[R] difference between lrm's Model L.R. and anova's Chi-Square
I am running lrm() with a single factor. I then run anova() on the fitted model to obtain a p-value associated with having that factor in the model. I am noticing that the Model L.R. in the lrm results is almost the same as the Chi-Square in the anova results, but not quite; the latter value is always slightly smaller. anova() calculates the p-value based on Chi-Square, but I have independent evidence that Model L.R. is the actual -2*log(LR), so should I be using that? Why are the values different? prob_a - inv.logit(rnorm(1,0,1)) prob_b - inv.logit(rnorm(1,0,1)) data - data.frame( factor=c(rep(a,500),rep(b,500)), outcome=c(sample(c(1,0),100,replace=T,prob=c(prob_a,1-prob_a)), sample(c(1,0),100,replace=T,prob=c(prob_b,1-prob_b fit - lrm(outcome~factor,data) fit # gives Model L.R. e.g. 8.23, 11.76, 6.89... anova(fit)# gives Chi-Square e.g. 8.19, 11.69, 6.85... __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] difference between lrm's Model L.R. and anova's Chi-Square
Quoting Frank E Harrell Jr [EMAIL PROTECTED]: anova (anova.Design) computes Wald statistics. When the log-likelihood is very quadratic, these statistics will be very close to log-likelihood ratio chi-square statistics. In general LR chi-square tests are better; we use Wald tests for speed. It's best to take the time and do lrtest(fit1,fit2) in Design, where one of the two fits is a subset of the other. Frank Harrell Thanks, this is great, but in my case, there's just one factor, fit1 - lrm(outcome~factor,data) and I'm having trouble constructing the subset 'null model', as e.g. fit2 - lrm(outcome~1,data) returns an error message. How do I construct a null model with lrm() so that I can use lrtest() to test a model with only one predictor? I apologize for asking what must be a very simple question but I have been unable to find the answer by searching R-help. Thanks, Dan P.S. Second point: I have another case where I use lmer(), and there the null model includes a random effect so I don't get the problem above. It looks like with lmer objects anova() uses LLR, not Wald. Is that right? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Poisson regression in R
glmstat wrote: I have these questions: (1) Use Poisson regression to estimate the main effects of car, age, and dist (each treated as categorical and modelled using indicator variables) and interaction terms. (2) It was determined by one study that all the interactions were unimportant and decided that age and car could be treated as though they were continuous variables. Fit a model incorporating these features and compare it with the best model obtained in (1). This looks like homework, so only hints are offered. You don't seem to be using n, consider incorporating an offset (I would expect most texts on Poison regr. to discuss this). n is the number of insurance policies y is the number of claims car is the car in an insurance category age is the age of policy holder dist is the district where the policy holder lived (1 for London and other major cities, and 0 otherwise) Data: car age disty n 1 1 0 65 317 1 2 0 65 476 1 3 0 52 486 1 4 0 310 3259 2 1 0 98 486 2 2 0 159 1004 2 3 0 175 1355 2 4 0 877 7660 3 1 0 41 223 3 2 0 117 539 3 3 0 137 697 3 4 0 477 3442 4 1 0 11 40 4 2 0 35 148 4 3 0 39 214 4 4 0 167 1019 1 1 1 2 20 1 2 1 5 33 1 3 1 4 40 1 4 1 36 316 2 1 1 7 31 2 2 1 10 81 2 3 1 22 122 2 4 1 102 724 3 1 1 5 18 3 2 1 7 39 3 3 1 16 68 3 4 1 63 344 4 1 1 0 3 4 2 1 6 16 4 3 1 8 25 4 4 1 33 114 I need help finding the correct R code to construct models. According to the previous study, the model in (2) is simpler than (1), fits well (deviance = 53.11, d.f. = 60, p-value = 0.72) and gives coefficients (standard errors): AGE, – 0.177 (0.018); CAR, 0.198 (0.021); DIST, 0.210 (0.059). As of the first model, I think that I should use this code, but not sure: firstmodel-glm(y~factor(age)*factor(car)*factor(dist),family=poisson) As of the second model, I used this code, but it produces results that contradict what the previous study says (and deleting intercept does not help): secondmodel-glm(y~age+car+factor(dist),family=poisson) summary(secondmodel) Call: glm(formula = y ~ age + car + factor(dist), family = poisson) Deviance Residuals: Min1QMedian3Q Max -14.0258 -3.3200 -0.62962.0575 18.1442 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)3.082220.08127 37.92 2e-16 *** age0.836640.02067 40.48 2e-16 *** car -0.167230.01612 -10.37 2e-16 *** factor(dist)1 -2.159370.05849 -36.92 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 5660.6 on 31 degrees of freedom Residual deviance: 1154.5 on 28 degrees of freedom AIC: 1330.8 Number of Fisher Scoring iterations: 5 -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Could not install aroma.affymetrix
Hi My, On Sat, Mar 1, 2008 at 8:04 PM, My Coyne [EMAIL PROTECTED] wrote: I don't know if this is the correct forum to ask the following question; however, when I search the aroma.affymetrix discussion group, it suggested that I should posted the question to r-help. Here it goes. I think you have misinterpreted some message(s), because I don't know of any place where we suggest that you should ask r-help list about installation problems on aroma.affymetrix. Let me know if I am wrong so the documentation can be corrected. I followed the instructions on aroma.affymetrix trying to install the packages; following are the steps: install.packages(c(R.oo, R.utils), contriburl=http://www.braju.com/R/repos/;) That explains how to install the R.oo and R.utils packages, but it is certainly not the suggested way to install aroma.affymetrix. *Please* let me know where you got this from. See the Page 'Installation, Updates Patches' on the aroma.affymetrix website (http://www.braju.com/R/aroma.affymetrix/) for installation instructions. /Henrik trying URL 'http://www.braju.com/R/repos//R.oo_1.4.1.zip' Content type 'application/zip' length 889981 bytes (869 Kb) opened URL downloaded 869 Kb trying URL 'http://www.braju.com/R/repos//R.utils_1.0.0.zip' Content type 'application/zip' length 893989 bytes (873 Kb) opened URL downloaded 873 Kb package 'R.oo' successfully unpacked and MD5 sums checked package 'R.utils' successfully unpacked and MD5 sums checked The downloaded packages are in C:\Documents and Settings\mcoyne\Local Settings\Temp\Rtmpjhahhp\downloaded_packages updating HTML package descriptions However, when I tried to load the package aroma.affymetrix I got the following error: library (aroma.affymetrix) Error in library(aroma.affymetrix) : there is no package called 'aroma.affymetrix' Any help is greatly appreciated. My D. Coyne [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie: Incorrect number of dimensions
Here is the structure str(all_differ) num [1:6280, 1:8] 2.22e-16 2.22e-16 2.22e-16 2.22e-16 2.22e-16 ... - attr(*, dimnames)=List of 2 ..$ : NULL ..$ : chr [1:8] rawp Bonferroni Holm Hochberg ... I tried, but it still giving me the same error message Incorrect Dimension probe.names-all_differ[[1]][all_differ[[6280]][,BY]=0.01] Am i using the wrong dimension? thanks, Keizer jholtman wrote: It would be helpful if you provided commented, minimal, self-contained, reproducible code. What does str(all_differ) say? That will tell you the structure of the object that you are trying to work with. On Sat, Mar 1, 2008 at 3:35 AM, Keizer_71 [EMAIL PROTECTED] wrote: dim(data.sub) [1] 1 140 #extracting all differentially express genes## library(multtest) two_side- (1-pt(abs(data.sub),50))*2 diff- mt.rawp2adjp(two_side) all_differ-diff[[1]][37211:1,] all_differ #list of differentially expressed genes## probe.names- + all_differ[[2]][all_differ[[1]][,BY]=0.01] Error in all_differ[[1]][, BY] : incorrect number of dimensions Hi, I am pretty new with R. What i am trying to do is to find all differentially express genes and list of differentially expressed genes. Am i doing something wrong? I keep getting incorrect number of dimensions. How do i find out the correct dimensions? thanks, Keizer -- View this message in context: http://www.nabble.com/Newbie%3A-Incorrect-number-of-dimensions-tp15773090p15773090.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Newbie%3A-Incorrect-number-of-dimensions-tp15773090p15785977.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie: Incorrect number of dimensions
You have a 'matrix' not a 'list', so I think you want something like this: all_differ[all_differ[, 'BY'] = 0.01, 'BY'], this will return the values where column is 'BY' in all rows which is less or equal to 0.01. On 02/03/2008, Keizer_71 [EMAIL PROTECTED] wrote: Here is the structure str(all_differ) num [1:6280, 1:8] 2.22e-16 2.22e-16 2.22e-16 2.22e-16 2.22e-16 ... - attr(*, dimnames)=List of 2 ..$ : NULL ..$ : chr [1:8] rawp Bonferroni Holm Hochberg ... I tried, but it still giving me the same error message Incorrect Dimension probe.names-all_differ[[1]][all_differ[[6280]][,BY]=0.01] Am i using the wrong dimension? thanks, Keizer jholtman wrote: It would be helpful if you provided commented, minimal, self-contained, reproducible code. What does str(all_differ) say? That will tell you the structure of the object that you are trying to work with. On Sat, Mar 1, 2008 at 3:35 AM, Keizer_71 [EMAIL PROTECTED] wrote: dim(data.sub) [1] 1 140 #extracting all differentially express genes## library(multtest) two_side- (1-pt(abs(data.sub),50))*2 diff- mt.rawp2adjp(two_side) all_differ-diff[[1]][37211:1,] all_differ #list of differentially expressed genes## probe.names- + all_differ[[2]][all_differ[[1]][,BY]=0.01] Error in all_differ[[1]][, BY] : incorrect number of dimensions Hi, I am pretty new with R. What i am trying to do is to find all differentially express genes and list of differentially expressed genes. Am i doing something wrong? I keep getting incorrect number of dimensions. How do i find out the correct dimensions? thanks, Keizer -- View this message in context: http://www.nabble.com/Newbie%3A-Incorrect-number-of-dimensions-tp15773090p15773090.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Newbie%3A-Incorrect-number-of-dimensions-tp15773090p15785977.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] jpeg and margin text
Thomas Schwander wrote:
Hi guys,
I use R 2.6.2 and Windows XP.
