Re: [R] randomForest
Duncan == Duncan Murdoch [EMAIL PROTECTED]
on Thu, 07 Jul 2005 15:44:38 -0400 writes:
Duncan On 7/7/2005 3:38 PM, Weiwei Shi wrote:
Hi there:
I have a question on random foresst:
recently i helped a friend with her random forest and i came with this
problem:
her dataset has 6 classes and since the sample size is pretty small:
264 and the class distr is like this (Diag is the response variable)
sample.size - lapply(1:6, function(i) sum(Diag==i))
sample.size
[[1]]
[1] 36
and later you get problems because you didn't know that a *list*
such as 'sample.size' should be made into a so called
*atomic vector* {and there's a function is.atomic(.) ! to test for it}
and Duncan and others told you about unlist().
Now there are two things I'd want to add:
1) If you had used
s.size - table(Diag)
you had used a faster and simpler expression with the same result.
Though in general (when there can be zero counts), to give the
same result, you'd need
s.size - table(factor(Diag, levels = 1:6))
Still a bit preferable to the lapply(.) IMO
2) You should get into the habit of using
sapply(.) rather than lapply(.).
sapply() originally was exactly devised for the above task:
and stands for ``[s]implified lapply''.
It always returns an ``unlisted'' result when appropriate.
Regards,
Martin Maechler, ETH Zurich
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[R] choice of graph
Hi, It's about 2 weeks that I think about a graph to translate my datas. But I don't have an really idea. I 'm going to expose you my problem: I have a questionnaire with 15 questions and you have more possibilties to answer to these. For example: The trainer is competent: Yes No I have learn a lot at the training: Bad Quit bad Medium Good Quit good etc I would like to represent all my question on the same plot and differentiate the different type of answer possible. On my first impression, I had doing a classification of the different type of answer and doing a barplot, but my responsable don't want a classification but want to see all the question with their own type of answer. I have attached my first idea! Can you please help me? Thanks a lot Sabine - __ 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] validation, calibration and Design
Hi R experts, I am trying to do a prognostic model validation study, using cancer survival data. There are 2 data sets - 1500 cases used to develop a nomogram, and another of 800 cases used as an independent validation cohort. I have validated the nomogram in the original data (easy with the Design tools), and then want to show that it also has good results with the independent data using 60 month survival. I would also like to show that the nomogram is significantly different to an existing model based on 60 month survival data generated by it (eg by McNemar's test). Hence, somewhat shortened: #using R 2.01 on Windows library(Hmisc) library(Design) data1 #dataframe with predictor variables A and B, cens and time columns (months) ddist1 - datadist(data1) options(datadist='ddist1') s1 - Surv(data1$time, data1$cens) cph.nomo - cph(s1 ~ A+B, surv=T, x=T, y=T, time.inc=60) survcph - Survival(cph.nomo, x=T, y=T, time.inc=60, surv=T) surv5 - function(lp) survcph(60, lp) nomogram(cph.nomo, lp=T, conf.int=F, fun=list(surv5, surv7), funlabel=c(5 yr DFS)) # now have a useful nomogram model, with good discrimination and #calibration when checked with validate and calibrate (not shown) #move on to validation cohort of n=800 Data2 #Validation data with same predictor variables A, B, cens, time # do I need to put data2 into datadist?? s2 - Surv(data2$time, data2$cens) #able to derive 60 month estimates of survival using data2.est5 - survest(cph.nomo, expand.grid(A=data2$A, B=data2$B), times=c(60), conf.int=0) rcorr.cens(data2.est5$surv, s2) # tests discrimination of the model #against the validation data observed censored data # I cant find a way to use calibrate in this setting though?? # Also, if I have the 5 year estimates for 2 different models, I can # use rcorr.cens to show discrimination, but which values are # suitable for a test of difference (eg with McNemars)? # I have tried predict / newdata function a number of ways but it # typically returns an error relating to unequal vector lengths What I cant work out is where to go now to derive a calibration curve of the predicted 5 year result (val.data5) and the observed (s2). Or can I do it another way? For example, could I merge the 2 data frames and use lines1:1500 to build the model and the last 800 lines to validate? Obviously I am a novice, and sure to be missing something simple. I have spent countless hours pouring over Prof Harrell's text (which is great but doesn't have a specific example of this) and Design Help plus the R news archive with no success, so any help is very much appreciated. Scott Williams MD Peter MacCallum Cancer Centre Melbourne Australia [[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] time series regression
On Fri, 8 Jul 2005, yyan liu wrote: Hi: I have two time series y(t) and x(t). I want to regress Y on X. Because Y is a time series and may have autocorrelation such as AR(p), so it is not efficient to use OLS directly. The model I am trying to fit is like Y(t)=beta0+beta1*X(t)+rho*Y(t-1)+e(t) e(t) is iid normal random error. Anybody know whether there is a function in R can fit such models? The function can also let me specify how many beta's and rho's I can have in the model. If you want to estimate the model by ML, you can use arima() and specify further regressors via the `xreg' argument. Estimation by OLS can be done via lm(), but that typically requires setting up the lags yourself. More convenient interfaces are provided in the `dyn' package by Gabor Grothendieck and my `dynlm' package. Z Thx a lot! liu __ 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] choice of graph
Hi, It's about 2 weeks that I think about a graph to translate my datas. But I don't have an really idea. I 'm going to expose you my problem: I have a questionnaire with 15 questions and you have more possibilties to answer to these. For example: The trainer is competent: Yes No I have learn a lot at the training: Bad Quit bad Medium Good Quit good etc I would like to represent all my question on the same plot and differentiate the different type of answer possible. On my first impression, I had doing a classification of the different type of answer and doing a barplot, Did you try a mosaicplot too? ?mosaicplot Best, Matthias but my responsable don't want a classification but want to see all the question with their own type of answer. I have attached my first idea! Can you please help me? Thanks a lot Sabine - __ 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] (no subject)
Hello, The estimate of glm dispersion can be based on the deviance or on the Pearson statistic. I have compared output from R glm() to another statastical package and it appears that R uses the Pearson statistic. I was wondering if it is possible to make use R the deviance instead by modifying the glm(...) function? Thanks for your attention. Kind regards, Robin Smit This e-mail and its contents are subject to the DISCLAIMER at http://www.tno.nl/disclaimer/email.html [[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] Problems with R on OS X
I used R on OS X 10.3x quite some time with no serious problems. Sometimes R stopped when I tried to execute a bigger program. After updating to OS X to version 10.4 R worked but I still had the problem with bigger programs. Therefore I re-installed R on top of the existing R version. The installation finished properly but suddenly R did not work. Then I reinstalled OS X 10.4 because I thought some left over registry information from R may caused the problem. R still crashed. I hoped the problem would be overcome with the new R version 1.12. Unfortunately this is not the case. The error report (see appendix) says that Thread 0 crashed. What can I do? Heinz Schild Date/Time: 2005-07-08 18:13:16.748 +0200 OS Version: 10.4.1 (Build 8B15) Report Version: 3 Command: R Path:/Applications/R.app/Contents/MacOS/R Parent: WindowServer [62] Version: 1.12 (1622) PID:269 Thread: 0 Exception: EXC_BREAKPOINT (0x0006) Code[0]:0x0001 Code[1]:0x92897200 Thread 0 Crashed: 0 com.apple.Foundation0x92897200 _NSRaiseError + 264 1 com.apple.Foundation0x92896f3c +[NSException raise:format:] + 40 2 com.apple.Foundation0x9286c120 -[NSString stringByAppendingString:] + 104 3 org.R-project.R 0x6ce0 -[RController doLoadHistory:] + 812 (crt.c:300) 4 org.R-project.R 0x3ce0 -[RController awakeFromNib] + 1980 (crt.c:300) 5 com.apple.Foundation0x92886788 -[NSSet makeObjectsPerformSelector:] + 164 6 com.apple.AppKit0x93636f94 -[NSIBObjectData nibInstantiateWithOwner:topLevelObjects:] + 864 7 com.apple.AppKit0x93623324 loadNib + 240 8 com.apple.AppKit0x93622d7c +[NSBundle(NSNibLoading) _loadNibFile:nameTable:withZone:ownerBundle:] + 716 9 com.apple.AppKit0x9367a05c +[NSBundle(NSNibLoading) loadNibFile:externalNameTable:withZone:] + 156 10 com.apple.AppKit0x93709e10 +[NSBundle(NSNibLoading) loadNibNamed:owner:] + 344 11 com.apple.AppKit0x93709bb0 NSApplicationMain + 344 12 org.R-project.R 0x2dd0 _start + 392 (crt.c:267) 13 dyld0x8fe01048 _dyld_start + 60 Thread 1: 0 libSystem.B.dylib 0x9000a778 mach_msg_trap + 8 1 libSystem.B.dylib 0x9000a6bc mach_msg + 60 2 com.apple.CoreFoundation0x9074a4d8 __CFRunLoopRun + 832 3 com.apple.CoreFoundation0x90749ddc CFRunLoopRunSpecific + 268 4 com.apple.Foundation0x928716b8 -[NSConnection sendInvocation:] + 1468 5 com.apple.Foundation0x92870154 -[NSObject(NSForwardInvocation) forward::] + 408 6 libobjc.A.dylib 0x909b10d0 _objc_msgForward + 176 7 org.R-project.R 0x4244 -[RController readThread:] + 244 (crt.c:300) 8 com.apple.Foundation0x9287c2b4 forkThreadForFunction + 108 9 libSystem.B.dylib 0x9002c3d4 _pthread_body + 96 Thread 0 crashed with PPC Thread State: srr0: 0x92897200 srr1: 0x0202f030vrsave: 0x cr: 0x24002422 xer: 0x0004 lr: 0x928971d8 ctr: 0x9285903c r0: 0x r1: 0xb1f0 r2: 0xa2856508 r3: 0xbfffeda0 r4: 0x r5: 0x92858964 r6: 0xbfffee24 r7: 0x00ff r8: 0xbfffee10 r9: 0x03c09f20 r10: 0x909aa668 r11: 0x24002422 r12: 0x9285903c r13: 0x r14: 0x r15: 0x r16: 0x0002 r17: 0x00043524 r18: 0x00043524 r19: 0x00043524 r20: 0x00043524 r21: 0x00043524 r22: 0x00043524 r23: 0x00043524 r24: 0xa365e07c r25: 0x000369b4 r26: 0x01175740 r27: 0x03c2e7d0 r28: 0x03c37c30 r29: 0xa285b7ec r30: 0x90a239c0 r31: 0x92897108 Binary Images Description: 0x1000 -0x38fff org.R-project.R 1.12 (1622) /Applications/R.app/Contents/MacOS/R 0x52000 -0x84fff libreadline.5.0.dylib /Library/Frameworks/R.framework/Resources/lib/libreadline.5.0.dylib 0x98000 -0xc3fff libncurses.5.dylib /usr/lib/libncurses.5.dylib 0x205000 - 0x392fff libR.dylib /Library/Frameworks/R.framework/Resources/lib/libR.dylib 0x8fe0 - 0x8fe50fff dyld 43 /usr/lib/dyld 0x9000 - 0x901a6fff libSystem.B.dylib /usr/lib/libSystem.B.dylib 0x901fe000 - 0x90202fff libmathCommon.A.dylib /usr/lib/system/libmathCommon.A.dylib 0x90204000 - 0x90257fff com.apple.CoreText 1.0.0 (???) /System/Library/Frameworks/ApplicationServices.framework/Versions/A/Frameworks/CoreText.framework/Versions/A/CoreText 0x90284000 - 0x90335fff ATS /System/Library/Frameworks/ApplicationServices.framework/Versions/A/Frameworks/ATS.framework/Versions/A/ATS 0x90364000 - 0x9069cfff com.apple.CoreGraphics 1.256.5 (???) /System/Library/Frameworks/ApplicationServices.framework/Versions/A/Frameworks/CoreGraphics.framework/Versions/A/CoreGraphics 0x90727000 - 0x90800fff com.apple.CoreFoundation 6.4.1 (368.1) /System/Library/Frameworks/CoreFoundation.framework/Versions/A/CoreFoundation 0x90849000 - 0x90849fff
[R] R on kubuntu
Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- Constant Depièreux Managing Director Applied QUality Technologies Europe sprl Rue des Déportés 123, B-4800 Verviers (Tel) +32 87 292175 - (Fax) +32 87 292171 - (Mobile) +32 475 555 818 (Web) http://www.aqte.be - (Courriel) [EMAIL PROTECTED] (Skype) cdepiereux __ 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] R on kubuntu
Constant Depièreux [EMAIL PROTECTED] writes: Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? I don't think it's any different than any other Debian-derived platform. -- 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] R on kubuntu
Constant Depièreux wrote: Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. Hi, I've just moved from SuSE 9.1 to Ubuntu 5.04. The instalation is clean and easy, although you have to waste some time selecting packages. After I got all the packages installed I just compiled R like usual and it is working very well. Regards EJ ps: I have to tell you that Ubuntu was a very good surprise for me. __ 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] plot(cox.zph()): customize xlab ylab
Hello, plot(cox.zph(my.ph),var=1,xlab=Year) gives the error: Error in plot.default(range(xx), yr, type = n, xlab = Time, ylab = ylab[i], : formal argument xlab matched by multiple actual arguments How can I customize the xlab and ylab for plots of cox.zph? Thanks, Dan Bebber Department of Plant Sciences University of Oxford UK ___ How much free photo storage do you get? Store your holiday __ 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] estVARXar parameter significance
Hi, Does anyone know how to check the significance (p-values/t-values) of the estimated parameters via the estVARXar function (in the dse package)? Thanks in advance, Sam. [[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] calculating dispersion formula using deviance ( was Re: (no subject) )
Please try to use a meaningful subject line. See below for comments. On Mon, 2005-07-11 at 11:30 +0200, Smit, R. (Robin) wrote: Hello, The estimate of glm dispersion can be based on the deviance or on the Pearson statistic. I have compared output from R glm() to another statastical package and it appears that R uses the Pearson statistic. A quick search would also highlight the following thread http://www.r-project.org/nocvs/mail/r-help/2002/6938.html I was wondering if it is possible to make use R the deviance instead by modifying the glm(...) function? I don't know what the formula for using the deviance is but _IF_ it is the square root of ratio of null deviance by its degrees of freedom, then sqrt( fit$deviance / fit$df.null ) should be useful. Thanks for your attention. Kind regards, Robin Smit This e-mail and its contents are subject to the DISCLAIMER at http://www.tno.nl/disclaimer/email.html [[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] plot(cox.zph()): customize xlab ylab
I am not sure if there is an easy way around this. An ugly hack is to make a copy the function survival:::plot.cox.zph and make your modified function. But there are others in the list who might know neater solutions. Regards, Adai On Mon, 2005-07-11 at 11:10 +0100, Dan Bebber wrote: Hello, plot(cox.zph(my.ph),var=1,xlab=Year) gives the error: Error in plot.default(range(xx), yr, type = n, xlab = Time, ylab = ylab[i], : formal argument xlab matched by multiple actual arguments How can I customize the xlab and ylab for plots of cox.zph? Thanks, Dan Bebber Department of Plant Sciences University of Oxford UK ___ How much free photo storage do you get? Store your holiday __ 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] R on kubuntu
No problem at all. If you allow universe packages, many binary R packages are available (from the GNU/Linux Debian sid). On Monday 11 July 2005 11:54, Constant Depièreux wrote: Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- Tristan LEFEBURE Laboratoire d'écologie des hydrosystèmes fluviaux (UMR 5023) Université Lyon I - Campus de la Doua Bat. Darwin C 69622 Villeurbanne - France Phone: (33) (0)4 26 23 44 02 Fax: (33) (0)4 72 43 15 23 __ 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] Sweave and complex numbers
Hi
When using Sweave, most of my functions get called with complex
arguments.
