:
Hi,
I tried to split a big file into some small files seperately by R. I can only
do that writing duplicated codes. When I tried to write a loop, I only got
one appned or destroyed exported file. For example:
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-subset(data1,select=c(V1,V2,V3))
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-subset(data1,select=c(V1,V4,V5))
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
or
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-data.frame(data1[,1],data1[,2],data1[,3]
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-data.frame(data1[,1],data1[,4],data1[,5]
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
I tried to write a loop for it like :
i - 1
while (i 3)
{
qq-data.frame(test[,1],test[,2[i],test[,2[i]+1])
write.table(qq,file=C:\\Alice\\bb.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
i - i+1;
}
But it's not right. I tried hard to revise it but I just can't make it. If
someone can help me out here, your help is really greatly appreciated. Thank
you sooo much!
Best,
Alice
-
[[alternative HTML version deleted]]
__
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-
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Re: [R] How to export multiple files using write.table in the loop?
Hello!
I've solved the problems so don't bother. I can use Paste function like this:
i - 1
while (i 15)
{
s-data.frame(data1[,1],data1[,(2*i)],data1[,(2*i+1)])
write.table(s,paste(F:\\s.txt,i,sep=.),quote=FALSE,row.names=FALSE,col.names=FALSE)
i - i+1;
}
Thank you very much and sorry for multiple e-mails!
Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
col.list - list(1:3, c(1,4:5))
for(cols in col.list) write.table(data1[,cols], ...whatever...)
On 11/10/05, alice.0309 wrote:
Hi,
I tried to split a big file into some small files seperately by R. I can only
do that writing duplicated codes. When I tried to write a loop, I only got
one appned or destroyed exported file. For example:
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-subset(data1,select=c(V1,V2,V3))
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-subset(data1,select=c(V1,V4,V5))
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
or
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t, check.names=FALSE)
a-data.frame(data1[,1],data1[,2],data1[,3]
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-data.frame(data1[,1],data1[,4],data1[,5]
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
I tried to write a loop for it like :
i - 1
while (i 3)
{
qq-data.frame(test[,1],test[,2[i],test[,2[i]+1])
write.table(qq,file=C:\\Alice\\bb.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
i - i+1;
}
But it's not right. I tried hard to revise it but I just can't make it. If
someone can help me out here, your help is really greatly appreciated. Thank
you sooo much!
Best,
Alice
-
[[alternative HTML version deleted]]
__
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Re: [R] How to export multiple files using write.table in the loop?
In that case you can simplify it a bit like this:
fn - /s.txt
for(i in 1:14)
write.table(data1[, c(1, 2*i, 2*i+1)], paste(fn, i, sep = .),
quote = FALSE, row.names = FALSE, col.names = FALSE)
On 11/10/05, alice.0309 [EMAIL PROTECTED] wrote:
Hello!
I've solved the problems so don't bother. I can use Paste function like
this:
i - 1
while (i 15)
{
s-data.frame(data1[,1],data1[,(2*i)],data1[,(2*i+1)])
write.table(s,paste(F:\\s.txt,i,sep=.),quote=FALSE,row.names=FALSE,col.names=FALSE)
i - i+1;
}
Thank you very much and sorry for multiple e-mails!
Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try this:
col.list - list(1:3, c(1,4:5))
for(cols in col.list) write.table(data1[,cols], ...whatever...)