I’ve got the following question: I wrote a lot of text into the margin of a
plot. When I use the “jpeg”-function, only a little part of the margin text
is displayed in the final jpeg? Any ideas to change it?
Source-Code:
datenbankdaten-data.frame(LETTERS[1:20],
c(8,4.8,3.8,2.7,2.6,2.1,2,1.9,1.6,1,0.99,0.98,0.98,0.97,0.96,.96,0.8,0.8,0.7
,0.6),
c(8,4.8,3.8,0,2.6,2.1,2,1.9,1.6,1.1,0.99,0.98,0.97,0.97,0.96,.95,0.8,0.8,0,0
.6))
Zugehoerigkeit-LETTERS[1:20]
Bezeichnung-letters[1:20]
colnames(datenbankdaten)-c(Bezeichnung,A,B)
# par = Setzt die Grafikparameter, die durch plot automatisch verwendet
werden
# mar = Anzahl der Textzeilen von den inneren Achsen des Schaubild zum Rand
# Zeilenvektor: c(bottom, left, top, right)
par(mar=c(3,4,4,25),family=serif)
Fontparameter-1
today-Sys.Date()
Vormonat-Sys.Date()-20
Stand_aktuell-format(today, %B %Y)
Stand_Vormonat-format(Vormonat, %B %Y)
jpeg(paste(C:\\TOP-20.jpeg,sep=),quality=100,width=1024, height=768)
plot(datenbankdaten[,2],ylim=c(0,max(datenbankdaten[,2],datenbankdaten[,3]))
,
xlab=, ylab=,
font.axis=Fontparameter,font.lab=Fontparameter,font.main=Fontparameter,font.
sub=Fontparameter,
bty=n,axes=F)
axis(1,at=1:20,label=rep(,20))
axis(1,at=1:9,label=sprintf(0%.0f,1:9))
axis(1,at=10:20,label=10:20)
Maximum-max(datenbankdaten[,2],datenbankdaten[,3])
axis(2,at=0:Maximum,label=sprintf(%.2f,0:Maximum))
for(i in 1:(length(datenbankdaten[,2])+2)){
abline(h=i-1)
}
for(i in 1:20){
if(datenbankdaten[i,2]=datenbankdaten[i,3]) farbe=red else
farbe=green
arrows(i,datenbankdaten[i,2],i,datenbankdaten[i,3],col=farbe,length=0,lwd=1.
5)
points(i,datenbankdaten[i,2],pch=22,col=light blue,bg=light
blue,cex=1.7)
points(i,datenbankdaten[i,3],pch=21,col=grey,bg=grey,cex=1.5)
}
for(i in 1:20){
if(datenbankdaten[i,2]datenbankdaten[i,3])
#dreieck nach oben
points(i,datenbankdaten[i,2]-0.1,pch=24,col=red,bg=red)
if(datenbankdaten[i,2]datenbankdaten[i,3])
#dreieck nach unten
points(i,datenbankdaten[i,2]+0.1,pch=25,col=green,bg=green)
}
# Legend muss noch in die Ränder geschrieben werden, würde besser aussehen
Verkleinern-0.8
for(i in 1:20){
if(i=9) text(par(usr)[2] + 0.5, Maximum-(i-1)*0.4, srt=0, adj = 0, labels
= sprintf(0%.0f,i),xpd = TRUE,cex=Verkleinern)
else text(par(usr)[2] + 0.5, Maximum-(i-1)*0.4, srt=0, adj = 0, labels =
i,xpd = TRUE,cex=Verkleinern)
text(par(usr)[2] + 1, Maximum-(i-1)*0.4, srt=0, adj = 0,
labels = Zugehoerigkeit[i],
xpd = TRUE,cex=Verkleinern)
text(par(usr)[2] + 3.6, Maximum-(i-1)*0.4, srt=0, adj = 0,
labels = Bezeichnung[i],
xpd = TRUE,cex=Verkleinern)
}
lines(c(par(usr)[2] + 0.4,par(usr)[2] +
0.4),c(Maximum+0.1,0.25),xpd=TRUE)
lines(c(par(usr)[2] + 11.5,par(usr)[2] +
11.5),c(Maximum+0.1,0.25),xpd=TRUE)
lines(c(par(usr)[2] + 0.9,par(usr)[2] +
0.9),c(Maximum+0.1,0.25),xpd=TRUE)
lines(c(par(usr)[2] + 3.45,par(usr)[2] +
3.45),c(Maximum+0.1,0.25),xpd=TRUE)
Abstand-Maximum-0.15
# aktueller Standpunkt
for(i in 1:20){
lines(c(par(usr)[2] + 0.4,par(usr)[2] +
11.5),c(Maximum+0.1-Abstand*(i-1)/20,Maximum+0.1-Abstand*(i-1)/20),xpd=TRUE)
}
lines(c(par(usr)[2] + 0.4,par(usr)[2] + 11.5),c(0.25,0.25),xpd=TRUE)
#Ende Test
legend(topright,legend=c(Stand_aktuell,Stand_Vormonat),pch=c(22,21),
bty=o,pt.bg=c(light blue,grey),cex=0.8,bg=white)
dev.off()
Hi Thomas,
You seem to have a very wide table on the right of the plot, try this:
x11(width=10,height=7)
par(mar=c(5,4,4,20))
plot(...