They get typeset in with additions that I don't want; 1+1i appears
as 1 + (0 + 1i)
and I would rather have plain old 1+1i.
Example follows:
\documentclass[a4paper]{article}
\title{A Test File}
\author{Robin Hankin}
\usepackage{a4wide}
\begin{document}
\maketitle
A simple example:
print=TRUE=
f - function(x){
x+c(1,1+1i)}
f(6+7i)
@
Question: why does \verb=c(1,1+1i)= get printed as
\verb=c(1,1 + (0+1i))= ?
And how can I stop it?
\end{document}
can anyone help me here?
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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[R] polr (MASS) link functions
When we analyse ordered categorical data, with categories II, should we expect an equivalent analysis if the ordering is changed to II ? Not with a cloglog link (see example below). However, I suspect the cloglog analysis on the reordered categories is equivalent to a loglog analysis on the original ordering. Perhaps for consistency polr should include a loglog link? Or warn that a different answer is available, possibly more appropriate? Cheese tasting example from McCullagh and Nelder: library(MASS) frqs - c(0,0,1,7,8,8,19,8,1, + 6,9,12,11,7,6,1,0,0, + 1,1,6,8,23,7,5,1,0, + 0,0,0,1,3,7,14,16,11) Y1 - ordered(rep(1:9,4)) Y2 - ordered(rep(9:1,4)) trt - factor(rep(c(A,B,C,D),each=9)) fit1 - polr(Y1 ~ trt, method=clog,weight=frqs) fit2 - polr(Y2 ~ trt, method=clog,weight=frqs) print(fit1) Call: polr(formula = Y1 ~ trt, weights = frqs, method = clog) Coefficients: trtB trtC trtD -1.809487 -0.879997 0.987389 Intercepts: 1|2 2|3 3|4 4|5 5|6 6|7 -2.50668784 -2.04055560 -1.42592187 -0.80166219 -0.01722569 0.55135471 7|8 8|9 1.55688848 2.84302173 Residual Deviance: 725.2341 AIC: 747.2341 print(fit2) Call: polr(formula = Y2 ~ trt, weights = frqs, method = clog) Coefficients: trtB trtC trtD 1.9513455 1.1033902 -0.9588354 Intercepts: 1|22|33|44|55|66|77|8 -1.3832854 -0.5216350 0.4225419 0.9994969 1.8317538 2.5777980 3.4532818 8|9 4.4026602 Residual Deviance: 718.0052 AIC: 740.0052 == I.White University of Edinburgh Ashworth Laboratories, West Mains Road Edinburgh EH9 3JT Tel: 0131 650 5490 Fax: 0131 650 6564 E-mail: [EMAIL PROTECTED] __ 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] plot(cox.zph()): customize xlab ylab
Adaikalavan Ramasamy wrote: I am not sure if there is an easy way around this. An ugly hack is to make a copy the function survival:::plot.cox.zph and make your modified function. But there are others in the list who might know neater solutions. This hack is uglier (and might not work properly on some devices, but you could try): plot(cox.zph(my.ph),var=1,col.lab=white) title(xlab=Year, ylab=Beta for blah blah blah) Duncan Murdoch Regards, Adai On Mon, 2005-07-11 at 11:10 +0100, Dan Bebber wrote: Hello, plot(cox.zph(my.ph),var=1,xlab=Year) gives the error: Error in plot.default(range(xx), yr, type = n, xlab = Time, ylab = ylab[i], : formal argument xlab matched by multiple actual arguments How can I customize the xlab and ylab for plots of cox.zph? Thanks, Dan Bebber Department of Plant Sciences University of Oxford UK ___ How much free photo storage do you get? Store your holiday __ 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] Sweave and complex numbers
Robin Hankin wrote:
Hi
When using Sweave, most of my functions get called with complex
arguments.
They get typeset in with additions that I don't want; 1+1i appears
as 1 + (0 + 1i)
and I would rather have plain old 1+1i.
Example follows:
\documentclass[a4paper]{article}
\title{A Test File}
\author{Robin Hankin}
\usepackage{a4wide}
\begin{document}
\maketitle
A simple example:
print=TRUE=
f - function(x){
x+c(1,1+1i)}
f(6+7i)
@
Question: why does \verb=c(1,1+1i)= get printed as
\verb=c(1,1 + (0+1i))= ?
And how can I stop it?
\end{document}
can anyone help me here?
The R parser only understands pure imaginary constants as complex
numbers. It parses 1+1i as the sum of the real constant 1 and the
complex constant 0+1i.
This isn't easy to work around. Sweave could special case these, or the
parser or deparser could, but it looks messy to me. I'd guess the best
solution would be if it happened in the parser (i.e. a real constant
plus an imaginary constant was folded into a complex constant), but
right now our parser doesn't do any sorts of optimizations like this.
It's not easy to add the first one, not least because the parser doesn't
know for sure that 1+1i really is a complex number: you might have
redefined the meaning of +.
Maybe someone else has an answer to your 2nd question.
Duncan Murdoch
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[R] Misbehaviour of DSE
Folks, I am finding problems with using dse: library(dse1) Loading required package: tframe Error: c(package '%s' required by '%s' could not be found, setRNG, dse1) library(dse2) Loading required package: setRNG Error: package 'setRNG' could not be loaded In addition: Warning message: there is no package called 'setRNG' in: library(pkg, character.only = TRUE, logical = TRUE, lib.loc = lib.loc) This is on R 2.1 on an Apple ibook (OS X) panther. Thanks, -ans. __ 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] Boxplot philosophy {was Boxplot in R}
AdaiR == Adaikalavan Ramasamy [EMAIL PROTECTED]
on Mon, 11 Jul 2005 03:04:44 +0100 writes:
AdaiR Just an addendum on the philosophical aspect of doing
AdaiR this. By selecting the 5% and 95% quantiles, you are
AdaiR always going to get 10% of the data as extreme and
AdaiR these points may not necessarily outliers. So when
AdaiR you are comparing information from multiple columns
AdaiR (i.e. boxplots), it is harder to say which column
AdaiR contains more extreme value compared to others etc.
Yes, indeed!
People {and software implementations} have several times provided
differing definitions of how the boxplot whiskers should be defined.
I strongly believe that this is very often a very bad idea!!
A boxplot should be a universal mean communication and so one
should be *VERY* reluctant redefining the outliers.
I just find that Matlab (in their statistics toolbox)
does *NOT* use such a silly 5% / 95% definition of the whiskers,
at least not according to their documentation.
That's very good (and I wonder where you, Larry, got the idea of
the 5 / 95 %).
Using such a fixed percentage is really a very inferior idea to
John Tukey's definition {the one in use in all implementations
of S (including R) probably for close to 20 years now}.
I see one flaw in Tukey's definition {which is shared of course
by any silly percentage based ``outlier'' definition}:
The non-dependency on the sample size.
If you have a 1000 (or even many more) points,
you'll get more and more `outliers' even for perfectly normal data.
But then, I assume John Tukey would have told us to do more
sophisticated things {maybe things like the violin plots} than
boxplot if you have really very many data points, you may want
to see more features -- or he would have agreed to use
boxplot(*, range = monotone_slowly_growing(n) )
for largish sample sizes n.
Martin Maechler, ETH Zurich
AdaiR Regards, Adai
AdaiR On Sun, 2005-07-10 at 18:10 -0500, Larry Xie wrote:
I am trying to draw a plot like Matlab does:
The upper extreme whisker represents 95% of the data;
The upper hinge represents 75% of the data;
The median represents 50% of the data;
The lower hinge represents 25% of the data;
The lower extreme whisker represents 5% of the data.
It looks like:
--- 95%
|
|
--- 75%
| |
|-| 50%
| |
| |
--- 25%
|
--- 5%
Anyone can give me some hints as to how to draw a boxplot like that?
What function does it? I tried boxplot() but couldn't figure it out.
If it's boxplot(), what arguments should I pass to the function? Thank
you for your help. I'd appreciate it.
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[R] class- vs. as()
Dear all, I would appreciate a lot, if someone could explain to me in a simple way, why the assignment class- is not always working and one has to take as() like in the example below. (v - matrix(1:9, 3)) [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 class(v) [1] matrix class(v) - integer class(v) [1] matrix v [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 (vi - as(v, integer)) [1] 1 2 3 4 5 6 7 8 9 class(vi) [1] integer With many thanks, Stefan [[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] validation, calibration and Design
Williams Scott wrote: Hi R experts, I am trying to do a prognostic model validation study, using cancer survival data. There are 2 data sets - 1500 cases used to develop a nomogram, and another of 800 cases used as an independent validation cohort. I have validated the nomogram in the original data (easy with the Design tools), and then want to show that it also has good results with the independent data using 60 month survival. I would also like to show that the nomogram is significantly different to an existing model based on 60 month survival data generated by it (eg by McNemar's test). Scott, A nomogram is a graphical device, not a model to validate. It merely represents a model. If the 800 subjects came from the same hospitals in roughly the same time era, you are doing an internal validation and this is an exceedingly inefficient way to do it. Not only is this wasting 800 subjects from developing the model, but the validation sample is not large enough to yield reliable accuracy estimates. And I don't see how McNemar's test applies as nothing is binary about this problem. If you have two models that have identical degrees of overfitting (e.g. were based on the same number of CANDIDATE degrees of freedom) you can use the rcorrp.cens function to test for differences in discrimination or paired predictions. If you really have an external sample (say 800 subjects from another country) you can use the groupkm function in Design to get a Kaplan-Meier-based calibration curve. Otherwise I would recombine the data, develop the model on all subjects you can get, and use the bootstrap to validate it. Frank Hence, somewhat shortened: #using R 2.01 on Windows library(Hmisc) library(Design) data1 #dataframe with predictor variables A and B, cens and time columns (months) ddist1 - datadist(data1) options(datadist='ddist1') s1 - Surv(data1$time, data1$cens) cph.nomo - cph(s1 ~ A+B, surv=T, x=T, y=T, time.inc=60) survcph - Survival(cph.nomo, x=T, y=T, time.inc=60, surv=T) surv5 - function(lp) survcph(60, lp) nomogram(cph.nomo, lp=T, conf.int=F, fun=list(surv5, surv7), funlabel=c(5 yr DFS)) # now have a useful nomogram model, with good discrimination and #calibration when checked with validate and calibrate (not shown) #move on to validation cohort of n=800 Data2 #Validation data with same predictor variables A, B, cens, time # do I need to put data2 into datadist?? s2 - Surv(data2$time, data2$cens) #able to derive 60 month estimates of survival using data2.est5 - survest(cph.nomo, expand.grid(A=data2$A, B=data2$B), times=c(60), conf.int=0) rcorr.cens(data2.est5$surv, s2) # tests discrimination of the model #against the validation data observed censored data # I cant find a way to use calibrate in this setting though?? # Also, if I have the 5 year estimates for 2 different models, I can # use rcorr.cens to show discrimination, but which values are # suitable for a test of difference (eg with McNemars)? # I have tried predict / newdata function a number of ways but it # typically returns an error relating to unequal vector lengths What I cant work out is where to go now to derive a calibration curve of the predicted 5 year result (val.data5) and the observed (s2). Or can I do it another way? For example, could I merge the 2 data frames and use lines1:1500 to build the model and the last 800 lines to validate? Obviously I am a novice, and sure to be missing something simple. I have spent countless hours pouring over Prof Harrell's text (which is great but doesn't have a specific example of this) and Design Help plus the R news archive with no success, so any help is very much appreciated. Scott Williams MD Peter MacCallum Cancer Centre Melbourne Australia [[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 -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Off topic -2 Ln Lambda and Chi square
Dear R : Sorry for the off topic question, but does anyone know the reference for the -2 Ln Lambda following a Chi Square distribution, please? Possibly one of Bartlett's? Thanks in advance! Sincerely, Laura Holt mailto: [EMAIL PROTECTED] __ 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] Sweave and complex numbers
Duncan Murdoch [EMAIL PROTECTED] writes: Question: why does \verb=c(1,1+1i)= get printed as \verb=c(1,1 + (0+1i))= ? The R parser only understands pure imaginary constants as complex numbers. It parses 1+1i as the sum of the real constant 1 and the complex constant 0+1i. This isn't easy to work around. Sweave could special case these, or the parser or deparser could, but it looks messy to me. I'd guess the best solution would be if it happened in the parser (i.e. a real constant plus an imaginary constant was folded into a complex constant), but right now our parser doesn't do any sorts of optimizations like this. It's not easy to add the first one, not least because the parser doesn't know for sure that 1+1i really is a complex number: you might have redefined the meaning of +. H... I'd say that the user would deserve what she gets in that case. I suspect that the real issue is that R's tokenizer is hand-written and not smart enough to recognize complex constants. Figuring out whether there is a good reason for that or whether we might actually use an automatic tokenizer like flex is somewhere on the far end of my want-to-do list... One thing that could be done quite immediately is to let the deparser forget about a zero real part and drop the parentheses in that case. After all 1i is perfectly legal R, and identical(1i,0+1i) [1] TRUE (Those parse/deparse asymmetries are maddening. *Next* time we redesign R we should get this straightened out) -- 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] plot(cox.zph()): customize xlab ylab
Duncan, your solution could be simplified using ann=FALSE in the plot fit - coxph( Surv(futime, fustat) ~ age + rx, ovarian) plot( cox.zph(fit), ann=F ) title( xlab=My own label, ylab=A new label, main=A clever title) Now, why did I not think of this before ? Regards, Adai On Mon, 2005-07-11 at 07:50 -0400, Duncan Murdoch wrote: Adaikalavan Ramasamy wrote: I am not sure if there is an easy way around this. An ugly hack is to make a copy the function survival:::plot.cox.zph and make your modified function. But there are others in the list who might know neater solutions. This hack is uglier (and might not work properly on some devices, but you could try): plot(cox.zph(my.ph),var=1,col.lab=white) title(xlab=Year, ylab=Beta for blah blah blah) Duncan Murdoch Regards, Adai On Mon, 2005-07-11 at 11:10 +0100, Dan Bebber wrote: Hello, plot(cox.zph(my.ph),var=1,xlab=Year) gives the error: Error in plot.default(range(xx), yr, type = n, xlab = Time, ylab = ylab[i], : formal argument xlab matched by multiple actual arguments How can I customize the xlab and ylab for plots of cox.zph? Thanks, Dan Bebber Department of Plant Sciences University of Oxford UK ___ How much free photo storage do you get? Store your holiday __ 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] plot(cox.zph()): customize xlab ylab
On 7/11/2005 9:10 AM, Adaikalavan Ramasamy wrote: Duncan, your solution could be simplified using ann=FALSE in the plot fit - coxph( Surv(futime, fustat) ~ age + rx, ovarian) plot( cox.zph(fit), ann=F ) title( xlab=My own label, ylab=A new label, main=A clever title) Now, why did I not think of this before ? That's nearly perfect. I think the only problem is that plot.cox.zph can do multiple frames, and this will only allow annotation on the last one. But if you put it in a loop and do them one at a time, this should be fine. Duncan Murdoch Regards, Adai On Mon, 2005-07-11 at 07:50 -0400, Duncan Murdoch wrote: Adaikalavan Ramasamy wrote: I am not sure if there is an easy way around this. An ugly hack is to make a copy the function survival:::plot.cox.zph and make your modified function. But there are others in the list who might know neater solutions. This hack is uglier (and might not work properly on some devices, but you could try): plot(cox.zph(my.ph),var=1,col.lab=white) title(xlab=Year, ylab=Beta for blah blah blah) Duncan Murdoch Regards, Adai On Mon, 2005-07-11 at 11:10 +0100, Dan Bebber wrote: Hello, plot(cox.zph(my.ph),var=1,xlab=Year) gives the error: Error in plot.default(range(xx), yr, type = n, xlab = Time, ylab = ylab[i], : formal argument xlab matched by multiple actual arguments How can I customize the xlab and ylab for plots of cox.zph? Thanks, Dan Bebber Department of Plant Sciences University of Oxford UK ___ How much free photo storage do you get? Store your holiday __ 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] misc3d package
Hi, I am trying to install the misc3d package on a Windows (XP) installation of R 2.0.1 using install.packages(misc3d) but with no success. I have used this approach with other packages OK, but for misc3d I get the following output... trying URL `http://cran.r-project.org/bin/windows/contrib/2.0/PACKAGES' Content type `text/plain; charset=iso-8859-1' length 27996 bytes opened URL downloaded 27Kb Warning message: No package misc3d on CRAN. in: download.packages(pkgs, destdir = tmpd, available = available, Any help would be much appreciated. Regards, Mark This message has been checked for viruses but the contents of an attachment may still contain software viruses, which could damage your computer system: you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation. __ 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] Problems with corARMA
Dear All,
I just came across the same error message (running a gnls{nlme}
model). [R-2.1.1 on Mac OS 10.4]
For a reproducible example, please download the file:
http://mit.edu/costas/www/GR.txt
and run the following:
GR - read.table(GR.txt,header=T)
attach(GR)
myyear - year-1969
library(nlme)
gnls4a - gnls(fatalities/vehicles1000~a0*(vehicles1000/
population1000)^a1,start=c
(a0=1,a1=1),data=GR,na.action=na.omit,correlation=corARMA(form=~myyear))
Thanks a lot,
Costas
On 10 Ιουν 2005, at 8:43 ΠΜ, Pamela McCaskie wrote:
Dear all
I am tryiing to fit the following lme with an ARMA correlation
structure:
test - lme(fixed=fev1f~year, random=~1|id2, data=pheno2,
correlation=corARMA(value=0.2, form=~year|id2), na.action=na.omit)
But I get the following error message:
Error in getGroupsFormula.default(correlation, asList = TRUE) :
Form argument must be a formula
I have used this same form argument with differerent correlation
structures and it has worked fine. Does anyone know why it won't
recognise ~year | id2 (or even ~ 1 | id2) as a formula?