On 11/10/05, alice.0309 wrote:
Hi,
I tried to split a big file into some small files seperately by R. I can
only do that writing duplicated codes. When I tried to write a loop, I only
got one appned or destroyed exported file. For example:
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t,
check.names=FALSE)
a-subset(data1,select=c(V1,V2,V3))
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-subset(data1,select=c(V1,V4,V5))
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
or
data1- read.table(file = C:\\Alice\\MBEI.txt, sep=\t,
check.names=FALSE)
a-data.frame(data1[,1],data1[,2],data1[,3]
write.table-(a,file=C:\\Alice\\aa1.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
a-data.frame(data1[,1],data1[,4],data1[,5]
write.table(a,file=C:\\Alice\\aa2.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
I tried to write a loop for it like :
i - 1
while (i 3)
{
qq-data.frame(test[,1],test[,2[i],test[,2[i]+1])
write.table(qq,file=C:\\Alice\\bb.txt,quote=FALSE,row.names=FALSE,col.name=FALSE,sep=\t)
i - i+1;
}
But it's not right. I tried hard to revise it but I just can't make it. If
someone can help me out here, your help is really greatly appreciated. Thank
you sooo much!
Best,
Alice
-
[[alternative HTML version deleted]]
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[R] undefined symbol in grDevices.so
Hello I'm trying to use rpy with latest R (2.2.0), but unfortunately it seems there is some kind of undefined symbol in grDevices.so (utf8locale) Within python, this message appears: import rpy Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/usr/local/lib/R/library/grDevices/libs/grDevices.so': /usr/local/lib/R/library/grDevices/libs/grDevices.so: undefined symbol: utf8locale Loading required package: grDevices Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/usr/local/lib/R/library/grDevices/libs/grDevices.so': /usr/local/lib/R/library/grDevices/libs/grDevices.so: undefined symbol: utf8locale In addition: Warning message: package grDevices in options(defaultPackages) was not found Error: package 'grDevices' could not be loaded Any suggestions? Thanks in advance. Juan Pablo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] strange classification behaviour
You could use cut. The key calculation would be:
w - .05; eps - 1e-5
breakpoints - seq(min(kk), max(kk), .05)
breakpoints - floor( (breakpoints + (w/2) + eps) / w) * w
values - cut(kk, c(breakpoints, Inf), right = FALSE)
values - ordered(values)
If you don't like the labels produced add lab = breakpoints as a cut arg.
On 11/10/05, RenE J.V. Bertin [EMAIL PROTECTED] wrote:
Hello,
I've written a routine that takes an input vector and returns a 'binned'
version with a requested bin width and converted to an ordered factor by
default. It also attempts to make sure that all factor levels intermediate to
the input range are present.
This is the code as I currently have it:
Classify - function( values, ClassWidth=0.05, ordered.factor=TRUE, all=TRUE )
{
valuesName - deparse(substitute(values))
if( is.numeric(values) ){
values - floor( (values+ (ClassWidth/2) ) / ClassWidth ) *
ClassWidth
# determine the numerical range of the input
levels - range( values, finite=TRUE )
if( ordered.factor ){
if( all ){
# if we want all levels, construct a levels vector that
can be passed to factor's levels argument:
levels - seq( levels[1], levels[2], by=ClassWidth )
values - factor(values, levels=levels, ordered=TRUE )
}
else{
values - factor(values, ordered=TRUE )
}
}
}
else{
levels - range( values, finite=TRUE )
if( all ){
levels - seq( levels[1], levels[2], by=ClassWidth )
values - factor( values, levels=levels, ordered=ordered.factor
)
}
else{
values - factor( values, ordered=ordered.factor )
}
}
comment(values) - paste( comment(values),
; Classify(, valuesName, , ClassWidth=, ClassWidth, ,
ordered.factor=, ordered.factor, ),
sep=)
values
}
This does work, but has some strange side-effects that I think might be due
to rounding errors:
## kk-c( 0.854189 0.374423 0.522893 0.670796 0.913540 0.979011
0.510378 0.320440 -0.576764 0.940343 )
## Classify( kk, ClassWidth=0.05, all=FALSE )
[1] 0.85 0.35 0.5 0.65 0.9 10.5 0.3 -0.6 0.95
Levels: -0.6 0.3 0.35 0.5 0.65 0.85 0.9 0.95 1
### result as expected, but using this on the hor. axis of a graph can be ...
surprising.