Jim
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] an efficient pairwise matrix cell's comparison function
Does this do what you want? A - matrix(sample(0:2, 25, TRUE), ncol=5) B - matrix(1:25, ncol=5) C - ifelse(A == 0, 0, B) A [,1] [,2] [,3] [,4] [,5] [1,]11121 [2,]10110 [3,]00102 [4,]01200 [5,]12122 B [,1] [,2] [,3] [,4] [,5] [1,]16 11 16 21 [2,]27 12 17 22 [3,]38 13 18 23 [4,]49 14 19 24 [5,]5 10 15 20 25 C [,1] [,2] [,3] [,4] [,5] [1,]16 11 16 21 [2,]20 12 170 [3,]00 130 23 [4,]09 1400 [5,]5 10 15 20 25 On Sun, Mar 2, 2008 at 7:11 AM, Diogo André Alagador [EMAIL PROTECTED] wrote: To all, I am undergoing an analysis involving big matrices of about 3x200 which I have to handle in a more efficient way. So I would like some advice to build such efficient function to deliver the following result: - starting with 2 matrices of the same dimension (eg. A and B) 0 0 3 5 6 0 0 5 A= 0 0 6 4 B= 0 4 3 5 0 0 5 0 1 0 0 9 - the function should deliver a C matrix (same dimension too), where at each position C(i,j), compares A and B. if A(i,j)=0, than C(i,j)=0, if A(i,j)!=0, than C(i,j)=B(i,j) 6 0 0 5 C= 0 0 3 5 0 0 0 0 Although not an expert I could build a function with 2 cycles (reading columns and rows) which is not quick. Maybe you can help me in this challenge. Much thanks in advance, Diogo André Alagador Biodiversity Global Change Lab, Museo Nacional de Ciencias Naturales, CSIC, Madrid, España Forest Research Centre, Instituto Superior de Agronomia, Universidade Técnica de Lisboa, Lisboa, Portugal [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] an efficient pairwise matrix cell's comparison function
C - B C[A==0] - 0 would be somewhat more efficient. On Sun, 2 Mar 2008, jim holtman wrote: Does this do what you want? A - matrix(sample(0:2, 25, TRUE), ncol=5) B - matrix(1:25, ncol=5) C - ifelse(A == 0, 0, B) A [,1] [,2] [,3] [,4] [,5] [1,]11121 [2,]10110 [3,]00102 [4,]01200 [5,]12122 B [,1] [,2] [,3] [,4] [,5] [1,]16 11 16 21 [2,]27 12 17 22 [3,]38 13 18 23 [4,]49 14 19 24 [5,]5 10 15 20 25 C [,1] [,2] [,3] [,4] [,5] [1,]16 11 16 21 [2,]20 12 170 [3,]00 130 23 [4,]09 1400 [5,]5 10 15 20 25 On Sun, Mar 2, 2008 at 7:11 AM, Diogo André Alagador [EMAIL PROTECTED] wrote: To all, I am undergoing an analysis involving big matrices of about 3x200 which I have to handle in a more efficient way. So I would like some advice to build such efficient function to deliver the following result: - starting with 2 matrices of the same dimension (eg. A and B) 0 0 3 5 6 0 0 5 A= 0 0 6 4 B= 0 4 3 5 0 0 5 0 1 0 0 9 - the function should deliver a C matrix (same dimension too), where at each position C(i,j), compares A and B. if A(i,j)=0, than C(i,j)=0, if A(i,j)!=0, than C(i,j)=B(i,j) 6 0 0 5 C= 0 0 3 5 0 0 0 0 Although not an expert I could build a function with 2 cycles (reading columns and rows) which is not quick. Maybe you can help me in this challenge. Much thanks in advance, Diogo André Alagador Biodiversity Global Change Lab, Museo Nacional de Ciencias Naturales, CSIC, Madrid, España Forest Research Centre, Instituto Superior de Agronomia, Universidade Técnica de Lisboa, Lisboa, Portugal [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [OT] normal (as in Guassian)
Hi Folks, Apologies to anyone who'd prefer not to see this query on this list; but I'm asking because it is probably the forum where I'm most likely to get a good answer! I'm interested in the provenance of the name normal distribution (for what I'd really prefer to call the Gaussian distribution). According to Wikipedia, The name normal distribution was coined independently by Charles S. Peirce, Francis Galton and Wilhelm Lexis around 1875. So be it, if that was the case -- but I would like to know why they chose the name normal: what did they intend to convey? As background: I'm reflecting a bit on the usage in statistics of everyday language as techincal terms, as in significantly different. This, for instance, is likely to be misunderstood by the general publidc when they encounter statements in the media. Likewise, normally distributed would probably be interpreted as distributed in the way one would normally expect or, perhaps, there was nothing unusual about the distribution. Comments welcome! With thanks, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 02-Mar-08 Time: 13:04:17 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie: Incorrect number of dimensions
This is exactly what i am looking for. many thanks Chris On 3/2/08, Henrique Dallazuanna [EMAIL PROTECTED] wrote: You have a 'matrix' not a 'list', so I think you want something like this: all_differ[all_differ[, 'BY'] = 0.01, 'BY'], this will return the values where column is 'BY' in all rows which is less or equal to 0.01. On 02/03/2008, Keizer_71 [EMAIL PROTECTED] wrote: Here is the structure str(all_differ) num [1:6280, 1:8] 2.22e-16 2.22e-16 2.22e-16 2.22e-16 2.22e-16 ... - attr(*, dimnames)=List of 2 ..$ : NULL ..$ : chr [1:8] rawp Bonferroni Holm Hochberg ... I tried, but it still giving me the same error message Incorrect Dimension probe.names-all_differ[[1]][all_differ[[6280]][,BY]=0.01] Am i using the wrong dimension? thanks, Keizer jholtman wrote: It would be helpful if you provided commented, minimal, self-contained, reproducible code. What does str(all_differ) say? That will tell you the structure of the object that you are trying to work with. On Sat, Mar 1, 2008 at 3:35 AM, Keizer_71 [EMAIL PROTECTED] wrote: dim(data.sub) [1] 1 140 #extracting all differentially express genes## library(multtest) two_side- (1-pt(abs(data.sub),50))*2 diff- mt.rawp2adjp(two_side) all_differ-diff[[1]][37211:1,] all_differ #list of differentially expressed genes## probe.names- + all_differ[[2]][all_differ[[1]][,BY]=0.01] Error in all_differ[[1]][, BY] : incorrect number of dimensions Hi, I am pretty new with R. What i am trying to do is to find all differentially express genes and list of differentially expressed genes. Am i doing something wrong? I keep getting incorrect number of dimensions. How do i find out the correct dimensions? thanks, Keizer -- View this message in context: http://www.nabble.com/Newbie%3A-Incorrect-number-of-dimensions-tp15773090p15773090.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Newbie%3A-Incorrect-number-of-dimensions-tp15773090p15785977.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- Christophe Lo (078) 8275 7029 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question on lag.zoo
Hi Guys, I'm using zoo package now. I found lag is not doing what I assumed. x - zoo(11:21) z - zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 lag(x) 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 z 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1961 Q3 1 2 3 4 5 6 7 8 9 10 lag(z) 1959 Q1 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1 2 3 4 5 6 7 8 9 10 Why z and lag (z) are of same length while lag(x) is shorter by one than x? I assume lag(z) would give me like this: 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1961 Q3 2 3 4 5 6 7 8 9 10 ie preserve the relationship between timestamp and the value. Same things applies to lag(x) but I guess both make sense: 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 or 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Any insight? Cheers, Bo _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] difference between lrm's Model L.R. and anova's Chi-Square
[EMAIL PROTECTED] wrote: Quoting Frank E Harrell Jr [EMAIL PROTECTED]: anova (anova.Design) computes Wald statistics. When the log-likelihood is very quadratic, these statistics will be very close to log-likelihood ratio chi-square statistics. In general LR chi-square tests are better; we use Wald tests for speed. It's best to take the time and do lrtest(fit1,fit2) in Design, where one of the two fits is a subset of the other. Frank Harrell Thanks, this is great, but in my case, there's just one factor, fit1 - lrm(outcome~factor,data) and I'm having trouble constructing the subset 'null model', as e.g. fit2 - lrm(outcome~1,data) returns an error message. How do I construct a null model with lrm() so that I can use lrtest() to test a model with only one predictor? The overall LR chi-square test statistic is in the standard output of lrm (which uses print.lrm). I apologize for asking what must be a very simple question but I have been unable to find the answer by searching R-help. Thanks, Dan P.S. Second point: I have another case where I use lmer(), and there the null model includes a random effect so I don't get the problem above. It looks like with lmer objects anova() uses LLR, not Wald. Is that right? Please check the lmer documentation. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question on lag.zoo