Any help would be great
Pam
--
Pamela A McCaskie
BSc(Mathematical Sciences)(Hons)
Western Australian Institute for Medical Research
University of Western Australia
SCGH Campus
Ground Floor, B Block
QE-II Medical Centre
Hospital Avenue, Nedlands
Western Australia 6009
AUSTRALIA
Email:[EMAIL PROTECTED]
Phone:+61-8-9346 1612
Mob: 0417 926 607
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Massachusetts Institute of Technology
Intelligent Transportation Systems Program
77 Massachusetts Ave., Rm. 1-249, Cambridge, MA 02139
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Re: [R] misc3d package
On Mon, 11 Jul 2005, Mark Edmondson-Jones wrote: Hi, I am trying to install the misc3d package on a Windows (XP) installation of R 2.0.1 using install.packages(misc3d) but with no success. I have used this approach with other packages OK, but for misc3d I get the following output... Try using a current version of R for which there is a pre-compiled windows binary available or install the source package. Best, Z trying URL `http://cran.r-project.org/bin/windows/contrib/2.0/PACKAGES' Content type `text/plain; charset=iso-8859-1' length 27996 bytes opened URL downloaded 27Kb Warning message: No package misc3d on CRAN. in: download.packages(pkgs, destdir = tmpd, available = available, Any help would be much appreciated. Regards, Mark This message has been checked for viruses but the contents of an attachment may still contain software viruses, which could damage your computer system: you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation. __ 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] misc3d package
On 7/11/2005 9:57 AM, Mark Edmondson-Jones wrote: Hi, I am trying to install the misc3d package on a Windows (XP) installation of R 2.0.1 using install.packages(misc3d) but with no success. I have used this approach with other packages OK, but for misc3d I get the following output... trying URL `http://cran.r-project.org/bin/windows/contrib/2.0/PACKAGES' Content type `text/plain; charset=iso-8859-1' length 27996 bytes opened URL downloaded 27Kb Warning message: No package misc3d on CRAN. in: download.packages(pkgs, destdir = tmpd, available = available, Any help would be much appreciated. The message should be read at face value. There is no version of misc3d for R 2.0.1. I think misc3d was released in June, well after 2.1.0 came out, and we don't do binary builds for old releases. If you want to use that package, you'll need to upgrade (or try to compile it yourself). 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
Re: [R] misc3d package
misc3d is a very new package. CRAN's windows binary repository for R-2.0.x is no longer updated, hence does not contain the package. The corresponding ReadMe tells us: Last update: 19.04.2005. Either upgrade to R-2.1.1 and try again, or compile the package from sources yourself. Uwe Ligges Mark Edmondson-Jones wrote: Hi, I am trying to install the misc3d package on a Windows (XP) installation of R 2.0.1 using install.packages(misc3d) but with no success. I have used this approach with other packages OK, but for misc3d I get the following output... trying URL `http://cran.r-project.org/bin/windows/contrib/2.0/PACKAGES' Content type `text/plain; charset=iso-8859-1' length 27996 bytes opened URL downloaded 27Kb Warning message: No package misc3d on CRAN. in: download.packages(pkgs, destdir = tmpd, available = available, Any help would be much appreciated. Regards, Mark This message has been checked for viruses but the contents of an attachment may still contain software viruses, which could damage your computer system: you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation. __ 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] plot(cox.zph()): customize xlab ylab
On Mon, 11 Jul 2005, Adaikalavan Ramasamy wrote: I am not sure if there is an easy way around this. An ugly hack is to make a copy the function survival:::plot.cox.zph and make your modified function. But there are others in the list who might know neater solutions. If you then send a patch to the package maintainer it stops being an ugly hack and turns into an example of collaborative open-source development :) -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] class- vs. as()
What do you want? Consider the following: v - matrix(1:9, 3) class(v[1,1]) [1] integer class(as.vector(v)) [1] integer v2 - v dim(v2) - NULL class(v2) [1] integer spencer graves [EMAIL PROTECTED] wrote: Dear all, I would appreciate a lot, if someone could explain to me in a simple way, why the assignment class- is not always working and one has to take as() like in the example below. (v - matrix(1:9, 3)) [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 class(v) [1] matrix class(v) - integer class(v) [1] matrix v [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 (vi - as(v, integer)) [1] 1 2 3 4 5 6 7 8 9 class(vi) [1] integer With many thanks, Stefan [[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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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] randomForest
Thanks.
Many people pointed that out. (It was due to that I only knew lappy by
that time :).
On 7/11/05, Martin Maechler [EMAIL PROTECTED] wrote:
Duncan == Duncan Murdoch [EMAIL PROTECTED]
on Thu, 07 Jul 2005 15:44:38 -0400 writes:
Duncan On 7/7/2005 3:38 PM, Weiwei Shi wrote:
Hi there:
I have a question on random foresst:
recently i helped a friend with her random forest and i came with this
problem:
her dataset has 6 classes and since the sample size is pretty small:
264 and the class distr is like this (Diag is the response variable)
sample.size - lapply(1:6, function(i) sum(Diag==i))
sample.size
[[1]]
[1] 36
and later you get problems because you didn't know that a *list*
such as 'sample.size' should be made into a so called
*atomic vector* {and there's a function is.atomic(.) ! to test for it}
and Duncan and others told you about unlist().
Now there are two things I'd want to add:
1) If you had used
s.size - table(Diag)
you had used a faster and simpler expression with the same result.
Though in general (when there can be zero counts), to give the
same result, you'd need
s.size - table(factor(Diag, levels = 1:6))
Still a bit preferable to the lapply(.) IMO
2) You should get into the habit of using
sapply(.) rather than lapply(.).
sapply() originally was exactly devised for the above task:
and stands for ``[s]implified lapply''.
It always returns an ``unlisted'' result when appropriate.
Regards,
Martin Maechler, ETH Zurich
--
Weiwei Shi, Ph.D
Did you always know?
No, I did not. But I believed...
---Matrix III
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Re: [R] R on kubuntu
However according to: http://packages.ubuntu.com/hoary/math/r-base you will have to live with R version 2.0.1 until the next version of Ubuntu is released. (If I understood the Ubuntu policy...) Or is there some other repository providing more updated binaries for Ubuntu 5.04? Cheers, Henrik Andersson Lefebure Tristan wrote: No problem at all. If you allow universe packages, many binary R packages are available (from the GNU/Linux Debian sid). On Monday 11 July 2005 11:54, Constant Depièreux wrote: Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- - Henrik Andersson Netherlands Institute of Ecology - Centre for Estuarine and Marine Ecology P.O. Box 140 4400 AC Yerseke Phone: +31 113 577473 [EMAIL PROTECTED] http://www.nioo.knaw.nl/ppages/handersson __ 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] plot(cox.zph()): customize xlab ylab
Dear all,
I've modified the plot.cox.zph function to allow
customized xlab and ylab (see below). Someone might
like to confirm that it works.
Thanks for all the assistance.
Dan
___
plot.cox.zph - function (x, resid = TRUE, se = TRUE,
df = 4, nsmo = 40, var,
xlab=Time,ylab = paste(Beta(t) for,
dimnames(yy)[[2]]),...)
{
xx - x$x
yy - x$y
d - nrow(yy)
df - max(df)
nvar - ncol(yy)
pred.x - seq(from = min(xx), to = max(xx), length
= nsmo)
temp - c(pred.x, xx)
lmat - ns(temp, df = df, intercept = TRUE)
pmat - lmat[1:nsmo, ]
xmat - lmat[-(1:nsmo), ]
qmat - qr(xmat)
if (se) {
bk - backsolve(qmat$qr[1:df, 1:df], diag(df))
xtx - bk %*% t(bk)
seval - d * ((pmat %*% xtx) * pmat) %*%
rep(1, df)
}
if (missing(var))
var - 1:nvar
else {
if (is.character(var))
var - match(var, dimnames(yy)[[2]])
if (any(is.na(var)) || max(var) nvar ||
min(var)
1)
stop(Invalid variable requested)
}
if (x$transform == log) {
xx - exp(xx)
pred.x - exp(pred.x)
}
else if (x$transform != identity) {
xtime - as.numeric(dimnames(yy)[[1]])
apr1 - approx(xx, xtime, seq(min(xx),
max(xx), length = 17)[2 *
(1:8)])
temp - signif(apr1$y, 2)
apr2 - approx(xtime, xx, temp)
xaxisval - apr2$y
xaxislab - rep(, 8)
for (i in 1:8) xaxislab[i] - format(temp[i])
}
for (i in var) {
y - yy[, i]
yhat - pmat %*% qr.coef(qmat, y)
if (resid)
yr - range(yhat, y)
else yr - range(yhat)
if (se) {
temp - 2 * sqrt(x$var[i, i] * seval)
yup - yhat + temp
ylow - yhat - temp
yr - range(yr, yup, ylow)
}
if (x$transform == identity)
plot(range(xx), yr, type = n, xlab =
xlab, ylab = ylab[i],
...)
else if (x$transform == log)
plot(range(xx), yr, type = n, xlab =
xlab, ylab = ylab[i],
log = x, ...)
else {
plot(range(xx), yr, type = n, xlab =
xlab, ylab = ylab[i],
axes = FALSE, ...)
axis(1, xaxisval, xaxislab)
axis(2)
box()
}
if (resid)
points(xx, y)
lines(pred.x, yhat)
if (se) {
lines(pred.x, yup, lty = 2)
lines(pred.x, ylow, lty = 2)
}
}
}
--- Thomas Lumley [EMAIL PROTECTED] wrote:
On Mon, 11 Jul 2005, Adaikalavan Ramasamy wrote:
I am not sure if there is an easy way around this.
An ugly hack is to
make a copy the function survival:::plot.cox.zph
and make your
modified function. But there are others in the
list who might know
neater solutions.