## Classify( kk, ClassWidth=0.05, all=TRUE )
[1] 0.85 NA 0.5 NA NA 10.5 NA -0.6 NA
33 Levels: -0.6 -0.55 -0.5 -0.45 -0.4 -0.35 -0.3 -0.25 -0.2
-0.15 -0.1 -0.05 0 ... 1
## summary( Classify( kk, ClassWidth=0.05, all=TRUE ) )
-0.6 -0.55 -0.5 -0.45
-0.4 -0.35
1 0 0 0
0 0
-0.3 -0.25 -0.2 -0.15
-0.1 -0.05
0 0 0 0
0 0
0 0.04990.1 0.15
0.2 0.25
0 0 0 0
0 0
0.3 0.350.4 0.45
0.5 0.55
0 0 0 0
2 0
0.6 0.650.7 0.75
0.8 0.85
0 0 0 0
0 1
0.9 0.95 1 NA's
0 0 1 5
### ???
What happens is probably that the value in my input that classify to 0.3 or
0.35 are not found in the list of levels that I calculate due to rounding
errors. Adding an element 0.05 to kk supports this idea.
Is there a way around this, for instance a more robust way to do what I'm
trying to do here (or a function provided by R)?
When I modify the relevant code above to
levels - floor( (seq( levels[1], levels[2], by=ClassWidth
) + (ClassWidth/2)) / ClassWidth ) * ClassWidth
values - factor( values, levels=levels, ordered=TRUE )
the result is as expected, but I find that not very elegant...
Thanks in advance,
RenE Bertin
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[R] R on Windows XP x64
Hi, I am running R 2.2.0 on the Windows XP x64. The mechanism of error hanlder seems different. It will take a very long time to pop up a error message diaglog box, even when some simple errors happen such as Syntax error or object not found. Does anybody have the similar experience? Thanks a lot. BTW: everything works fine under 32-bit Windows XP, error messages come out immediately. Feng __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] error in rowSums:'x' must be numeric
On Thu, 2005-11-10 at 16:49 +0100, Illyes Eszter wrote: Dear All, It's Eszter again from Hungary. I could not solve my problem form yesterday, so I still have to ask your help. I have a binary dataset of vegetation samples and species as a comma separated file. I would like to calculate the Jaccard distance of the dataset. I have the following error message: Error in rowSums(x, prod(dn), p, na.rm) : 'x' must be numeric In addition: Warning message: results may be meaningless because input data have negative entries in: vegdist(t2, method = jaccard, binary = FALSE, diag = FALSE, Do you have any idea what can be the problem? I have only 0 and 1 in the dataset. Eszter, An old truth is that if The Computer is always right and you are wrong when The Computer says that you have non-numeric data. Check your data first. An obvious way of checking this is to repeat the command that found the problem: rowSums(t2). After that (probably) reports the same error, you can check your data sayin, e.g., str(t2) which displays you the variables in a very compact form. Now some wild speculation. When you read your data as comma separated file, very often the column names are taken as the first variable. Check this and remove the first column if needed. This is so common that I've even thought that I perhaps need to write a sanitizing function for cvs files to do the following: rownames(x) - x[,1] x - x[,-1] or to take the first column as rownames and then remove the non-numeric first column. A pertinent problem in R communication with cvs files is that R assumes that with header=TRUE the header line has one entry less than data rows. However, popular software (read Excel) refuses to write data so: even if you make the first column empty in your header line, the popular software adds a comma before the first intended entry, and so you have the same number of entries in the header line and in the data. The result is that the column that intended as rownames is taken as a non-numeric variable. This is such a common problem that an innocent user (not me: I'm no more innocent) would expect R cope with that kind of input format. cheers, jari oksanen -- Jari Oksanen -- Dept Biology, Univ Oulu, 90014 Oulu, Finland email [EMAIL PROTECTED], homepage http://cc.oulu.fi/~jarioksa/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