On Sun, Mar 2, 2008 at 9:28 AM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Its a FAQ. x is of class zoo but z is of class zooreg so one is using That should be lag.zooreg zoo.lag and the is using zooreg.lag. See question #6 in the zoo FAQ: vignette(zoo-faq) and also ?lag.zoo On Sun, Mar 2, 2008 at 9:12 AM, Bo Zhou [EMAIL PROTECTED] wrote: Hi Guys, I'm using zoo package now. I found lag is not doing what I assumed. x - zoo(11:21) z - zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 lag(x) 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 z 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1961 Q3 1 2 3 4 5 6 7 8 9 10 lag(z) 1959 Q1 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1 2 3 4 5 6 7 8 9 10 Why z and lag (z) are of same length while lag(x) is shorter by one than x? I assume lag(z) would give me like this: 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4 1961 Q1 1961 Q2 1961 Q3 2 3 4 5 6 7 8 9 10 ie preserve the relationship between timestamp and the value. Same things applies to lag(x) but I guess both make sense: 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 or 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Any insight? Cheers, Bo _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Idioms for a timeseries operation - moving window
See ?embed and from zoo see: ?rollapply ?rollmean ?rollmax There is a function coded in C in the caTools package for speed. On Sun, Mar 2, 2008 at 9:24 AM, Bo Zhou [EMAIL PROTECTED] wrote: Hi Guys, Need your wisdom on this. Say I have a time series (in zoo format) like this x - zoo(11:21) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 I want to do a moving window sampling of it. The result can either be a matrix or a dataframe like this my.super.moving.window(x, length=3, by=1) 11 12 13 12 13 14 13 14 15 14 15 16 18 19 20 19 20 21 This shouldn't be new. Many many people must have done this before. Any idea what's the best(efficient and elegant) way to do this? Cheers, Bo _ Climb to the top of the charts! Play the word scramble challenge with star power. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Idioms for a timeseries operation - moving window
Hi Guys, Need your wisdom on this. Say I have a time series (in zoo format) like this x - zoo(11:21) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 I want to do a moving window sampling of it. The result can either be a matrix or a dataframe like this my.super.moving.window(x, length=3, by=1) 11 12 13 12 13 14 13 14 15 14 15 16 18 19 20 19 20 21 This shouldn't be new. Many many people must have done this before. Any idea what's the best(efficient and elegant) way to do this? Cheers, Bo _ Climb to the top of the charts! Play the word scramble challenge with star power. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Idioms for a timeseries operation - moving window
Time to upgrade to the latest version of zoo. You can also read it online here: http://cran.r-project.org/web/packages/zoo On Sun, Mar 2, 2008 at 9:47 AM, Bo Zhou [EMAIL PROTECTED] wrote: Ah! Gabor is the man. BTW I don't have zoo-faq. (I actually googled it out) This is what I have: Vignettes in package 'zoo': zoo-quickrefzoo Quick Reference (source, pdf) zoo zoo: An S3 Class and Methods for Indexed Totally Ordered Observations (source, pdf) It's a typical windows xp install of R 2.6.1. Any idea why? Cheers, Bo Date: Sun, 2 Mar 2008 09:40:09 -0500 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Idioms for a timeseries operation - moving window CC: r-help@r-project.org See ?embed and from zoo see: ?rollapply ?rollmean ?rollmax There is a function coded in C in the caTools package for speed. On Sun, Mar 2, 2008 at 9:24 AM, Bo Zhou [EMAIL PROTECTED] wrote: Hi Guys, Need your wisdom on this. Say I have a time series (in zoo format) like this x - zoo(11:21) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 I want to do a moving window sampling of it. The result can either be a matrix or a dataframe like this my.super.moving.window(x, length=3, by=1) 11 12 13 12 13 14 13 14 15 14 15 16 18 19 20 19 20 21 This shouldn't be new. Many many people must have done this before. Any idea what's the best(efficient and elegant) way to do this? Cheers, Bo _ Climb to the top of the charts! Play the word scramble challenge with star power. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Connect and share in new ways with Windows Live. Get it now! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Newbie:Export Data into Excel from R
Hi, All i want is to export my list into c: drive and save it as csv file and manually import into Excel. I have the read the article but i am having issues http://pbil.univ-lyon1.fr/library/base/html/write.table.html excel-write.table(probe_gene, file = c:\foo.csv, sep = ,, col.names = NA) Error in file(file, ifelse(append, a, w)) : unable to open connection In addition: Warning message: cannot open file 'c:\foo.csv', reason 'Invalid argument' any suggestions? thanks, chris -- View this message in context: http://www.nabble.com/Newbie%3AExport-Data-into-Excel-from-R-tp15788950p15788950.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Idioms for a timeseries operation - moving window
Thanks. Problem solved. rollapply(x,3,force, align=left) 1 1 2 3 2 2 3 4 3 3 4 5 4 4 5 6 5 5 6 7 6 6 7 8 7 7 8 9 8 8 9 10 x-1:10 embed(x, 3)[, 3:1] [,1] [,2] [,3] [1,]123 [2,]234 [3,]345 [4,]456 [5,]567 [6,]678 [7,]789 [8,]89 10 Date: Sun, 2 Mar 2008 10:04:38 -0500 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Idioms for a timeseries operation - moving window CC: r-help@r-project.org Try: rollapply(x, 3, force) or embed(x, 3)[, 3:1] On Sun, Mar 2, 2008 at 9:53 AM, Bo Zhou [EMAIL PROTECTED] wrote: I needed that matrix for following calculations. So I don't think roll* would work for me. Re: embed x - 1:10 embed (x, 3) [,1] [,2] [,3] [1,]321 [2,]432 [3,]543 [4,]654 [5,]765 [6,]876 [7,]987 [8,] 1098 x [1] 1 2 3 4 5 6 7 8 9 10 The matrix is in revered order. Anyway to put the matrix in the original order? Ie 1 2 3 2 3 4 ... Cheers, Bo Date: Sun, 2 Mar 2008 09:40:09 -0500 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Idioms for a timeseries operation - moving window CC: r-help@r-project.org See ?embed and from zoo see: ?rollapply ?rollmean ?rollmax There is a function coded in C in the caTools package for speed. On Sun, Mar 2, 2008 at 9:24 AM, Bo Zhou [EMAIL PROTECTED] wrote: Hi Guys, Need your wisdom on this. Say I have a time series (in zoo format) like this x - zoo(11:21) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 I want to do a moving window sampling of it. The result can either be a matrix or a dataframe like this my.super.moving.window(x, length=3, by=1) 11 12 13 12 13 14 13 14 15 14 15 16 18 19 20 19 20 21 This shouldn't be new. Many many people must have done this before. Any idea what's the best(efficient and elegant) way to do this? Cheers, Bo _ Climb to the top of the charts! Play the word scramble challenge with star power. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Shed those extra pounds with MSN and The Biggest Loser! Learn more. _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Idioms for a timeseries operation - moving window
Ah! Gabor is the man. BTW I don't have zoo-faq. (I actually googled it out) This is what I have: Vignettes in package zoo: zoo-quickrefzoo Quick Reference (source, pdf) zoo zoo: An S3 Class and Methods for Indexed Totally Ordered Observations (source, pdf) It's a typical windows xp install of R 2.6.1. Any idea why? Cheers, Bo Date: Sun, 2 Mar 2008 09:40:09 -0500 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Idioms for a timeseries operation - moving window CC: r-help@r-project.org See ?embed and from zoo see: ?rollapply ?rollmean ?rollmax There is a function coded in C in the caTools package for speed. On Sun, Mar 2, 2008 at 9:24 AM, Bo Zhou [EMAIL PROTECTED] wrote: Hi Guys, Need your wisdom on this. Say I have a time series (in zoo format) like this x - zoo(11:21) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 I want to do a moving window sampling of it. The result can either be a matrix or a dataframe like this my.super.moving.window(x, length=3, by=1) 11 12 13 12 13 14 13 14 15 14 15 16 18 19 20 19 20 21 This shouldn't be new. Many many people must have done this before. Any idea what's the best(efficient and elegant) way to do this? Cheers, Bo _ Climb to the top of the charts! Play the word scramble challenge with star power. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Idioms for a timeseries operation - moving window
I needed that matrix for following calculations. So I don't think roll* would work for me. Re: embed x - 1:10 embed (x, 3) [,1] [,2] [,3] [1,]321 [2,]432 [3,]543 [4,]654 [5,]765 [6,]876 [7,]987 [8,] 1098 x [1] 1 2 3 4 5 6 7 8 9 10 The matrix is in revered order. Anyway to put the matrix in the original order? Ie 1 2 3 2 3 4 ... Cheers, Bo Date: Sun, 2 Mar 2008 09:40:09 -0500 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Idioms for a timeseries operation - moving window CC: r-help@r-project.org See ?embed and from zoo see: ?rollapply ?rollmean ?rollmax There is a function coded in C in the caTools package for speed. On Sun, Mar 2, 2008 at 9:24 AM, Bo Zhou [EMAIL PROTECTED] wrote: Hi Guys, Need your wisdom on this. Say I have a time series (in zoo format) like this x - zoo(11:21) x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 I want to do a moving window sampling of it. The result can either be a matrix or a dataframe like this my.super.moving.window(x, length=3, by=1) 11 12 13 12 13 14 13 14 15 14 15 16 18 19 20 19 20 21 This shouldn't be new. Many many people must have done this before. Any idea what's the best(efficient and elegant) way to do this? Cheers, Bo _ Climb to the top of the charts! Play the word scramble challenge with star power. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Make plots with GNUplot. Have anyone tried that?