If you then send a patch to the package maintainer
it stops being an ugly
hack and turns into an example of collaborative
open-source development :)
-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] Problems with R on OS X
Hi Heinz, Can you send me your version of: '~/Library/Preferences/org.R- project.R.plist'? Most Mac OS questions are posted/answered on R-SIG-Mac. Thanks, Rob On Jul 11, 2005, at 2:42 AM, Heinz Schild wrote: I used R on OS X 10.3x quite some time with no serious problems. Sometimes R stopped when I tried to execute a bigger program. After updating to OS X to version 10.4 R worked but I still had the problem with bigger programs. Therefore I re-installed R on top of the existing R version. The installation finished properly but suddenly R did not work. Then I reinstalled OS X 10.4 because I thought some left over registry information from R may caused the problem. R still crashed. I hoped the problem would be overcome with the new R version 1.12. Unfortunately this is not the case. The error report (see appendix) says that Thread 0 crashed. What can I do? Heinz Schild Errror Report.txt __ 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] Off topic -2 Ln Lambda and Chi square
If you meant the lambda as the likelihood ratio test statistic, the asymptotic chi-squared distribution comes from the asymptotic normality of the MLEs. The proof is in a paper by Abraham Wald in 1943. See Stuart Ord (Kendall's Advanced Statistics) for discussion (e.g., vol. 2, 5th edition). Andy From: Laura Holt Dear R : Sorry for the off topic question, but does anyone know the reference for the -2 Ln Lambda following a Chi Square distribution, please? Possibly one of Bartlett's? Thanks in advance! Sincerely, Laura Holt mailto: [EMAIL PROTECTED] __ 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] R on kubuntu
On 7/11/05, Henrik Andersson [EMAIL PROTECTED] wrote: However according to: http://packages.ubuntu.com/hoary/math/r-base you will have to live with R version 2.0.1 until the next version of Ubuntu is released. (If I understood the Ubuntu policy...) Or is there some other repository providing more updated binaries for Ubuntu 5.04? It's possible to install the latest packages from the Debian repository but awkward. You would need to add a debian/testing or debian/unstable repository to /etc/apt/sources.list and configure apt to use the Ubuntu archives unless you explicitly require the testing or unstable version of a package. I think that trying to install such a version will require you to update your entire compiler tool chain, which may be more effort than you want to make. I just recently installed Kubuntu on a machine and will, if I manage to get the hardware working, do another today. If that works I try instaling r-base-dev/unstabl and report back to the r-sig-debian mailing list. (Note that there is a special mailing list for Debian-related questions and I am moving this thread to that list.) Cheers, Henrik Andersson Lefebure Tristan wrote: No problem at all. If you allow universe packages, many binary R packages are available (from the GNU/Linux Debian sid). On Monday 11 July 2005 11:54, Constant Depièreux wrote: Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- - Henrik Andersson Netherlands Institute of Ecology - Centre for Estuarine and Marine Ecology P.O. Box 140 4400 AC Yerseke Phone: +31 113 577473 [EMAIL PROTECTED] http://www.nioo.knaw.nl/ppages/handersson __ 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] R on kubuntu
On Monday 11 July 2005 16:44, Henrik Andersson wrote: Or is there some other repository providing more updated binaries for Ubuntu 5.04? I don't think, but I'm fine with R 2.0.1 ... The next ubuntu release, Ubuntu 5.10 (The Breezy Badger), is for October 2005 and will include R 2.1.1 (http://packages.ubuntu.com/breezy/math/r-base). -- Tristan LEFEBURE Laboratoire d'écologie des hydrosystèmes fluviaux (UMR 5023) Université Lyon I - Campus de la Doua Bat. Darwin C 69622 Villeurbanne - France Phone: (33) (0)4 26 23 44 02 Fax: (33) (0)4 72 43 15 23 __ 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] Off topic -2 Ln Lambda and Chi square
On Mon, 11 Jul 2005 11:18:55 -0400, Liaw, Andy wrote: LA If you meant the lambda as the likelihood ratio test statistic, the LA asymptotic chi-squared distribution comes from the asymptotic LA normality of the MLEs. The proof is in a paper by Abraham Wald in LA 1943. See Stuart Ord (Kendall's Advanced Statistics) for LA discussion (e.g., vol. 2, 5th edition). LA LA Andy LA LA From: Laura Holt LA LA Dear R : LA LA Sorry for the off topic question, but does anyone know the LA reference for LA the -2 Ln Lambda following a Chi Square distribution, please? LA see S. S Wilks ( 1938) The large-sample distribution of the likelihood ratio for testing composite hypotheses - Ann. Math. Stat, 9, 60-62 -- Adelchi Azzalini [EMAIL PROTECTED] Dipart.Scienze Statistiche, Università di Padova, Italia tel. +39 049 8274147, http://azzalini.stat.unipd.it/ __ 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] Isolating string containing only file name from complete path
Hi all, What I'd like to do is to is to be able to extract a string corresponding to only the file name from a string containing the complete path, i.e. from the following path string: /Users/ken/Desktop/test/runs/file1 I would like to end up with: file1 This would be most ideally done in a platform-independent way. Thanks in advance, -Ken __ 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] plot(cox.zph()): customize xlab ylab
Dan, I think this works fine now.
I think the 'ylab' argument in other plot function take only a single
value whereas in the case of plot.cox.zph, it needs a vector. It is
especially confusing when this function does plots everything on a
single page by default and all you see is the last plot.
Thus, can I suggest that you check that the length of the ylab is the
same as the number of terms in the coxph call. To do this you could put
something along the following anywhere after the line yy - x$y
n - length(coefficients(fit))
if( length(ylab) != n){
warning( paste('ylab' is of length, length(ylab),
and was expecting a length of, n, elements.) )
}
Regards, Adai
On Mon, 2005-07-11 at 16:04 +0100, Dan Bebber wrote:
Dear all,
I've modified the plot.cox.zph function to allow
customized xlab and ylab (see below). Someone might
like to confirm that it works.
Thanks for all the assistance.
Dan
___
plot.cox.zph - function (x, resid = TRUE, se = TRUE,
df = 4, nsmo = 40, var,
xlab=Time,ylab = paste(Beta(t) for,
dimnames(yy)[[2]]),...)
{
xx - x$x
yy - x$y
d - nrow(yy)
df - max(df)
nvar - ncol(yy)
pred.x - seq(from = min(xx), to = max(xx), length
= nsmo)
temp - c(pred.x, xx)
lmat - ns(temp, df = df, intercept = TRUE)
pmat - lmat[1:nsmo, ]
xmat - lmat[-(1:nsmo), ]
qmat - qr(xmat)
if (se) {
bk - backsolve(qmat$qr[1:df, 1:df], diag(df))
xtx - bk %*% t(bk)
seval - d * ((pmat %*% xtx) * pmat) %*%
rep(1, df)
}
if (missing(var))
var - 1:nvar
else {
if (is.character(var))
var - match(var, dimnames(yy)[[2]])
if (any(is.na(var)) || max(var) nvar ||
min(var)
1)
stop(Invalid variable requested)
}
if (x$transform == log) {
xx - exp(xx)
pred.x - exp(pred.x)
}
else if (x$transform != identity) {
xtime - as.numeric(dimnames(yy)[[1]])
apr1 - approx(xx, xtime, seq(min(xx),
max(xx), length = 17)[2 *
(1:8)])
temp - signif(apr1$y, 2)
apr2 - approx(xtime, xx, temp)
xaxisval - apr2$y
xaxislab - rep(, 8)
for (i in 1:8) xaxislab[i] - format(temp[i])
}
for (i in var) {
y - yy[, i]
yhat - pmat %*% qr.coef(qmat, y)
if (resid)
yr - range(yhat, y)
else yr - range(yhat)
if (se) {
temp - 2 * sqrt(x$var[i, i] * seval)
yup - yhat + temp
ylow - yhat - temp
yr - range(yr, yup, ylow)
}
if (x$transform == identity)
plot(range(xx), yr, type = n, xlab =
xlab, ylab = ylab[i],
...)
else if (x$transform == log)
plot(range(xx), yr, type = n, xlab =
xlab, ylab = ylab[i],
log = x, ...)
else {
plot(range(xx), yr, type = n, xlab =
xlab, ylab = ylab[i],
axes = FALSE, ...)
axis(1, xaxisval, xaxislab)
axis(2)
box()
}
if (resid)
points(xx, y)
lines(pred.x, yhat)
if (se) {
lines(pred.x, yup, lty = 2)
lines(pred.x, ylow, lty = 2)
}
}
}
--- Thomas Lumley [EMAIL PROTECTED] wrote:
On Mon, 11 Jul 2005, Adaikalavan Ramasamy wrote:
I am not sure if there is an easy way around this.
An ugly hack is to
make a copy the function survival:::plot.cox.zph
and make your
modified function. But there are others in the
list who might know
neater solutions.
If you then send a patch to the package maintainer
it stops being an ugly
hack and turns into an example of collaborative
open-source development :)
-thomas
___
Yahoo! Messenger - NEW crystal clear PC to PC calling worldwide with
voicemail http://uk.messenger.yahoo.com
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Isolating string containing only file name from complete path
Ken Termiso [EMAIL PROTECTED] wrote: Hi all, What I'd like to do is to is to be able to extract a string corresponding to only the file name from a string containing the complete path, i.e. from the following path string: /Users/ken/Desktop/test/runs/file1 I would like to end up with: file1 ?basename -- Sebastian P. Luque __ 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] Isolating string containing only file name from complete path
x - /Users/ken/Desktop/test/runs/file1.txt basename(x) [1] file1.txt On Mon, 2005-07-11 at 16:13 +, Ken Termiso wrote: Hi all, What I'd like to do is to is to be able to extract a string corresponding to only the file name from a string containing the complete path, i.e. from the following path string: /Users/ken/Desktop/test/runs/file1 I would like to end up with: file1 This would be most ideally done in a platform-independent way. Thanks in advance, -Ken __ 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] small first graph of par(3,2), other 5 are correct
Hi, I'm trying to produce 6 graphs on a single page using code I've borrowed from an example by Paul Murrell: (http://www.stat.auckland.ac.nz/~paul/RGraphics/custombase-xmastree.R). It involves placing 6 horizontal barplots on one page and adding common labels. The problem is the first graph in my figure (the one in the (1,1) position) is smaller than the other 5. A toy example is included below. When I compare the par() options set after producing each graph, all of the parameters that change look like what I would expect: names(after.g1[after.g1%in%after.g2==FALSE]) [1] fig mai mar mfg plt what.chgd-names(after.g1[after.g1%in%after.g2==FALSE]) after.g1[what.chgd] $fig [1] 0.000 0.500 0.667 1.000 $mai [1] 0.0309375 0.0618750 0.3093750 0.3093750 $mar [1] 0.5 1.0 5.0 5.0 $mfg [1] 1 1 3 2 $plt [1] 0.02357143 0.88214285 0.01437097 0.85629032 after.g2[what.chgd] $fig [1] 0.500 1.000 0.667 1.000 $mai [1] 0.0309375 0.3093750 0.3093750 0.0618750 $mar [1] 0.5 5.0 5.0 1.0 $mfg [1] 1 2 3 2 $plt [1] 0.11785715 0.97642857 0.01437097 0.85629032 One other aspect of this is that the problem does not occur if after I create all 6 graphs once, I rerun the code again, but omit only the par(mfrow=c(3,2)) statement. If I don't reset par(), then the first graph is then the identical size of the other five. Up to now I've gotten around this by simply running the code twice, omitting the par(mfrow=c(3,2)) the second time and printing the result. Now, however, I'm sending the graphs to a pdf file and must set the par(mfrow=c(3,2)) after opening the pdf device, so the problem shows up with each time. Any help on fixing this problem so that all 6 graphs appear the same size on the figure would be most welcome. Thanks, Scot Here is the toy example that shows the layout and plots in my figure: par(mfrow=c(3,2)) groups-LETTERS[1:5] ### for graph 1 ### # data leftci - c(1:5) rightci - c(2:6) # left column graph so: par(mar=c(0.5, 1, 5, 5)) # right column graph so: #par(mar=c(0.5, 5, 5, 1)) plot.new() title(main = Graph 1) plot.window(xlim=c(0, 8), ylim=c(-1.5, 5.5)) ticks - seq(0, 8, 1) y - 1:5 # how many spaces on y axis: 1 for each group h - 0.2 # height? a function of y? segments(0, y, 8, y, lty=dotted) # dotted segments on which bar lies (like a grid) rect(leftci, y-h, rightci, y+h, col=dark grey) #mtext(groups, at=y, adj=1, side=2, las=2, cex=.75) par(cex.axis=1.0, mex=0.5) axis(1, at=ticks, labels=abs(ticks), pos=0, ) box(inner, col=grey) after.g1-show(par()) ### for graph 2 ### # data: leftci - c(1:5) rightci - c(2:6) # left column graph so: #par(mar=c(0.5, 1, 5, 5)) # right column graph so: par(mar=c(0.5, 5, 5, 1)) plot.new() title(main = Graph 2) plot.window(xlim=c(0, 8), ylim=c(-1.5, 5.5)) ticks - seq(0, 8, 1) y - 1:5 # how many spaces on y axis: 1 for each group h - 0.2 # height? a function of y? segments(0, y, 8, y, lty=dotted) # dotted segments on which bar lies (like a grid) rect(leftci, y-h, rightci, y+h, col=dark grey) mtext(groups, at=y, adj=.5, side=2, las=2, cex=.75, line = 5) # this line only for right column graphs par(cex.axis=1.0, mex=0.5) axis(1, at=ticks, labels=abs(ticks), pos=0, ) box(inner, col=grey) # end graph 2 after.g2-show(par()) ### graph 3### # data leftci - c(1:5) rightci - c(2:6) # left column graph so: par(mar=c(0.5, 1, 5, 5)) # right column graph so: #par(mar=c(0.5, 5, 5, 1)) plot.new() title(main = Graph 3) plot.window(xlim=c(0, 8), ylim=c(-1.5, 5.5)) ticks - seq(0, 8, 1) y - 1:5 # how many spaces on y axis: 1 for each group h - 0.2 # height? a function of y? segments(0, y, 8, y, lty=dotted) # dotted segments on which bar lies (like a grid) rect(leftci, y-h, rightci, y+h, col=dark grey) #mtext(groups, at=y, adj=1, side=2, las=2, cex=.75) par(cex.axis=1.0, mex=0.5) axis(1, at=ticks, labels=abs(ticks), pos=0, ) box(inner, col=grey) after.g3-show(par()) ### for graph 4 ### # data: leftci - c(1:5) rightci - c(2:6) # left column graph so: #par(mar=c(0.5, 1, 5, 5)) # right column graph so: par(mar=c(0.5, 5, 5, 1)) plot.new() title(main = Graph 4) plot.window(xlim=c(0, 8), ylim=c(-1.5, 5.5)) ticks - seq(0, 8, 1) y - 1:5 # how many spaces on y axis: 1 for each group h - 0.2 # height? a function of y? segments(0, y, 8, y, lty=dotted) # dotted segments on which bar lies (like a grid) rect(leftci, y-h, rightci, y+h, col=dark grey) mtext(groups, at=y, adj=.5, side=2, las=2, cex=.75, line = 5) # this line only for right column graphs par(cex.axis=1.0, mex=0.5) axis(1, at=ticks, labels=abs(ticks), pos=0, ) box(inner, col=grey) after.g4-show(par()) # end graph 4 ### graph 5### # data leftci - c(1:5) rightci - c(2:6) # left column graph so: par(mar=c(0.5, 1, 5, 5)) # right column graph so: #par(mar=c(0.5, 5, 5, 1)) plot.new() title(main = Graph 5) plot.window(xlim=c(0, 8), ylim=c(-1.5, 5.5)) ticks - seq(0, 8, 1) y - 1:5 # how many spaces on y axis: 1 for each
[R] exact values for p-values
Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? stephen __ 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] Projection Pursuit
Hello, Just a quick question about ppr in library modreg. I have looked at Ripley and Venables 2002 and it says that projection pursuit works by projecting X in M carefully chosen directions I want to know how it choses the directions? I presume it moves around the high-dimensional space of unit vectors finding ones that separate the response variables, but how. I looked at the code, but can't get deeper in than the Fortran call to Smart. I tried to look up Friedman 87, but can't find an electronic reference. Can anyone help, or point me in the direction of an online reference? Thanks in advance, Oliver Lyttelton __ 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] exact values for p-values - more information.
Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? stephen __ 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] R on kubuntu
You can compile the newest version... EJ Henrik Andersson wrote: However according to: http://packages.ubuntu.com/hoary/math/r-base you will have to live with R version 2.0.1 until the next version of Ubuntu is released. (If I understood the Ubuntu policy...) Or is there some other repository providing more updated binaries for Ubuntu 5.04? Cheers, Henrik Andersson Lefebure Tristan wrote: No problem at all. If you allow universe packages, many binary R packages are available (from the GNU/Linux Debian sid). On Monday 11 July 2005 11:54, Constant Depièreux wrote: Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. __ 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] exact values for p-values - more information.
On Mon, 11 Jul 2005, S.O. Nyangoma wrote: Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? You can always extract the `exact' p-value from the summary.lm object or you can compute it by hand via pf(39540, df1 = 1, df2 = 7025, lower.tail = FALSE) For all practical purposes, the above means that the p-value is 0. I guess you are on a 32-bit machine, then it also means that the p-value is smaller than the Machine epsilon .Machine$double.eps So if you want to report the p-value somewhere, I think R's output should be more than precise enough. If you want to compute some other values that depend on such a p-value, then it is probably wiser to compute on a log scale, i.e. instead pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE) use pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE, log.p = TRUE) However, don't expect to be able to evaluate it at such extreme values such as 39540. Z __ 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] demo(scoping)
Why have you sent a message about this, which no indication except the non-English word `entercount'? Note that the message came from within try(), so it was intentional. If you look at the source it says try(ross$withdraw(500)) # no way.. It is helpful to learn how the code being demonstrated (here try) works. If you don't understand it. please explain your difficulty in as much detail as you can. On Mon, 4 Jul 2005, ronggui wrote: entercount an error with demo(scoping). demo(scoping) demo(scoping) ~~~ ___snip_ ross$balance() Your balance is 120 try(ross$withdraw(500)) Error in ross$withdraw(500) : You don't have that much money! version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status beta major2 minor1.1 year 2005 month06 day 13 language R -- Department of Sociology Fudan University,Shanghai Blog:http://sociology.yculblog.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 -- 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] timezone problems
For the record, this was covered by an answer to your bug report.
The problem is your OS which mishandles a timezone of `GMT', so
Sys.getenv(TZ) as `GMT' is not actually setting your OS to GMT.
Hence NA is the correct answer.
I know no way to set Windows to GMT as distinct from the timezone of
London (with summer time).
On Thu, 7 Jul 2005, Martin Keller-Ressel wrote:
Thank you Don for your hints. I have checked my environment vairable TZ
again. But everything is set correctly. I think the problem is with
Sys.timezone(). Maybe it is a conflict between how my system formats the
time/date and what Sys.timezone() expects.
This is what I get on my system:
Sys.getenv(TZ)
TZ
GMT
Sys.time()
[1] 2005-07-07 07:32:39 GMT
## everything fine so far
Sys.timezone()
[1] NA
## This is what Sys.timezone looks like:
Sys.timezone
function ()
{
z - as.POSIXlt(Sys.time())
attr(z, tzone)[2 + z$isdst]
}
environment: namespace:base
z - as.POSIXlt(Sys.time())
attributes(z)
$names
[1] sec min hour mday mon year wday yday isdst
$class
[1] POSIXt POSIXlt
$tzone
[1] GMT
attr(z,tzone)
[1] GMT
z$isdst
[1] 0
attr(z,tzone)[2]
[1] NA
I dont understand why Sys.timezone doesn't use attr(z,tzone) but tries
to read its (2+z$isdst)-th element.
Of course it would be easy to write a workaround, but I wonder why nobody
else is having this problem.
best regards,
Martin Keller-Ressel
On Wed, 06 Jul 2005 14:45:25 -, Don MacQueen [EMAIL PROTECTED] wrote:
How did you set the TZ system variable?
If you did not use Sys.putenv(), try using it instead.
Otherwise, I think you have to ask the package maintainer.
You may be misleading yourself by using Sys.time() to test whether TZ is
set.
What does Sys.getenv() tell you?
I get a timezone code from Sys.time() even when TZ is not defined (see
example below).
(but I do have a different OS)
Sys.timezone()
[1]
Sys.time()
[1] 2005-07-06 07:34:15 PDT
Sys.getenv('TZ')
TZ
Sys.putenv(TZ='US/Pacific')
Sys.timezone()
[1] US/Pacific
Sys.getenv('TZ')
TZ
US/Pacific
Sys.time()
[1] 2005-07-06 07:34:38 PDT
Sys.putenv(TZ='GMT')
Sys.time()
[1] 2005-07-06 14:35:45 GMT
version
_ platform powerpc-apple-darwin7.9.0
arch powerpc os darwin7.9.0
system powerpc, darwin7.9.0status
major2 minor1.1
year 2005month06
day 20 language R
At 9:55 AM + 7/5/05, Martin Keller-Ressel wrote:
Hi,
Im using R 2.1.1 and running Code that previously worked (on R 2.1.0 I
believe) using the 'timeDate' function from the fCalendar package. The
code now throws an error:
Error in if (Sys.timezone() != GMT) warning(Set timezone to GMT!)
However I have read the documentation of the fCalendar package and I
have set my system variable TZ to GMT.
I tracked the error down to the function Sys.timezone() which returns
NA in spite of what Sys.time() returns.
Sys.timezone()
[1] NA
Sys.time()
[1] 2005-07-05 08:41:53 GMT
My version:
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor1.1
year 2005
month06
day 20
language R
Any help is appreciated,
Martin Keller-Ressel
---
Martin Keller-Ressel
Research Unit of Financial and Actuarial Mathematics
TU Vienna
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Research Unit of Financial and Actuarial Mathematics
TU Vienna
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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] Weighted nls
Please do your homework, as the posting guide asks you to.
1. From the Example section of ?nls:
## weighted nonlinear regression
Treated - Puromycin[Puromycin$state == treated, ]
weighted.MM - function(resp, conc, Vm, K)
{
## Purpose: exactly as white book p.451 -- RHS for nls()
## Weighted version of Michaelis-Menten model
##
## Arguments: 'y', 'x' and the two parameters (see book)
##
## Author: Martin Maechler, Date: 23 Mar 2001, 18:48
pred - (Vm * conc)/(K + conc)
(resp - pred) / sqrt(pred)
}
Pur.wt - nls( ~ weighted.MM(rate, conc, Vm, K), data = Treated,
start = list(Vm = 200, K = 0.1),
trace = TRUE)
2. This has been covered several times on this list. Search the archive.
Andy
From: Rundle, Daniel
Dear R Community,
I am attempting to perform a weighted non-linear least
squares fit. It has already been noted that the weights
option is not yet implemented for the nls function, but no
one seems to offer any suggestions for getting around this
problem. I am still curious if a) anyone has code they have
written which includes a weight options for nls, or b) if
there is another model which would allow least squares
fitting to a non-linear function.
Thank you for reading this,
Dan
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Re: [R] exact values for p-values - more information.
I just checked: pf(39540, 1, 7025, lower.tail=FALSE, log.p=TRUE) [1] -Inf This is not correct. With 7025 denominator degrees of freedom, we might use the chi-square approximation to the F distribution: pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE) [1] -19775.52 In sum, my best approximation to pf(39540, 1, 7025, lower.tail=FALSE, log.p=TRUE), given only a minute to work on this, is exp(pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE)) = exp(-19775.52). I'm confident that many violations of assumptions would likely be more important than the differences between p-value: 2.2e-16 and these other two answers. However, I have also used numbers like exp(-19775.52) to guestimate relative degrees of plausibility for different alternatives. That doesn't mean they are right, only the best I can get with the available resources. spencer graves Achim Zeileis wrote: On Mon, 11 Jul 2005, S.O. Nyangoma wrote: Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? You can always extract the `exact' p-value from the summary.lm object or you can compute it by hand via pf(39540, df1 = 1, df2 = 7025, lower.tail = FALSE) For all practical purposes, the above means that the p-value is 0. I guess you are on a 32-bit machine, then it also means that the p-value is smaller than the Machine epsilon .Machine$double.eps So if you want to report the p-value somewhere, I think R's output should be more than precise enough. If you want to compute some other values that depend on such a p-value, then it is probably wiser to compute on a log scale, i.e. instead pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE) use pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE, log.p = TRUE) However, don't expect to be able to evaluate it at such extreme values such as 39540. Z __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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] Projection Pursuit
Google can be helpful. The 10th hit I got from projection pursuit regression is http://www.scs.gmu.edu/~jgentle/csi991/03f/ppreg1024.rtf, which gives some rough outline of the algorithm. If you want more detail, Ripley (1996) PRNN would suffice, I believe. Andy From: Oliver Lyttelton Hello, Just a quick question about ppr in library modreg. I have looked at Ripley and Venables 2002 and it says that projection pursuit works by projecting X in M carefully chosen directions I want to know how it choses the directions? I presume it moves around the high-dimensional space of unit vectors finding ones that separate the response variables, but how. I looked at the code, but can't get deeper in than the Fortran call to Smart. I tried to look up Friedman 87, but can't find an electronic reference. Can anyone help, or point me in the direction of an online reference? Thanks in advance, Oliver Lyttelton __ 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] Weighted nls
Rundle, Daniel [EMAIL PROTECTED] writes: Dear R Community, I am attempting to perform a weighted non-linear least squares fit. It has already been noted that the weights option is not yet implemented for the nls function, but no one seems to offer any suggestions for getting around this problem. I am still curious if a) anyone has code they have written which includes a weight options for nls, or b) if there is another model which would allow least squares fitting to a non-linear function. gnls in the nlme package. Or follow the example on help(nls) immediately after the ## weighted nonlinear regression (!) -- 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] small first graph of par(3,2), other 5 are correct
Scot, Here is your toy example in more condensed form: x11() par(mar=c(0.5, 1, 5, 5)) par(mfrow=c(3,2)) plot(1:10) par(mex) # mex=1.0 here par(cex.axis=1.0, mex=0.5) # Now you change it for (i in 1:5) plot(1:10) When you build your first plot (effectively at the plot.new() command), your mex parameter is 1, and that leaves lots of space in the margins. Then you set mex=0.5, which holds for all subsequent plots, and they get less margin space. -- David Brahm ([EMAIL PROTECTED]) __ 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] exact values for p-values - more information.
Compare the following t.test( 1:100, 101:200 )$p.value t.test( 1:100, 101:200 ) In the latter, the print method truncates to 2.2e-16. You can go as far as (depending on your machine) .Machine$double.xmin [1] 2.225074e-308 before it becomes indistinguishable from zero. But there are good reasons to truncate it at 2.2e-16 such as the difficulty in trying to accurately estimate the extreme tail probabilities. Regards, Adai On Mon, 2005-07-11 at 18:52 +0200, S.O. Nyangoma wrote: Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? stephen __ 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] indexing into and modifying dendrograms
I would like to be able to exert certain types of control over the
plotting of dendrograms (representing hierarchical clusterings) that I
think is best achieved by modifying the dendrogram object
prior to plotting. I am using the dendrogram class and associated
methods.
Define the cluster number of each cluster formed as the corresponding
row of the merge object. So, if you are clustering m objects, the
cluster numbers range from 1 to m-1, with cluster m-1 containing all m
objects, by definition. I basically want a way to index into an
object of class dendrogram using the above definition of cluster
number and/or to act on a dendrogram, where I specify the target node
using cluster number.
The first application would be to 'flip' the two elements in target
node of the dendrogram (made clear in the small example below). (The
setting is genomics and I have applications where I want to man-handle
my dendrograms to make certain features of the clustering more obvious
to the naked eye.) I could imagine other, related actions that would
be useful in decorating dendrograms.
I think I need a function that takes a dendrogram and cluster
number(s) as input and returns the relevant part(s) of the dendrogram
object -- but in a form that makes it easy to then, say, set certain
attributes (perhaps recursively) for the target nodes (and perhaps
those contained in it). I'm including a small example below that
hopefully illustrates this (it looks long, but it's mostly comments!).
Any help would be appreciated.
Jenny Bryan
## get ready for step-by-step figures
par(mfrow = c(2,2))
## get 5 objects, with 2-dimensional features
pts - rbind(c(2,1.6),
c(1.8,2.4),
c(2.1, 2.7),
c(5,2.6),
c(4.7,3.1))
plot(pts, xlim = c(0,6), ylim = c(0,4),type = n,
xlab = Feature 1, ylab = Feature 2)
points(pts,pch = as.character(1:5))
## build a hierarhical tree, store as a dendrogram
aggTree - hclust(dist(pts), method = single)
(dend1 - JB.as.dendrogram.hclust(aggTree))
## NOTE: only thing I added to official version of
## as.dendrogram.hclust:
## each node has an attribute cNum, which gives
## the merge step at which it was formed,
## i.e. gives the row of the merge object which
## describes the formation of that node
## one new line near end of nMerge loop:
## ***
## *** 51,56
## --- 51,60
## attr(z[[x[2]]], midpoint))/2
## }
## attr(zk, height) - oHgt[k]
## +
## + ## JB added July 6 2005
## + attr(zk, cNum) - k
## +
## z[[k - as.character(k)]] - zk
## }
## z - z[[k]]
attributes(dend1)
attributes(dend1[[1]])
## here's a table relating dend1 and the cNum attribute
## dend1 cNum
## -
## dend14
## dend1[[1]] 2
## dend1[[2]] 3
## dend1[[2]][[1]] not set
## dend1[[2]][[1]] 1
## use cNum attribute in edgetext
## following example in dendrogram documentation
## would really rather associate with the node than the edge
## but current plotting function has no notion of nodetext
addE - function(n) {
if(!is.leaf(n)) {
attr(n, edgePar) - list(p.col=plum)
attr(n, edgetext) - attr(n,cNum)
}
n
}
dend2 - dendrapply(dend1, addE)
## overlays the cNum (cluster number) attribute on dendrogram
plot(dend2, main = dend2)