There is a very basic interface between R and gnuplot in the TeachingDemos
package. Look at the help for gp.plot.
[snip]
This looks mighty interesting =)
Is it possible to plot with lines (gnuplot syntex) so all the data
points are connected?
Also is it possible to make it write the output to a pdf file? (I
using the gnuplot cvs which have pdfcairo support)
When I try this, it doesn't write the pdf file. it works with gnuplot cvs.
library(TeachingDemos)
fuelData-read.table('fuel.csv',header=TRUE, sep=',')
attach(fuelData)
gp.open(where='/usr/local/bin/gnuplot')
gp.send('set terminal pdfcairo font 'cmr10' size 8cm,4.6cm')
gp.send('set output '../figures/q1-raw-data-gp.pdf'')
gp.send('unset key')
gp.send('set xlabel 'rtime'')
gp.send('set ylabel 'FPI'')
gp.send('set xrange [1979:2005]')
gp.plot(rtime,fpi)
gp.send('unset output')
gp.close()
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie:Export Data into Excel from R
R uses unix type slashes that is / rather than \ even if you're using Windows You probably need either to write file =c:/foo.csv or escape the \ thusly file = c:\\foo.csv Either way should work. Personally find it easier to use '/. --- Keizer_71 [EMAIL PROTECTED] wrote: Hi, All i want is to export my list into c: drive and save it as csv file and manually import into Excel. I have the read the article but i am having issues http://pbil.univ-lyon1.fr/library/base/html/write.table.html excel-write.table(probe_gene, file = c:\foo.csv, sep = ,, col.names = NA) Error in file(file, ifelse(append, a, w)) : unable to open connection In addition: Warning message: cannot open file 'c:\foo.csv', reason 'Invalid argument' any suggestions? thanks, chris -- View this message in context: http://www.nabble.com/Newbie%3AExport-Data-into-Excel-from-R-tp15788950p15788950.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie:Export Data into Excel from R
You will likely need to escape that backslash, i.e. c:\\foo.csv. Haris Skiadas Department of Mathematics and Computer Science Hanover College On Mar 2, 2008, at 10:12 AM, Keizer_71 wrote: Hi, All i want is to export my list into c: drive and save it as csv file and manually import into Excel. I have the read the article but i am having issues http://pbil.univ-lyon1.fr/library/base/html/write.table.html excel-write.table(probe_gene, file = c:\foo.csv, sep = ,, col.names = NA) Error in file(file, ifelse(append, a, w)) : unable to open connection In addition: Warning message: cannot open file 'c:\foo.csv', reason 'Invalid argument' any suggestions? thanks, chris -- View this message in context: http://www.nabble.com/Newbie%3AExport- Data-into-Excel-from-R-tp15788950p15788950.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [OT] normal (as in Guassian)
I'm not a statistician, but do i remember well that among all distributions with a given mean and variance, the normal distribution has the highest entropy? This is good enough for me to call it normal Gabor On Sun, Mar 02, 2008 at 10:10:21AM -0600, roger koenker wrote: A nice survey of this territory is: http://books.google.com/books?id=TN3_d7ibo30Cpg=PA85lpg=PA85dq=stigler+normal+oxymoronsource=webots=OwGhmnDk3Osig=J7ou_L8-_Mu4L14c3KJAhefrD4Ihl=en I particularly like the phrase: [normal] is in this respect a rare one-word oxymoron. url:www.econ.uiuc.edu/~rogerRoger Koenker email [EMAIL PROTECTED] Department of Economics vox:217-333-4558University of Illinois fax:217-244-6678Champaign, IL 61820 On Mar 2, 2008, at 7:33 AM, (Ted Harding) wrote: Hi Folks, Apologies to anyone who'd prefer not to see this query on this list; but I'm asking because it is probably the forum where I'm most likely to get a good answer! I'm interested in the provenance of the name normal distribution (for what I'd really prefer to call the Gaussian distribution). According to Wikipedia, The name normal distribution was coined independently by Charles S. Peirce, Francis Galton and Wilhelm Lexis around 1875. So be it, if that was the case -- but I would like to know why they chose the name normal: what did they intend to convey? As background: I'm reflecting a bit on the usage in statistics of everyday language as techincal terms, as in significantly different. This, for instance, is likely to be misunderstood by the general publidc when they encounter statements in the media. Likewise, normally distributed would probably be interpreted as distributed in the way one would normally expect or, perhaps, there was nothing unusual about the distribution. Comments welcome! With thanks, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 02-Mar-08 Time: 13:04:17 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]UNIL DGM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [OT] normal (as in Guassian)
A nice survey of this territory is: http://books.google.com/books?id=TN3_d7ibo30Cpg=PA85lpg=PA85dq=stigler+normal+oxymoronsource=webots=OwGhmnDk3Osig=J7ou_L8-_Mu4L14c3KJAhefrD4Ihl=en I particularly like the phrase: [normal] is in this respect a rare one-word oxymoron. url:www.econ.uiuc.edu/~rogerRoger Koenker email [EMAIL PROTECTED] Department of Economics vox:217-333-4558University of Illinois fax:217-244-6678Champaign, IL 61820 On Mar 2, 2008, at 7:33 AM, (Ted Harding) wrote: Hi Folks, Apologies to anyone who'd prefer not to see this query on this list; but I'm asking because it is probably the forum where I'm most likely to get a good answer! I'm interested in the provenance of the name normal distribution (for what I'd really prefer to call the Gaussian distribution). According to Wikipedia, The name normal distribution was coined independently by Charles S. Peirce, Francis Galton and Wilhelm Lexis around 1875. So be it, if that was the case -- but I would like to know why they chose the name normal: what did they intend to convey? As background: I'm reflecting a bit on the usage in statistics of everyday language as techincal terms, as in significantly different. This, for instance, is likely to be misunderstood by the general publidc when they encounter statements in the media. Likewise, normally distributed would probably be interpreted as distributed in the way one would normally expect or, perhaps, there was nothing unusual about the distribution. Comments welcome! With thanks, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 02-Mar-08 Time: 13:04:17 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie:Export Data into Excel from R
Keizer_71 [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: All i want is to export my list into c: drive and save it as csv file and manually import into Excel. I have the read the article but i am having issues http://pbil.univ-lyon1.fr/library/base/html/write.table.html excel-write.table(probe_gene, file = c:\foo.csv, sep = ,, col.names = NA) Error in file(file, ifelse(append, a, w)) : unable to open connection In addition: Warning message: cannot open file 'c:\foo.csv', reason 'Invalid argument' any suggestions? One can use the file.choose() function in place of a quoted file name. I do so because it gets around the \, /, \\ confusion in my brain, ... and it's faster. R does not use solitary \'s in its filenames for directory separation. It's the escape character for regular expressions, and R's development in the *nix world trumped any need to be consistent with the dark side's use of \ as a directory separator. You can use single / as well as \\ -- David Winsemius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] COMPAR.GEE Output
Charles Willis willis.charlie at gmail.com writes: Hello, I am running the program COMPAR.GEE within the package APE. My dependent variable is binomial, while my independent variable is a multi-state categorical variable. The output reports an estimate for each state of the independent variable except the first one. For example, for the variable X with 3 states, the output is: intercept (estimate) X2 (estimate) X3 (estimate) I have two questions: 1) Why does it not give me an intercept for the first variable and how do i get it; 2) can a get a general estimate of correlation, like a wald's statistic for the variable? 1. It's useful to provide a small self-contained example, as recommended by the posting guide (referred to after every r-help message). 2. I would suggest asking this kind of question on the r-sig-phylo mailing list, which is devoted to phylogenetic and comparative analyses in R. 3. In order to understand the output, you have to understand the way in which R parameterizes statistical models. I'm guessing that you specified family=binomial in your compar.gee call, which would mean by default assuming binomial error structure and a logit link (if you don't know what that means, you should probably read up on generalized linear models a bit ...). X1 is then the estimated logit-probability for state 1, X2 is the estimated *difference* in logit-prob between states 1 and 2, similarly for X3. If you just want the estimated probabilities for the three groups you can fit a model without an intercept using something like response~predictor-1 as your formula. 4. When you print the model object (e.g. c1=compar.gee(...); c1) it gives you standard errors and t statistics for each parameter which are (I believe) essentially Wald statistics, although you should certainly check the reference given in ?compar.gee ... good luck, Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] summarizing matrix data