## why does no plum polygon appear around the '4' for the root
## edge?
## swap order of clusters 2 and 3,
## i.e. 'flip' cluster 4
dend3 - dend2
dend3[[1]] - dend2[[2]]
dend3[[2]] - dend2[[1]]
plot(dend3, main = dend3)
## wish I could achieve with 'dend3 - flip(dend2, cNum = 4)
## swap order of cluster 1 and object 1,
## i.e. 'flip' cluster 3
dend4 - dend2
dend4[[2]][[1]] - dend2[[2]][[2]]
dend4[[2]][[2]] - dend2[[2]][[1]]
plot(dend4, main = dend4)
## wish I could achieve with 'dend4 - flip(dend2, cNum = 3)
## finally, it's clear that the midpoint attribute would also
## need to be modified by 'flip'
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[R] building packages on Windows
Hi, all, I just recently upgraded my computer though I'm using the same OS (XP). But now I'm having difficulty building packages and I cannot seem to solve the problem. I'm using R-2.1.1pat on Windows XP. Here is what I tried: D:\Users\sundard\slib\sundar\RR CMD CHECK sundar * checking for working latex ... OK * using log directory 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck' * using R version 2.1.1, 2005-06-21 * checking for file 'sundar/DESCRIPTION' ... OK * this is package 'sundar' version '1.1' * checking if this is a source package ... OK installing R.css in D:/Users/sundard/slib/sundar/R/sundar.Rcheck -- Making package sundar adding build stamp to DESCRIPTION installing NAMESPACE file and metadata Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar/NAMESPACE' Execution halted make[2]: *** [nmspace] Error 1 make[1]: *** [all] Error 2 make: *** [pkg-sundar] Error 2 *** Installation of sundar failed *** Removing 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar' ERROR Installation failed. I've also tried to remove the NAMESPACE which then passes `check' but fails on `build' with the following: * checking for file 'sundar/DESCRIPTION' ... OK * preparing 'sundar': * checking DESCRIPTION meta-information ... OK * removing junk files Error: cannot open file 'sundar/DESCRIPTION' for reading If this is relevant, here is my PATH: C:\WINDOWS\system32; C:\WINDOWS; .; D:\R\rw2011pat\bin; D:\R\tools\bin; D:\Perl\bin; D:\Tcl\bin; D:\TeXLive\bin\win32; D:\Mingw\bin; C:\Program Files\Insightful\splus62\cmd; C:\Program Files\HTML Help Workshop Here, the D:\R\tools\bin directory contains the utilities for building R from source, downloaded today. Am I missing something obvious? Thanks, --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] building packages on Windows
On 7/11/2005 3:21 PM, Sundar Dorai-Raj wrote: Hi, all, I just recently upgraded my computer though I'm using the same OS (XP). But now I'm having difficulty building packages and I cannot seem to solve the problem. I'm using R-2.1.1pat on Windows XP. Here is what I tried: D:\Users\sundard\slib\sundar\RR CMD CHECK sundar * checking for working latex ... OK * using log directory 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck' * using R version 2.1.1, 2005-06-21 * checking for file 'sundar/DESCRIPTION' ... OK * this is package 'sundar' version '1.1' * checking if this is a source package ... OK installing R.css in D:/Users/sundard/slib/sundar/R/sundar.Rcheck -- Making package sundar adding build stamp to DESCRIPTION installing NAMESPACE file and metadata Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar/NAMESPACE' Execution halted make[2]: *** [nmspace] Error 1 make[1]: *** [all] Error 2 make: *** [pkg-sundar] Error 2 *** Installation of sundar failed *** Removing 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar' ERROR Installation failed. I've also tried to remove the NAMESPACE which then passes `check' but fails on `build' with the following: * checking for file 'sundar/DESCRIPTION' ... OK * preparing 'sundar': * checking DESCRIPTION meta-information ... OK * removing junk files Error: cannot open file 'sundar/DESCRIPTION' for reading If this is relevant, here is my PATH: C:\WINDOWS\system32; C:\WINDOWS; .; D:\R\rw2011pat\bin; D:\R\tools\bin; D:\Perl\bin; D:\Tcl\bin; D:\TeXLive\bin\win32; D:\Mingw\bin; C:\Program Files\Insightful\splus62\cmd; C:\Program Files\HTML Help Workshop Here, the D:\R\tools\bin directory contains the utilities for building R from source, downloaded today. Am I missing something obvious? Yes, you need to put the Windows directories later in your path. There are some programs in the toolset which have like-named commands in Windows; you need to use ours, not theirs. There are other differences between your path and the one recommended in the Admin and Installation manual; they may also be causing problems. 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
Re: [R] building packages on Windows
Duncan Murdoch wrote: On 7/11/2005 3:21 PM, Sundar Dorai-Raj wrote: Hi, all, I just recently upgraded my computer though I'm using the same OS (XP). But now I'm having difficulty building packages and I cannot seem to solve the problem. I'm using R-2.1.1pat on Windows XP. Here is what I tried: D:\Users\sundard\slib\sundar\RR CMD CHECK sundar * checking for working latex ... OK * using log directory 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck' * using R version 2.1.1, 2005-06-21 * checking for file 'sundar/DESCRIPTION' ... OK * this is package 'sundar' version '1.1' * checking if this is a source package ... OK installing R.css in D:/Users/sundard/slib/sundar/R/sundar.Rcheck -- Making package sundar adding build stamp to DESCRIPTION installing NAMESPACE file and metadata Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar/NAMESPACE' Execution halted make[2]: *** [nmspace] Error 1 make[1]: *** [all] Error 2 make: *** [pkg-sundar] Error 2 *** Installation of sundar failed *** Removing 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar' ERROR Installation failed. I've also tried to remove the NAMESPACE which then passes `check' but fails on `build' with the following: * checking for file 'sundar/DESCRIPTION' ... OK * preparing 'sundar': * checking DESCRIPTION meta-information ... OK * removing junk files Error: cannot open file 'sundar/DESCRIPTION' for reading If this is relevant, here is my PATH: C:\WINDOWS\system32; C:\WINDOWS; .; D:\R\rw2011pat\bin; D:\R\tools\bin; D:\Perl\bin; D:\Tcl\bin; D:\TeXLive\bin\win32; D:\Mingw\bin; C:\Program Files\Insightful\splus62\cmd; C:\Program Files\HTML Help Workshop Here, the D:\R\tools\bin directory contains the utilities for building R from source, downloaded today. Am I missing something obvious? Yes, you need to put the Windows directories later in your path. There are some programs in the toolset which have like-named commands in Windows; you need to use ours, not theirs. There are other differences between your path and the one recommended in the Admin and Installation manual; they may also be causing problems. Duncan Murdoch Hi, Duncan, Thanks for the reply. I moved things around so that my PATH now looks like: .; D:\R\tools\bin; D:\Perl\bin; D:\Mingw\bin; D:\TeXLive\bin\win32; C:\Program Files\HTML Help Workshop; D:\R\rw2011pat\bin; D:\Tcl\bin; C:\Program Files\Insightful\splus62\cmd; C:\WINDOWS; C:\WINDOWS\system32 However, the problem was actually to do with permissions. Apparently when I when I do the following: chmod 666 DESCRIPTION NAMESPACE The permissions must have been changed when upgrading my computer since I didn't have any problem on my previous machine. However, now another problem has arisen: D:\Users\sundard\slib\sundar\RR CMD install sundar -- Making package sundar adding build stamp to DESCRIPTION installing NAMESPACE file and metadata output snipped Compiling d:\Users\sundard\slib\sundar\R\sundar\chm\sundar.chm HHC5003: Error: Compilation failed while compiling logo.jpg. HHC5003: Error: Compilation failed while compiling Rchm.css. The following files were not compiled: logo.jpg Rchm.css adding MD5 sums * DONE (sundar) This still installs fine except the chmhelp file is not created as it should be. I've deleted the chm subdirectory and tried to reinstall and I get the same error. This seems not to be an R problem, but possibly some configuration I'm not aware of, but I'd be grateful for any insight. Thanks, --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] building packages on Windows
On 7/11/05, Duncan Murdoch [EMAIL PROTECTED] wrote: On 7/11/2005 3:21 PM, Sundar Dorai-Raj wrote: Hi, all, I just recently upgraded my computer though I'm using the same OS (XP). But now I'm having difficulty building packages and I cannot seem to solve the problem. I'm using R-2.1.1pat on Windows XP. Here is what I tried: D:\Users\sundard\slib\sundar\RR CMD CHECK sundar * checking for working latex ... OK * using log directory 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck' * using R version 2.1.1, 2005-06-21 * checking for file 'sundar/DESCRIPTION' ... OK * this is package 'sundar' version '1.1' * checking if this is a source package ... OK installing R.css in D:/Users/sundard/slib/sundar/R/sundar.Rcheck -- Making package sundar adding build stamp to DESCRIPTION installing NAMESPACE file and metadata Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar/NAMESPACE' Execution halted make[2]: *** [nmspace] Error 1 make[1]: *** [all] Error 2 make: *** [pkg-sundar] Error 2 *** Installation of sundar failed *** Removing 'D:/Users/sundard/slib/sundar/R/sundar.Rcheck/sundar' ERROR Installation failed. I've also tried to remove the NAMESPACE which then passes `check' but fails on `build' with the following: * checking for file 'sundar/DESCRIPTION' ... OK * preparing 'sundar': * checking DESCRIPTION meta-information ... OK * removing junk files Error: cannot open file 'sundar/DESCRIPTION' for reading If this is relevant, here is my PATH: C:\WINDOWS\system32; C:\WINDOWS; .; D:\R\rw2011pat\bin; D:\R\tools\bin; D:\Perl\bin; D:\Tcl\bin; D:\TeXLive\bin\win32; D:\Mingw\bin; C:\Program Files\Insightful\splus62\cmd; C:\Program Files\HTML Help Workshop Here, the D:\R\tools\bin directory contains the utilities for building R from source, downloaded today. Am I missing something obvious? Yes, you need to put the Windows directories later in your path. There are some programs in the toolset which have like-named commands in Windows; you need to use ours, not theirs. There are other differences between your path and the one recommended in the Admin and Installation manual; they may also be causing problems. In http://cran.r-project.org/contrib/extra/batchfiles/ there is a batch file called Rfind.bat that searches your system and lists the paths of the various tools used by R. It does not actually change any environment variables nor does it change any other aspect of your system -- its display only so its generally safe to run. This might help you in configuration. __ 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] indexing into and modifying dendrograms
Probably not answering your questions here, but have you considered the
functions prune.tree and snip.tree from package tree or prune.rpart and
snip.rpart from the package rpart ?
On Mon, 2005-07-11 at 11:48 -0700, Jenny Bryan wrote:
I would like to be able to exert certain types of control over the
plotting of dendrograms (representing hierarchical clusterings) that I
think is best achieved by modifying the dendrogram object
prior to plotting. I am using the dendrogram class and associated
methods.
Define the cluster number of each cluster formed as the corresponding
row of the merge object. So, if you are clustering m objects, the
cluster numbers range from 1 to m-1, with cluster m-1 containing all m
objects, by definition. I basically want a way to index into an
object of class dendrogram using the above definition of cluster
number and/or to act on a dendrogram, where I specify the target node
using cluster number.
The first application would be to 'flip' the two elements in target
node of the dendrogram (made clear in the small example below). (The
setting is genomics and I have applications where I want to man-handle
my dendrograms to make certain features of the clustering more obvious
to the naked eye.) I could imagine other, related actions that would
be useful in decorating dendrograms.
I think I need a function that takes a dendrogram and cluster
number(s) as input and returns the relevant part(s) of the dendrogram
object -- but in a form that makes it easy to then, say, set certain
attributes (perhaps recursively) for the target nodes (and perhaps
those contained in it). I'm including a small example below that
hopefully illustrates this (it looks long, but it's mostly comments!).
Any help would be appreciated.
Jenny Bryan
## get ready for step-by-step figures
par(mfrow = c(2,2))
## get 5 objects, with 2-dimensional features
pts - rbind(c(2,1.6),
c(1.8,2.4),
c(2.1, 2.7),
c(5,2.6),
c(4.7,3.1))
plot(pts, xlim = c(0,6), ylim = c(0,4),type = n,
xlab = Feature 1, ylab = Feature 2)
points(pts,pch = as.character(1:5))
## build a hierarhical tree, store as a dendrogram
aggTree - hclust(dist(pts), method = single)
(dend1 - JB.as.dendrogram.hclust(aggTree))
## NOTE: only thing I added to official version of
## as.dendrogram.hclust:
## each node has an attribute cNum, which gives
## the merge step at which it was formed,
## i.e. gives the row of the merge object which
## describes the formation of that node
## one new line near end of nMerge loop:
## ***
## *** 51,56
## --- 51,60
## attr(z[[x[2]]], midpoint))/2
##}
## attr(zk, height) - oHgt[k]
## +
## + ## JB added July 6 2005
## + attr(zk, cNum) - k
## +
##z[[k - as.character(k)]] - zk
## }
## z - z[[k]]
attributes(dend1)
attributes(dend1[[1]])
## here's a table relating dend1 and the cNum attribute
## dend1 cNum
## -
## dend14
## dend1[[1]] 2
## dend1[[2]] 3
## dend1[[2]][[1]] not set
## dend1[[2]][[1]] 1
## use cNum attribute in edgetext
## following example in dendrogram documentation
## would really rather associate with the node than the edge
## but current plotting function has no notion of nodetext
addE - function(n) {
if(!is.leaf(n)) {
attr(n, edgePar) - list(p.col=plum)
attr(n, edgetext) - attr(n,cNum)
}
n
}
dend2 - dendrapply(dend1, addE)
## overlays the cNum (cluster number) attribute on dendrogram
plot(dend2, main = dend2)
## why does no plum polygon appear around the '4' for the root
## edge?
## swap order of clusters 2 and 3,
## i.e. 'flip' cluster 4
dend3 - dend2
dend3[[1]] - dend2[[2]]
dend3[[2]] - dend2[[1]]
plot(dend3, main = dend3)
## wish I could achieve with 'dend3 - flip(dend2, cNum = 4)
## swap order of cluster 1 and object 1,
## i.e. 'flip' cluster 3
dend4 - dend2
dend4[[2]][[1]] - dend2[[2]][[2]]
dend4[[2]][[2]] - dend2[[2]][[1]]
plot(dend4, main = dend4)
## wish I could achieve with 'dend4 - flip(dend2, cNum = 3)
## finally, it's clear that the midpoint attribute would also
## need to be modified by 'flip'
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Re: [R] exact values for p-values - more information.
Hi there, Actually my aim was to compare anumber of extreme values (e.g. 39540) with df1=1, df2=7025 via p-values. Spencer mentions that However, I have also used numbers like exp(-19775.52) to guestimate relative degrees of plausibility for different alternatives. Can someone point to me an article using this method? Regards. Stephen. - Original Message - From: Spencer Graves [EMAIL PROTECTED] Date: Monday, July 11, 2005 7:39 pm Subject: Re: [R] exact values for p-values - more information. I just checked: pf(39540, 1, 7025, lower.tail=FALSE, log.p=TRUE) [1] -Inf This is not correct. With 7025 denominator degrees of freedom, we might use the chi-square approximation to the F distribution: pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE) [1] -19775.52 In sum, my best approximation to pf(39540, 1, 7025, lower.tail=FALSE, log.p=TRUE), given only a minute to work on this, is exp(pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE)) = exp(-19775.52). I'm confident that many violations of assumptions would likely be more important than the differences between p-value: 2.2e-16 and That doesn't mean they are right, only the best I can get with the available resources. spencer graves Achim Zeileis wrote: On Mon, 11 Jul 2005, S.O. Nyangoma wrote: Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? You can always extract the `exact' p-value from the summary.lm object or you can compute it by hand via pf(39540, df1 = 1, df2 = 7025, lower.tail = FALSE) For all practical purposes, the above means that the p-value is 0. I guess you are on a 32-bit machine, then it also means that the p-value is smaller than the Machine epsilon .Machine$double.eps So if you want to report the p-value somewhere, I think R's output should be more than precise enough. If you want to compute some other values that depend on such a p-value, then it is probably wiser to compute on a log scale, i.e. instead pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE) use pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE, log.p = TRUE) However, don't expect to be able to evaluate it at such extreme values such as 39540. Z __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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] Dependence of bundle dse on setRNG (was Misbehaviour of DSE)
As the posting guide says, there is no R 2.1. The first message suggests this is R 2.1.0, and the posting guide does ask you to use the latest version (and to quote versions accurately). The dse bundle depends on package setRNG, which you have not installed, so you need to do that. Look at e.g. library(help=dse2) which the posting guide actually asks you to do and include in your posting. (And the message you got is pretty clearcut.) (Suggestion to Paul Gilbert: it would work better to have the setRNG dependence in the top-level DESCRIPTION and not in the dependence for each package. That way install.packages() can work out it is required from the information on CRAN, which is only at bundle level.) On Mon, 11 Jul 2005, Ajay Shah wrote: I am finding problems with using dse: library(dse1) Loading required package: tframe Error: c(package '%s' required by '%s' could not be found, setRNG, dse1) library(dse2) Loading required package: setRNG Error: package 'setRNG' could not be loaded In addition: Warning message: there is no package called 'setRNG' in: library(pkg, character.only = TRUE, logical = TRUE, lib.loc = lib.loc) This is on R 2.1 on an Apple ibook (OS X) panther. -- 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] Boxplot philosophy {was Boxplot in R}
On 11-Jul-05 Martin Maechler wrote:
AdaiR == Adaikalavan Ramasamy [EMAIL PROTECTED]
on Mon, 11 Jul 2005 03:04:44 +0100 writes:
AdaiR Just an addendum on the philosophical aspect of doing
AdaiR this. By selecting the 5% and 95% quantiles, you are
AdaiR always going to get 10% of the data as extreme and
AdaiR these points may not necessarily outliers. So when
AdaiR you are comparing information from multiple columns
AdaiR (i.e. boxplots), it is harder to say which column
AdaiR contains more extreme value compared to others etc.
Yes, indeed!
People {and software implementations} have several times provided
differing definitions of how the boxplot whiskers should be defined.
I strongly believe that this is very often a very bad idea!!
A boxplot should be a universal mean communication and so one
should be *VERY* reluctant redefining the outliers.
I just find that Matlab (in their statistics toolbox)
does *NOT* use such a silly 5% / 95% definition of the whiskers,
at least not according to their documentation.
That's very good (and I wonder where you, Larry, got the idea of
the 5 / 95 %).
Using such a fixed percentage is really a very inferior idea to
John Tukey's definition {the one in use in all implementations
of S (including R) probably for close to 20 years now}.
I see one flaw in Tukey's definition {which is shared of course
by any silly percentage based ``outlier'' definition}:
The non-dependency on the sample size.
If you have a 1000 (or even many more) points,
you'll get more and more `outliers' even for perfectly normal data.
But then, I assume John Tukey would have told us to do more
sophisticated things {maybe things like the violin plots} than
boxplot if you have really very many data points, you may want
to see more features -- or he would have agreed to use
boxplot(*, range = monotone_slowly_growing(n) )
for largish sample sizes n.
Martin Maechler, ETH Zurich
I happily agree with Martin's essay on Boxplot philiosophy.
It would cerainly confuse boxplot watchers if the interpretation
of what they saw had to vary from case to case. The fact that
careful (and necessarily detailed) explanations of what was
different this time would be necessary in the text would not
help much, and would defeat the primary objective of the boxplot
which is to present a summary of features of the data in a form
which can be grasped visually very quickly indeed.
I'm sure many of us have at times felt some frustration at the
rigidly precise numerical interpretations which Tukey imposed
on the elements of his many EDA techniques; but this did ensure
that the viewer really knew, at a glance, what he was looking at.
EDA brilliantly combined several aspects of looking at data:
selection of features of the data; highly efficient encoding of
these, and of their inter-relationships, into a medium directly
adapted to visual perception; robustness (so that the perceptions
were not unstable with respect to wondering just what the underlying
distribution might be); accessibility (in the sense of being truly
understood) to non-theoreticians; and capacity to be implemented on
primitive information technology.
Indeed, one might say that the core team of EDA consists of the
techniques for which you need only pencil and paper.
Nevertheless, Tukey was no rigid dogmatist. His objective was
always to give a good representation of the data, and he would
happily shift his ground, or adapt a technique (albeit probably
giving it a different name), or devise a new one, if that would
be useful for the case in hand.
Best wishes to all,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 11-Jul-05 Time: 22:19:47
-- XFMail --
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Re: [R] R on kubuntu
Dear R users, It's possible to install newer versions of R from the Ubuntu depositories, but the new version of R can only be found in 'breezy', the 'unstable' version of Ubuntu. You can however create a mixed system with most of your packages from the old 'hoary' and only some (like R) from the unstable 'breezy' by using the pinning mechanism in apt-get. Check out the instructions on pinning in the Ubuntu wiki: https://wiki.ubuntu.com/PinningHowto?highlight=%28apt-get+-t%29 In short it works like this: 1. Put in links to other depositories in /etc/apt/sources.list 2. Set the pinning priorities in /etc/apt/preferences 3. Install newer R-base with apt-get -t breezy install r-base Cheers, Hans Gardfjell Dept. of Ecology and Environmental Science Umeå University, Sweden Hello all, I am planning to redeploy my workstation under KUBUNTU. Does any body has any r experience installing/using r on this platform? Best regards. -- Constant Depièreux Managing Director Applied QUality Technologies Europe sprl Rue des Déportés 123, B-4800 Verviers (Tel) +32 87 292175 - (Fax) +32 87 292171 - (Mobile) +32 475 555 818 (Web) http://www.aqte.be - (Courriel) constant.depiereux at aqte.be https://stat.ethz.ch/mailman/listinfo/r-help (Skype) cdepiereux __ 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] Boxplot philosophy {was Boxplot in R}
FWIW:
I have been an enthusiastic user of boxplots for decades. Of course, the
issue of how to handle the whiskers (outliers] is a valid one, and indeed
sample size related. Dogma is always dangerous. I got to know John Tukey
somewhat (I used to chauffer him to and from meetings with a group of Merck
statisticians), and I,too,think he would have been the first to agree that
some flexibility here is wise.
HOWEVER, the chief advantage of boxplots is their simplicity at displaying
simultaneously and easily **several** important aspects of the data, of
which outliers are probably the most problematic (as they often result in
severe distortion of the plots without careful scaling). Even with dozens of
boxplots, center, scale, and skewness are easy to discern and compare. I
think this would NOT be true of violin plots and other more complex
versions -- simplicity can be a virtue.
Finally, a tidbit for boxplot afficianados: how does one detect bimodality
from a boxplot?
-- 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 Ted Harding
Sent: Monday, July 11, 2005 2:52 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] Boxplot philosophy {was Boxplot in R}
On 11-Jul-05 Martin Maechler wrote:
AdaiR == Adaikalavan Ramasamy [EMAIL PROTECTED]
on Mon, 11 Jul 2005 03:04:44 +0100 writes:
AdaiR Just an addendum on the philosophical aspect of doing
AdaiR this. By selecting the 5% and 95% quantiles, you are
AdaiR always going to get 10% of the data as extreme and
AdaiR these points may not necessarily outliers. So when
AdaiR you are comparing information from multiple columns
AdaiR (i.e. boxplots), it is harder to say which column
AdaiR contains more extreme value compared to others etc.
Yes, indeed!
People {and software implementations} have several times provided
differing definitions of how the boxplot whiskers should be defined.
I strongly believe that this is very often a very bad idea!!
A boxplot should be a universal mean communication and so one
should be *VERY* reluctant redefining the outliers.
I just find that Matlab (in their statistics toolbox)
does *NOT* use such a silly 5% / 95% definition of the whiskers,
at least not according to their documentation.
That's very good (and I wonder where you, Larry, got the idea of
the 5 / 95 %).
Using such a fixed percentage is really a very inferior idea to
John Tukey's definition {the one in use in all implementations
of S (including R) probably for close to 20 years now}.
I see one flaw in Tukey's definition {which is shared of course
by any silly percentage based ``outlier'' definition}:
The non-dependency on the sample size.
If you have a 1000 (or even many more) points,
you'll get more and more `outliers' even for perfectly normal data.
But then, I assume John Tukey would have told us to do more
sophisticated things {maybe things like the violin plots} than
boxplot if you have really very many data points, you may want
to see more features -- or he would have agreed to use
boxplot(*, range = monotone_slowly_growing(n) )
for largish sample sizes n.
Martin Maechler, ETH Zurich
I happily agree with Martin's essay on Boxplot philiosophy.
It would cerainly confuse boxplot watchers if the interpretation
of what they saw had to vary from case to case. The fact that
careful (and necessarily detailed) explanations of what was
different this time would be necessary in the text would not
help much, and would defeat the primary objective of the boxplot
which is to present a summary of features of the data in a form
which can be grasped visually very quickly indeed.
I'm sure many of us have at times felt some frustration at the
rigidly precise numerical interpretations which Tukey imposed
on the elements of his many EDA techniques; but this did ensure
that the viewer really knew, at a glance, what he was looking at.
EDA brilliantly combined several aspects of looking at data:
selection of features of the data; highly efficient encoding of
these, and of their inter-relationships, into a medium directly
adapted to visual perception; robustness (so that the perceptions
were not unstable with respect to wondering just what the underlying
distribution might be); accessibility (in the sense of being truly
understood) to non-theoreticians; and capacity to be implemented on
primitive information technology.
Indeed, one might say that the core team of EDA consists of the
techniques for which you need only pencil and paper.
Nevertheless, Tukey was no rigid dogmatist. His objective was
always to give a good representation of the data, and he would
happily shift his ground, or adapt a
[R] CIs in predict?
Dear All,
I am trying to put some Confidence intervals on some regressions from a linear
model with no luck. I can extract the fitted values using 'predict', but am
having difficulty in getting at the confidence intervals, or the standard
errors.
Any suggestions would be welcome
Cheers
Guy
Using Version 2.1.0 (2005-04-18) on a PC
vol.mod3 - lm(log.volume~log.area*lake,data=vol)
summary(vol.mod3)
plot(c(1.3,2.5),c(-0.7,0.45),type=n,xlab=Log area,ylab=Log volume)
areapred.a - seq(min(vol$log.area[vol$lake==a]),
max(vol$log.area[vol$lake==a]), length=100)
areapred.b - seq(min(vol$log.area[vol$lake==b]),
max(vol$log.area[vol$lake==b]), length=100)
preda - predict(vol.mod3, data.frame(log.area=areapred.a,interval=confidence
,lake=rep(a,100)))
#This gives the fitted values as predicted, but no CIs
preda
123456
789
-0.562577529 -0.553263576 -0.543949624 -0.534635671 -0.525321718 -0.516007765
-0.506693813 -0.497379860 -0.488065907
10 11 12 13 14 15
16 17 18
-0.478751955 -0.469438002 -0.460124049 -0.450810097 -0.441496144 -0.432182191
-0.422868239 -0.413554286 -0.404240333
19 20 21 22 23 24
25 26 27
-0.394926380 -0.385612428 -0.376298475 ETC ETC
#As does this, but with no SEs
preda - predict(vol.mod3, data.frame(log.area=areapred.a,se.fit=T
,lake=rep(a,100)))
preda
123456
789 10
-0.562577529 -0.553263576 -0.543949624 -0.534635671 -0.525321718 -0.516007765
-0.506693813 -0.497379860 -0.488065907 -0.478751955
11 12 13 14 15 16
17 18 19 20
-0.469438002 -0.460124049 -0.450810097 ETC ETC
Guy J Forrester
Biometrician
Manaaki Whenua - Landcare Research
PO Box 69, Lincoln, New Zealand.
Tel. +64 3 325 6701 x3738
Fax +64 3 325 2418
E-mail [EMAIL PROTECTED]
www.LandcareResearch.co.nz
WARNING: This email and any attachments may be confidential ...{{dropped}}
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Re: [R] CIs in predict?
Guy Forrester [EMAIL PROTECTED] writes: Dear All, I am trying to put some Confidence intervals on some regressions from a linear model with no luck. I can extract the fitted values using 'predict', but am having difficulty in getting at the confidence intervals, or the standard errors. Any suggestions would be welcome help(predict.lm) should get you there soon enough. -- 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] CIs in predict?
At 08:40 AM 12/07/2005, Guy Forrester wrote:
Dear All,
I am trying to put some Confidence intervals on some regressions from a
linear model with no luck. I can extract the fitted values using
'predict', but am having difficulty in getting at the confidence
intervals, or the standard errors.
Any suggestions would be welcome
Cheers
Guy
Using Version 2.1.0 (2005-04-18) on a PC
vol.mod3 - lm(log.volume~log.area*lake,data=vol)
summary(vol.mod3)
plot(c(1.3,2.5),c(-0.7,0.45),type=n,xlab=Log area,ylab=Log volume)
areapred.a - seq(min(vol$log.area[vol$lake==a]),
max(vol$log.area[vol$lake==a]), length=100)
areapred.b - seq(min(vol$log.area[vol$lake==b]),
max(vol$log.area[vol$lake==b]), length=100)
preda - predict(vol.mod3,
data.frame(log.area=areapred.a,interval=confidence ,lake=rep(a,100)))
You have interval=confidence inside your call to data.frame, not inside
your call to predict. Hence you are creating a data frame with a variable
called interval, with one level called confidence, and predict does not see
interval=confidence at all! See ?predict.lm.
HTH,
Simon.
#This gives the fitted values as predicted, but no CIs
preda
12345
6789
-0.562577529 -0.553263576 -0.543949624 -0.534635671 -0.525321718
-0.516007765 -0.506693813 -0.497379860 -0.488065907
10 11 12 13 14
15 16 17 18
-0.478751955 -0.469438002 -0.460124049 -0.450810097 -0.441496144
-0.432182191 -0.422868239 -0.413554286 -0.404240333
19 20 21 22 23
24 25 26 27
-0.394926380 -0.385612428 -0.376298475 ETC ETC
#As does this, but with no SEs
preda - predict(vol.mod3, data.frame(log.area=areapred.a,se.fit=T
,lake=rep(a,100)))
preda
12345
6789 10
-0.562577529 -0.553263576 -0.543949624 -0.534635671 -0.525321718
-0.516007765 -0.506693813 -0.497379860 -0.488065907 -0.478751955
11 12 13 14 15
16 17 18 19 20
-0.469438002 -0.460124049 -0.450810097 ETC ETC
Guy J Forrester
Biometrician
Manaaki Whenua - Landcare Research
PO Box 69, Lincoln, New Zealand.
Tel. +64 3 325 6701 x3738
Fax +64 3 325 2418
E-mail [EMAIL PROTECTED]
www.LandcareResearch.co.nz
WARNING: This email and any attachments may be confidential ...{{dropped}}
__
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Re: [R] Problems with R on OS X
On 11/07/2005, at 8:00 PM, Heinz Schild wrote: I used R on OS X 10.3x quite some time with no serious problems. Sometimes R stopped when I tried to execute a bigger program. After updating to OS X to version 10.4 R worked but I still had the problem with bigger programs. Therefore I re-installed R on top of the existing R version. The installation finished properly but suddenly R did not work. Then I reinstalled OS X 10.4 because I thought some left over registry information from R may caused the problem. R still crashed. I hoped the problem would be overcome with the new R version 1.12. Unfortunately this is not the case. The error report (see appendix) says that Thread 0 crashed. What can I do? Heinz Schild The best place for MacOS issues is the r-sig-mac list to which I have copied this mesage. MacOS X does not have a registry! Configuration files for MacOS applications live in ~/Library/Preferences. The file for the R GUI app is called org.R-project.R.plist. The rest of R on a Mac is just like any other UNIX platform with settings and history in ~/.Rdata and ~/.Rhistory. (Files starting with a period '.' are invisible in the MacOS desktop.) If you want to trash the configuration just delete these three files. The version 1.12 you quote is only for the GUI, which is just a front end for the R framework. The current version of the framework binary is 2.1.1, which you will see on the 'R Console' window when you start the application. If you try the 1.12 GUI with a version of the framework other than 2.1.1 it is very likely to crash. If you install the complete package from CRAN it will install both the GUI application and the framework. BIll Northcott __ 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] nlme plot
Hello,
I am running this script from Pinheiro Bates book in R Version 2.1.1 (WinXP).