Hi everyone, I'm sure this is simple, but I can't seem to figure this out. Situation. 3 different groups of subjects each submit n X n matrices of scores. What I want to do is aggregate each group of scores into a summary n X n matrix. I need the result to be a matrix so that I can calculate a dissimilarity structure on it. So I thought I would create a multi-dimensional array and store the data that way. The first two dimensions represent the data, the third the group membership (say 3 groups), and forth represents subjects(30 subjects per group)... subject.data - array(NA, dim=c(3,3,3,10)) data.1 - matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3) data.2 - matrix(seq(from=11,to=19,by=1),nrow=3,ncol=3) data.3 - matrix(seq(from=21,to=29,by=1),nrow=3,ncol=3) data.4 - matrix(seq(from=31,to=39,by=1),nrow=3,ncol=3) data.5 - matrix(seq(from=41,to=49,by=1),nrow=3,ncol=3) data.6 - matrix(seq(from=51,to=59,by=1),nrow=3,ncol=3) subject.data[,,1,1] - data.1 # subject 1 group 1 subject.data[,,2,1] - data.2 # subject 1 group 2 subject.data[,,3,1] - data.3 # subject 1 group 3 subject.data[,,3,2] - data.4 # subject 2 group 3 subject.data[,,2,2] - data.5 # subject 2 group 2 subject.data[,,1,2] - data.6 # subject 2 group 1 what I want is conceptually: array.group1 - apply(subject.data[,,1,],mean) array.group2 - apply(subject.data[,,2,],mean) array.group3 - apply(subject.data[,,3,],mean) each of the summary matrices then contain the mean for each cell across all subjects within a group so that it has the same size as the subject data. Also, after I've populated subject.data, there will be segments that will only contain NAs, how do you grab just the portions of the array that contain data? I know the function is.na() will get me an index I just don't know how to write it correctly. something like only.data - subject.data[!is.na(subject.data)] Does that make? I would appreciate any help regarding a better way to store the data and obviously how to get back these summary matrices... Thanks very much for your help. emilio [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] elegant way to minus on each row of a matrix
How to do this in an elegant way formatrix/data frame/zoo? mat= 1 2 3 4 5 6 7 8 9 vector= 1 2 3 result= 0 1 2 2 3 4 4 5 6 ie 1-1 2-1 3-1 4-2 5-2 6-2 7-3 8-3 9-3 Thanks in advance. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variance Calculation in R
sorry...in step 4-i need the R code to output in this format when i export to excel. ProbeID Variance 1 224588_at 21.58257457 thanks Keizer_71 wrote: Hello, Thanks everyone for helping me with the previous queries. step 1: Here is the orginal data: short sample ProbeID Sample_1_D Sample_1_CSample_2_D Sample_2_C 1 224588_at 2.425509867 11.34031409 11.46868531 11.75741478 step 2: i calculate the variance of the sample using this R code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) step 3: however, when i type in variableprobe, it gives me this. 1 21.58257457 step 4: I need the code to output this: ProbeID Variance 1 224588_at 21.58257457 What do i need to do to modify the code to give me better description like the one above? thank. Keizer -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15790768.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] regression output to latex
hello everybody i was seeking a converter beetween R regression output (eg with summary) and the conventional way to present regression output in paper: every model as a vertical vector with \beta, t beetween parenthesis below the first, and other statistics (R^2 etc) . I've seen hmisc and xtable, and if I didn't miss something, they don't include something like that. Thank you Luca -- Luca Braglia, aka Bragliozzo http://bragliozzo.altervista.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Variance Calculation in R
Hello, Thanks everyone for helping me with the previous queries. step 1: Here is the orginal data: short sample ProbeID Sample_1_D Sample_1_CSample_2_D Sample_2_C 1 224588_at 2.425509867 11.34031409 11.46868531 11.75741478 step 2: i calculate the variance of the sample using this R code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) step 3: however, when i type in variableprobe, it gives me this. 1 21.58257457 step 4: I need the code to output this: ProbeID Variance 1 224588_at 21.58257457 What do i need to do to modify the code to give me better description like the one above? thank. Keizer -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15790621.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Imputation Packages
Hi everyone, I am looking for a package in R which can help me in using the imputation technique to find the missing values for my regression analysis. Any help would be appreciated. Cheers Arun -- View this message in context: http://www.nabble.com/Imputation-Packages-tp15790749p15790749.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] elegant way to minus on each row of a matrix
Brilliant! t1=matrix(1:15,5,3) t1 [,1] [,2] [,3] [1,]16 11 [2,]27 12 [3,]38 13 [4,]49 14 [5,]5 10 15 t2=1:5 t2 [1] 1 2 3 4 5 sweep(t1, 1, t2) [,1] [,2] [,3] [1,]05 10 [2,]05 10 [3,]05 10 [4,]05 10 [5,]05 10 Date: Sun, 2 Mar 2008 14:56:22 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] elegant way to minus on each row of a matrix CC: r-help@r-project.org Try this: sweep(mat, 1, vec) On 02/03/2008, Bo Zhou [EMAIL PROTECTED] wrote: How to do this in an elegant way formatrix/data frame/zoo? mat= 1 2 3 4 5 6 7 8 9 vector= 1 2 3 result= 0 1 2 2 3 4 4 5 6 ie 1-1 2-1 3-1 4-2 5-2 6-2 7-3 8-3 9-3 Thanks in advance. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O _ Need to know the score, the latest news, or you need your Hotmail®-get your fix. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variance Calculation in R
Try this: write.table(cbind(data.matrix[1], Variance = apply(data.matrix[,-1], 1, var)),file='file.xls') On 02/03/2008, Keizer_71 [EMAIL PROTECTED] wrote: sorry...in step 4-i need the R code to output in this format when i export to excel. ProbeID Variance 1 224588_at 21.58257457 thanks Keizer_71 wrote: Hello, Thanks everyone for helping me with the previous queries. step 1: Here is the orginal data: short sample ProbeID Sample_1_D Sample_1_CSample_2_D Sample_2_C 1 224588_at 2.425509867 11.34031409 11.46868531 11.75741478 step 2: i calculate the variance of the sample using this R code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) step 3: however, when i type in variableprobe, it gives me this. 1 21.58257457 step 4: I need the code to output this: ProbeID Variance 1 224588_at 21.58257457 What do i need to do to modify the code to give me better description like the one above? thank. Keizer -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15790768.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] elegant way to minus on each row of a matrix
Hi Dimitris and everyone I tried this but now I know why it didn't work out for me initially. t1=matrix(1:10,5,2) t2=matrix(1,5,1) t2 [,1] [1,]1 [2,]1 [3,]1 [4,]1 [5,]1 t1 [,1] [,2] [1,]16 [2,]27 [3,]38 [4,]49 [5,]5 10 t1-as.vector(t2) [,1] [,2] [1,]05 [2,]16 [3,]27 [4,]38 [5,]49 t1-t2 Error in t1 - t2 : non-conformable arrays I'm too used to Matlab I guess. Cheers, Bo Date: Sun, 2 Mar 2008 18:56:15 +0100 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] CC: r-help@r-project.org Subject: Re: [R] elegant way to minus on each row of a matrix try this: mat - matrix(1:9, 3, 3, TRUE) dat - as.data.frame(mat) vec - 1:3 result.mat - mat - vec result.dat - dat - vec result.mat result.dat I hope it helps. Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Quoting Bo Zhou [EMAIL PROTECTED]: How to do this in an elegant way formatrix/data frame/zoo? mat= 1 2 3 4 5 6 7 8 9 vector= 1 2 3 result= 0 1 2 2 3 4 4 5 6 ie 1-1 2-1 3-1 4-2 5-2 6-2 7-3 8-3 9-3 Thanks in advance. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] elegant way to minus on each row of a matrix
Try this: sweep(mat, 1, vec) On 02/03/2008, Bo Zhou [EMAIL PROTECTED] wrote: How to do this in an elegant way formatrix/data frame/zoo? mat= 1 2 3 4 5 6 7 8 9 vector= 1 2 3 result= 0 1 2 2 3 4 4 5 6 ie 1-1 2-1 3-1 4-2 5-2 6-2 7-3 8-3 9-3 Thanks in advance. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] summarizing matrix data