But, I can't plot Figure 2.3.
What's wrong?
TIA.
Rod.
-
library(nlme)
names( Orthodont )
[1] distance age Subject Sex
levels( Orthodont$Sex )
[1] Male Female
OrthoFem - Orthodont[ Orthodont$Sex == Female, ]
fm1OrthF - lme( distance ~ age, data = OrthoFem, random = ~ 1 | Subject )
fm2OrthF - update( fm1OrthF, random = ~ age | Subject )
orthLRTsim - simulate.lme( fm1OrthF, fm2OrthF, nsim = 1000 )
plot( orthLRTsim, df = c(1, 2) )# produces Figure 2.3
Error in if ((dfType - as.double(names(x)[1])) == 1) { :
argument is of length zero
Execution halted
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[R] simulate.lme plot
Hello,
I am running this script from Pinheiro Bates book in R (Version 2.1.1,WinXP).
But, I can't plot Figure 2.3.
What's wrong?
Thanks, Rod.
-
library(nlme)
names( Orthodont )
[1] distance age Subject Sex
levels( Orthodont$Sex )
[1] Male Female
OrthoFem - Orthodont[ Orthodont$Sex == Female, ]
fm1OrthF - lme( distance ~ age, data = OrthoFem, random = ~ 1 | Subject )
fm2OrthF - update( fm1OrthF, random = ~ age | Subject )
orthLRTsim - simulate.lme( fm1OrthF, fm2OrthF, nsim = 1000 )
plot( orthLRTsim, df = c(1, 2) )# produces Figure 2.3
Error in if ((dfType - as.double(names(x)[1])) == 1) { :
argument is of length zero
Execution halted
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[R] transition matrix and discretized data
Hi there, I have data on earnings of 12000 individuals at two points in time. I intend to construct a transition matrix, where the typical element, p_ij, gives the probability that an individual ends at the j-th decile of the earnings distribution given that he was was initially at the i-th decile. Thus, this is a bi-stochastic matrix. The problem is that the income data is nearly discrete in the sense that many individuals hold the same income level at each point. For instance, there are 1400 individuals who earned in period-one the minimum positive income (say, $100). Therefore, in the first decile there will be more than 10% of the individuals. This happens for both periods, and for a few income levels. As a result, the transition matrix won't have both rows and columns summing to one. The solution I've found for this problem was to generate a uniform random vector, with entries ranging from, say, -.0001 to .0001, and ad it to both earnings vectors and compute the transition matrix. Repeat the procedure 1000 times and get the mean of the resulting matrices. The thing is I'm totally new to simulations. Here's part of what I'm trying to do: # X is a two-comun data frame. column 1 is the period-one individuals' earnings and column two is the period-two. n - nrow(X)#12000 sim - runif(n, -.0001, .0001) X - X + sim q - 10# in order to compute deciles. it could be quintiles, quartiles, whatever p - seq(0,1,1/q) f.x - quantile(X[,1], p, names=F) f.y - quantile(X[,2], p, names=F) f.x[1] - 0; f.y[1] - 0 a - cut(X[,1], f.x, right=T) b - cut(X[,2], f.y, right=T) P - table(a,b) P - P/rowSums(P) P The point is that I don't know how to store the matrix P efficiently so that it can be averaged with the remianing 999. Also, suggestions on how to solve the problem of discretized data are welcome. Thanks a lot, Dimitri [[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] nlme plot
On 7/11/05, R V [EMAIL PROTECTED] wrote:
Hello,
I am running this script from Pinheiro Bates book in R Version 2.1.1
(WinXP).
But, I can't plot Figure 2.3.
What's wrong?
There was a change in the way that R handles assignments of names of
components and that affected the construction of this plot. I've
rewritten the plot construction code (the logic in the current version
was bizarre) and will upload a new version of nlme after I test this
out.
By the way, there is a simpler way of reproducing the examples in our book. Use
source(system.file(scripts/ch02.R, package = nlme))
but don't try that with the current version of the nlme package.
There is another infelicity (to use Bill Venables' term) that I will
correct.
TIA.
Rod.
-
library(nlme)
names( Orthodont )
[1] distance age Subject Sex
levels( Orthodont$Sex )
[1] Male Female
OrthoFem - Orthodont[ Orthodont$Sex == Female, ]
fm1OrthF - lme( distance ~ age, data = OrthoFem, random = ~ 1 | Subject )
fm2OrthF - update( fm1OrthF, random = ~ age | Subject )
orthLRTsim - simulate.lme( fm1OrthF, fm2OrthF, nsim = 1000 )
plot( orthLRTsim, df = c(1, 2) )# produces Figure 2.3
Error in if ((dfType - as.double(names(x)[1])) == 1) { :
argument is of length zero
Execution halted
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Re: [R] exact values for p-values - more information.
I don't have a reference, but you could look under data mining and p values / Bonferroni. spencer graves S.O. Nyangoma wrote: Hi there, Actually my aim was to compare anumber of extreme values (e.g. 39540) with df1=1, df2=7025 via p-values. Spencer mentions that However, I have also used numbers like exp(-19775.52) to guestimate relative degrees of plausibility for different alternatives. Can someone point to me an article using this method? Regards. Stephen. - Original Message - From: Spencer Graves [EMAIL PROTECTED] Date: Monday, July 11, 2005 7:39 pm Subject: Re: [R] exact values for p-values - more information. I just checked: pf(39540, 1, 7025, lower.tail=FALSE, log.p=TRUE) [1] -Inf This is not correct. With 7025 denominator degrees of freedom, we might use the chi-square approximation to the F distribution: pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE) [1] -19775.52 In sum, my best approximation to pf(39540, 1, 7025, lower.tail=FALSE, log.p=TRUE), given only a minute to work on this, is exp(pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE)) = exp(-19775.52). I'm confident that many violations of assumptions would likely be more important than the differences between p-value: 2.2e-16 and That doesn't mean they are right, only the best I can get with the available resources. spencer graves Achim Zeileis wrote: On Mon, 11 Jul 2005, S.O. Nyangoma wrote: Hi there, If I do an lm, I get p-vlues as p-value: 2.2e-16 This is obtained from F =39540 with df1 = 1, df2 = 7025. Suppose am interested in exact value such as p-value = 1.6e-16 (note = and not ) How do I go about it? You can always extract the `exact' p-value from the summary.lm object or you can compute it by hand via pf(39540, df1 = 1, df2 = 7025, lower.tail = FALSE) For all practical purposes, the above means that the p-value is 0. I guess you are on a 32-bit machine, then it also means that the p-value is smaller than the Machine epsilon .Machine$double.eps So if you want to report the p-value somewhere, I think R's output should be more than precise enough. If you want to compute some other values that depend on such a p-value, then it is probably wiser to compute on a log scale, i.e. instead pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE) use pf(70, df1 = 1, df2 = 7025, lower.tail = FALSE, log.p = TRUE) However, don't expect to be able to evaluate it at such extreme values such as 39540. Z __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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 -- Spencer Graves, PhD Senior Development Engineer PDF Solutions, Inc. 333 West San Carlos Street Suite 700 San Jose, CA 95110, USA [EMAIL PROTECTED] www.pdf.com http://www.pdf.com Tel: 408-938-4420 Fax: 408-280-7915 __ 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] Boxplot philosophy {was Boxplot in R}
I'll bite: How does one detect bimodalidty from a boxplot?
spencer graves
Berton Gunter wrote:
FWIW:
I have been an enthusiastic user of boxplots for decades. Of course, the
issue of how to handle the whiskers (outliers] is a valid one, and indeed
sample size related. Dogma is always dangerous. I got to know John Tukey
somewhat (I used to chauffer him to and from meetings with a group of Merck
statisticians), and I,too,think he would have been the first to agree that
some flexibility here is wise.
HOWEVER, the chief advantage of boxplots is their simplicity at displaying
simultaneously and easily **several** important aspects of the data, of
which outliers are probably the most problematic (as they often result in
severe distortion of the plots without careful scaling). Even with dozens of
boxplots, center, scale, and skewness are easy to discern and compare. I
think this would NOT be true of violin plots and other more complex
versions -- simplicity can be a virtue.
Finally, a tidbit for boxplot afficianados: how does one detect bimodality
from a boxplot?
-- 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 Ted Harding
Sent: Monday, July 11, 2005 2:52 PM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] Boxplot philosophy {was Boxplot in R}
On 11-Jul-05 Martin Maechler wrote:
AdaiR == Adaikalavan Ramasamy [EMAIL PROTECTED]
on Mon, 11 Jul 2005 03:04:44 +0100 writes:
AdaiR Just an addendum on the philosophical aspect of doing
AdaiR this. By selecting the 5% and 95% quantiles, you are
AdaiR always going to get 10% of the data as extreme and
AdaiR these points may not necessarily outliers. So when
AdaiR you are comparing information from multiple columns
AdaiR (i.e. boxplots), it is harder to say which column
AdaiR contains more extreme value compared to others etc.
Yes, indeed!
People {and software implementations} have several times provided
differing definitions of how the boxplot whiskers should be defined.
I strongly believe that this is very often a very bad idea!!
A boxplot should be a universal mean communication and so one
should be *VERY* reluctant redefining the outliers.
I just find that Matlab (in their statistics toolbox)
does *NOT* use such a silly 5% / 95% definition of the whiskers,
at least not according to their documentation.
That's very good (and I wonder where you, Larry, got the idea of
the 5 / 95 %).
Using such a fixed percentage is really a very inferior idea to
John Tukey's definition {the one in use in all implementations
of S (including R) probably for close to 20 years now}.
I see one flaw in Tukey's definition {which is shared of course
by any silly percentage based ``outlier'' definition}:
The non-dependency on the sample size.
If you have a 1000 (or even many more) points,
you'll get more and more `outliers' even for perfectly normal data.
But then, I assume John Tukey would have told us to do more
sophisticated things {maybe things like the violin plots} than
boxplot if you have really very many data points, you may want
to see more features -- or he would have agreed to use
boxplot(*, range = monotone_slowly_growing(n) )
for largish sample sizes n.
Martin Maechler, ETH Zurich
I happily agree with Martin's essay on Boxplot philiosophy.
It would cerainly confuse boxplot watchers if the interpretation
of what they saw had to vary from case to case. The fact that
careful (and necessarily detailed) explanations of what was
different this time would be necessary in the text would not
help much, and would defeat the primary objective of the boxplot
which is to present a summary of features of the data in a form
which can be grasped visually very quickly indeed.
I'm sure many of us have at times felt some frustration at the
rigidly precise numerical interpretations which Tukey imposed
on the elements of his many EDA techniques; but this did ensure
that the viewer really knew, at a glance, what he was looking at.
EDA brilliantly combined several aspects of looking at data:
selection of features of the data; highly efficient encoding of
these, and of their inter-relationships, into a medium directly
adapted to visual perception; robustness (so that the perceptions
were not unstable with respect to wondering just what the underlying
distribution might be); accessibility (in the sense of being truly
understood) to non-theoreticians; and capacity to be implemented on
primitive information technology.
Indeed, one might say that the core team of EDA consists of the
techniques for which you need only pencil and paper.
Nevertheless, Tukey was no rigid dogmatist. His objective was
always to give a good representation of the data, and he would
happily shift his
[R] R CMD INSTALL use differenct c++ compiler
Dear R Users, When I installed e1071 use R CMD INSTALL, I got configure: WARNING: g++ 2.96 cannot reliably be used with this package. configure: error: Please use a different C++ compiler. But how to let R CMD INSTALL use a different C++ compiler? and which C++ compiler is good? -Luke __ 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] ---- Message Quarantined! ----
MailMarshal has quarantined the following message you sent: Message: B42d3485f0001.0001.0002.mml From:[EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Good day With the proliferation of mass mailer type viruses, Massey University has had to implement a policy that quarantines emails with attachments that could be potential viruses, e.g. VBS, SCR and EXE. However, this message is eligible for Self-Service Message Release. If this attachment is genuine and you are certain that it does not contain any viruses, please simply reply to this message without editing it. The quarantined email will be automatically released to [EMAIL PROTECTED] upon receipt of your reply, else it will be deleted after 14 days. Please note that your reply will be processed by the automated release system. If you have any queries or comments, please send them to [EMAIL PROTECTED] Message release code: MMRelCode ZYoJxwg/WQoP+jKD7y9MEPkkJa7kC4lbnw7+GjUyOmMbCbdfbt MkS6FoVQwZpNHhLosQRcgvZoUocQ+3XTWkjwMxfFQw0Ea0zz/z VtE7ZIuQXVj8ZKBR0Q4YvUEN2Qa/6Q5POCRmm/qTc6d+Wg== MMEndRelCode __ 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] How to obtain Frequency Tables for Histogram outputs and Frequency Polygons
Couple of questions: 1. How can I obtain the frequency tables for a histogram chart? 2. Is there a short cut to obtain the frequency polygons directly without having to generate the frequency table and doing a line plot? Thanks ina dvance for any reply. j. Example data: 0.3,0.229,0.218,0.211,0.207,0.21,0.212,0.232,0.312,0.289,0.486,0.475,0.274,0.478,0.284,0.477,0.281,0.29,0.29,0.32,0.28,0.404,0.489,0.406,0.267,0.456,0.443,0.457,0.411,0.444,0.401,0.468,0.309,0.425,0.312,0.3,0.464,0.457,0.442,0.469,0.423,0.403,0.432,0.379,0.361,0.37,0.47,0.374,0.387,0.381,0.311,0.257,0.381,0.162,0.229,0.221,0.25,0.164,0.368,0.367,0.332,0.165,0.152,0.478,0.489,0.482,0.167,0.164,0.164,0.15,0.149,0.368,0.366,0.361,0.359,0.359,0.364,0.364 __ 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 obtain Frequency Tables for Histogram outputs and Frequency Polygons
On 7/12/05, Tan Hui Hui Jenny [EMAIL PROTECTED] wrote: Couple of questions: 1. How can I obtain the frequency tables for a histogram chart? 2. Is there a short cut to obtain the frequency polygons directly without having to generate the frequency table and doing a line plot? res - hist(x) # res contains various data about histogram # create red polygons using rect plot(range(res$breaks), range(res$counts), type=n) rect(res$breaks[-length(res$breaks)],0,res$breaks[-1],res$counts, col=red) __ 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