One options is: data.1 - matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3) data.2 - matrix(seq(from=11,to=19,by=1),nrow=3,ncol=3) data.3 - matrix(seq(from=21,to=29,by=1),nrow=3,ncol=3) data.4 - matrix(seq(from=31,to=39,by=1),nrow=3,ncol=3) data.5 - matrix(seq(from=41,to=49,by=1),nrow=3,ncol=3) data.6 - matrix(seq(from=51,to=59,by=1),nrow=3,ncol=3) patt - ls(patt='^data\\.[0-9]') nsubs - 2 ngroups - 3 apply(array(as.vector(sapply(patt, get)), dim = c(dim(get(patt[1])), ngroups, nsubs)), 3, mean) On 02/03/2008, Emilio Gagliardi [EMAIL PROTECTED] wrote: Hi everyone, I'm sure this is simple, but I can't seem to figure this out. Situation. 3 different groups of subjects each submit n X n matrices of scores. What I want to do is aggregate each group of scores into a summary n X n matrix. I need the result to be a matrix so that I can calculate a dissimilarity structure on it. So I thought I would create a multi-dimensional array and store the data that way. The first two dimensions represent the data, the third the group membership (say 3 groups), and forth represents subjects(30 subjects per group)... subject.data - array(NA, dim=c(3,3,3,10)) data.1 - matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3) data.2 - matrix(seq(from=11,to=19,by=1),nrow=3,ncol=3) data.3 - matrix(seq(from=21,to=29,by=1),nrow=3,ncol=3) data.4 - matrix(seq(from=31,to=39,by=1),nrow=3,ncol=3) data.5 - matrix(seq(from=41,to=49,by=1),nrow=3,ncol=3) data.6 - matrix(seq(from=51,to=59,by=1),nrow=3,ncol=3) subject.data[,,1,1] - data.1 # subject 1 group 1 subject.data[,,2,1] - data.2 # subject 1 group 2 subject.data[,,3,1] - data.3 # subject 1 group 3 subject.data[,,3,2] - data.4 # subject 2 group 3 subject.data[,,2,2] - data.5 # subject 2 group 2 subject.data[,,1,2] - data.6 # subject 2 group 1 what I want is conceptually: array.group1 - apply(subject.data[,,1,],mean) array.group2 - apply(subject.data[,,2,],mean) array.group3 - apply(subject.data[,,3,],mean) each of the summary matrices then contain the mean for each cell across all subjects within a group so that it has the same size as the subject data. Also, after I've populated subject.data, there will be segments that will only contain NAs, how do you grab just the portions of the array that contain data? I know the function is.na() will get me an index I just don't know how to write it correctly. something like only.data - subject.data[!is.na(subject.data)] Does that make? I would appreciate any help regarding a better way to store the data and obviously how to get back these summary matrices... Thanks very much for your help. emilio [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] elegant way to minus on each row of a matrix
Hi Mark, (CC'ing r-help) Only need to change the parameter from 1 to 2 after that it worked great. Thanks t1=matrix(1:15,5,3) t1 [,1] [,2] [,3] [1,]16 11 [2,]27 12 [3,]38 13 [4,]49 14 [5,]5 10 15 t2=1:5 t2 [1] 1 2 3 4 5 apply(t1,2, function(x) x - t2) [,1] [,2] [,3] [1,]05 10 [2,]05 10 [3,]05 10 [4,]05 10 [5,]05 10 Date: Sun, 2 Mar 2008 12:51:53 -0500 From: [EMAIL PROTECTED] Subject: RE: [R] elegant way to minus on each row of a matrix To: [EMAIL PROTECTED] try result-apply(mat,1, function(.row) .row - vector) but I don't have R here so make sure it works. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bo Zhou Sent: Sunday, March 02, 2008 12:43 PM To: r-help@r-project.org Subject: [R] elegant way to minus on each row of a matrix How to do this in an elegant way formatrix/data frame/zoo? mat= 1 2 3 4 5 6 7 8 9 vector= 1 2 3 result= 0 1 2 2 3 4 4 5 6 ie 1-1 2-1 3-1 4-2 5-2 6-2 7-3 8-3 9-3 Thanks in advance. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] elegant way to minus on each row of a matrix
try this: mat - matrix(1:9, 3, 3, TRUE) dat - as.data.frame(mat) vec - 1:3 result.mat - mat - vec result.dat - dat - vec result.mat result.dat I hope it helps. Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Quoting Bo Zhou [EMAIL PROTECTED]: How to do this in an elegant way formatrix/data frame/zoo? mat= 1 2 3 4 5 6 7 8 9 vector= 1 2 3 result= 0 1 2 2 3 4 4 5 6 ie 1-1 2-1 3-1 4-2 5-2 6-2 7-3 8-3 9-3 Thanks in advance. _ 08 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variance Calculation in R
Hi Henrique, It is definitely better, but it doesn't show me the ProbeID which identify the probes name Here was the result when i export to excel with your code. Variance 1 2.425509867 21.6216446425273 any suggestions? thanks, Kei Keizer_71 wrote: Hello, Thanks everyone for helping me with the previous queries. step 1: Here is the orginal data: short sample ProbeID Sample_1_D Sample_1_CSample_2_D Sample_2_C 1 224588_at 2.425509867 11.34031409 11.46868531 11.75741478 step 2: i calculate the variance of the sample using this R code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) step 3: however, when i type in variableprobe, it gives me this. 1 21.58257457 step 4: I need the code to output this: ProbeID Variance 1 224588_at 21.58257457 What do i need to do to modify the code to give me better description like the one above? thank. Keizer -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15791115.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Make plots with GNUplot. Have anyone tried that?
Hi Louise Hoffman wrote: [snip] Seriously. Be specific if you have a problem. (read the posting guide). R can also plot. If you don't like R's plots (which I could not understand) you can export data and import them to gnuplot. So what? Okay, my post was not very good. The reason (I think) I need GNUplot, is that I would like to include the plots from R in a Latex report, where I would like to have all the text and equations in the plots with the same font as used in Latex. Take a look at Non-standard fonts in PostScript and PDF graphics in http://cran.r-project.org/doc/Rnews/Rnews_2006-2.pdf, plus http://www.stat.auckland.ac.nz/~paul/R/CM/CMR.html These describe one way to use LaTeX fonts in R plots. Paul So when I read about opening and closing dev for making a pdf I figured that the plots that R produces are like the once Matlab makes; shows what they ought to, nothing more, nothing less. So I was wondering if anyone know of an GNUplot friendly format and the code that would produce that text file. I am new to both R and GNUplot, so I am pure ears if someone knows how to make such plots in R. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] summarizing matrix data
Then you can try this:
apply(array(as.vector(sapply(patt, get)), dim = c(dim(get(patt[1])),
ngroups, nsubs)), 1:3, mean)
For create the matrices in a loop one option is:
lapply(seq(1, 51, by=10), function(x)matrix(seq(x, x+8), nrow=3, ncol=3))
On 02/03/2008, Emilio Gagliardi [EMAIL PROTECTED] wrote:
Hi Henrique,
Thank you for your help, but it doesn't produce what I'm looking for...
On Sun, Mar 2, 2008 at 10:50 AM, Henrique Dallazuanna [EMAIL PROTECTED]
wrote:
One options is:
data.1 - matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3)
data.2 - matrix(seq(from=11,to=19,by=1),nrow=3,ncol=3)
data.3 - matrix(seq(from=21,to=29,by=1),nrow=3,ncol=3)
data.4 - matrix(seq(from=31,to=39,by=1),nrow=3,ncol=3)
data.5 - matrix(seq(from=41,to=49,by=1),nrow=3,ncol=3)
data.6 - matrix(seq(from=51,to=59,by=1),nrow=3,ncol=3)
patt - ls(patt='^data\\.[0-9]')
nsubs - 2
ngroups - 3
apply(array(as.vector(sapply(patt, get)), dim = c(dim(get(patt[1])),
ngroups, nsubs)), 3, mean)
generated [1] 20 30 40
but what I need is an average per cell...so in the above, there are 6 (3 x
3) matrices across 3 groups ( 2 subjects per group)..what I need is 1 matrix
per group, not one mean. Moreover, I'm storing the data in a single
multi-dimensional array, not as a collection of arrays since I'm creating
each data matrix within a loop (I read a file, create the data matrix, store
it in a larger structure) so that I can perform operations on all the data
outside of the loop...
for(i in 1:6)
{
str - paste(test,i,sep=.)
str - matrix(seq(from=1,to=9,by=1),nrow=3,ncol=3) # works, creates 6
matrices
}
pattern - ls(patt='^test\\.[0-9]') # doesn't work variables only exist in
the loop
which is why I had:
subject.data[,,1,1] - data.1 # subject 1 group 1
subject.data[,,2,1] - data.2 # subject 1 group 2
subject.data[,,3,1] - data.3 # subject 1 group 3
subject.data[,,3,2] - data.4 # subject 2 group 3
subject.data[,,2,2] - data.5 # subject 2 group 2
subject.data[,,1,2] - data.6 # subject 2 group 1
to demonstrate all the data was in a meta structure so that I could access
them later...am I making this more complicated than it needs to be? How can
I create the matrices inside a loop and have them available outside of it?
thanks again,
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Variance Calculation in R
Then you can try: rownames(data.matrix) - as.character(data.matrix$ProbeID) data.matrix - data.matrix[-1] as.matrix(apply(data.matrix1, 1, var)) or out - apply(data.matrix1, 1, var) data.frame(ProbeID = names(out), Variance = unname(out)) Works for me On 02/03/2008, Keizer_71 [EMAIL PROTECTED] wrote: Hi Henrique, It is definitely better, but it doesn't show me the ProbeID which identify the probes name Here was the result when i export to excel with your code. Variance 1 2.425509867 21.6216446425273 any suggestions? thanks, Kei Keizer_71 wrote: Hello, Thanks everyone for helping me with the previous queries. step 1: Here is the orginal data: short sample ProbeID Sample_1_D Sample_1_CSample_2_D Sample_2_C 1 224588_at 2.425509867 11.34031409 11.46868531 11.75741478 step 2: i calculate the variance of the sample using this R code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) step 3: however, when i type in variableprobe, it gives me this. 1 21.58257457 step 4: I need the code to output this: ProbeID Variance 1 224588_at 21.58257457 What do i need to do to modify the code to give me better description like the one above? thank. Keizer -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15791115.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Imputation Packages
ArunPrasad agurubar at uark.edu writes: I am looking for a package in R which can help me in using the imputation technique to find the missing values for my regression analysis. The recommended search for imputation in r-project gave me 400 citations. I assume you have tried this. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] new to latex to pdf
Dear All,
I'm trying to teach myself latex along with the latex function in Hmisc
and have hit a roadblock that I can't seem to get around. I'd greatly
appreciate any pointers.
I'm running R 2.6.0 on Windows XP and have Miktex 2.7 installed.
I've reproduced the code below, taken from Frank Harrell's latexsummary
introduction. My question relates to getting a pdf version of the table
from the following code. The pdfs of the graphics (f1a and f1b)
generated by setpdf are fine. However, after a number of attempts using
different methods, I don't seem to be able to get a pdf of the table
from the s1 object (I see the right table in my previewer just fine).
I've tried texi2dvi (texfilename, pdf=T) but get only a series of errors
and an unformatted table in the pdf.
I'm sure I'm missing some fundamental concept here, but I'm afraid I'm
not seeing it. I'd appreciate if anyone could point me in the right
direction. ( I have no trouble writing my own simple latex code and
converting it to pdf using pdftex in miktex).
Thanks,
Mike Babyak
Department of Psychiatry and Behavioral Science
Duke University Medical Center
**
Here's the code:
library(Hmisc)
library(survival)
getHdata(pbc)
pbc-upData(pbc, moveUnits=TRUE,
labels=c(stage='Histologic Stage\nLudwig Criteria'))
kmsurv - function(S, times) {
f - survfit.km(factor(rep(1,nrow(S))), S)
tt - c(0, f$time)
ss - c(1, f$surv) # add first point to survival curve
approx(tt, ss, xout=times, method='constant', f=0)$y
}
describe.survival - function(y) {
km - kmsurv(y, c(2,5))
c('2 Year'=km[1], '5 Year'=km[2], 'Mean, y'=sum(y[,1])/sum(y[,2]))
}
S - with(pbc, Surv(fu.days/365.25, status))
s1 - summary(S ~ age + albumin + ascites + bili + drug + edema + chol,
fun=describe.survival, data=pbc)
for(w in 1:2) {
if(w==1) setpdf(f1a,sublines=1,h=5.25) else
setpdf(f1b,sublines=1,h=5)
plot(s1, which=if(w==1)1:2 else 3,
cex.labels=.7, cex.group.labels=.7*1.15, subtitles=T, main='',
pch=if(w==2) 16 else c('2','5'), # 16=solid circle
xlab=if(w==2)'Survival Time' else 'Survival Probability')
dev.off()
}
w - latex(s1, cdec=c(2,2,1), ctable=TRUE, caption='Survival')
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[R] About R-Project
Hello, My name is Ion Andronache and I am a candidate for a doctor's degree on geomorphology in Braila, Romania. I would like to ask for some advice. I would like to make an analysis on the Danube's flow capacity and to determine the circles and periodicity using the Schuster system. I understood that your soft can do these operations but I don't know how. Would you be so kind to guide me? Best wishes, Ion Andronache. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [OT] normal (as in Guassian)
Am 02.03.2008 um 17:44 schrieb Gabor Csardi: I'm not a statistician, but do i remember well that among all distributions with a given mean and variance, the normal distribution has the highest entropy? This is good enough for me to call it normal There's more. Among all rotation-symmetric distributions, the standard bivariate normal is the only one where x and y are independent. Also, the formula for the standard normal distribution is the only one that is its own Fourier transform. So, if we assume the same distribution for a momentum and a location of a physical object, according to Heisenberg's Law it has to be the normal. Whereas we ought to be wary about assumption of normality for the distribution of phenomena in nature, the normal and its henchmen play a defendable role when describing summaries of phenomena, like arithmetic means. I'd even go as far as buy into Youden's hype described in that Kruskal and Stigler essay. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Need help to locate my mistake
Dear readers
I would like to make General Linear Model (GLM) for the following data set
http://louise.hoffman.googlepages.com/fuel.csv
The code I have written is
fuelData-read.table('fuel.csv',header=TRUE, sep=',')
n-dim(fuelData)[1]
xOnes- matrix(1,nrow=n,ncol=1)
x-cbind(xOnes,fuelData[,3])
y-fuelData[,4]
theta-((t(x)%*%x)^(-1))%*%t(x)%*%y
which gives
theta
[,1]
[1,] 215.8374077
[2,] 0.1083491
When I do it in Matlab I get theta to be
79.69
0.18
which is correct. ~79 is the crossing of the y-axis.
Have I perhaps written theta wrong? The formula for theta is
(alpha,beta)^T = (x^T * x)^(-1) * x^T * Y
where ^T means transposed.
Can someone see where the mistake is?
Hugs,
Louise
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Re: [R] summarizing matrix data
Henrique, Again, thank you very much for your code snippets, I am learning from them, but they are also creating as many questions as answers... On Sun, Mar 2, 2008 at 11:48 AM, Henrique Dallazuanna [EMAIL PROTECTED] wrote: Then you can try this: apply(array(as.vector(sapply(patt, get)), dim = c(dim(get(patt[1])), ngroups, nsubs)), 1:3, mean) For create the matrices in a loop one option is: lapply(seq(1, 51, by=10), function(x)matrix(seq(x, x+8), nrow=3, ncol=3)) if I use the above code, I get back a list of matrices, which is awesome! The problem is that your code to get the means doesn't work on it ( apply(array(as.vector(...))...additionally, I'm not creating the data, but reading it in, file by file and I don't know what group to assign a file until I open it (group2, group4, group4,group1,group3,group2,group1,etc) so I have 120 files to open and process and each file can be assigned to 1 of 4 groups(or lists as your code demonstrates)...but I don't know until I open the file which group hence my loop and meta structure approach. So I need to open the file, create a matrix from it (which I have working), store it along with other data from the same group, and then after I've read all the files and associated the data with a group, then I can generate the means per cell. So I can create 4 lists of matrices, which is fine by me but I don't see how I extend your code snippets to the case of reading the data from files...I currently have 1 master matrix which holds all the subject data as I mentioned in my previous post. data[3,3,1,nsubjects] represents all the subject data in group 1 data[3,3,2,nsubjects] represents all the subject data in group 2 data[3,3,3,nsubjects] represents all the subject data in group 3 how do i pullout just one group's worth of data and apply the mean function to each cell ? group1 - get(data[,,1,] ? group2 - get(data[,,2,] ? mean1 - apply(group1,1:3,mean) # mean per cell, so nine means (3 x 3) for each group... mean2 - apply(group2,1:3,mean) # mean per cell, so nine means (3 x 3) for each group... thanks so much for your patience! emilio [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variance Calculation in R
unfortunately, it is not showing probeID Henrique Dallazuanna wrote: Try this: write.table(cbind(data.matrix[1], Variance = apply(data.matrix[,-1], 1, var)),file='file.xls') On 02/03/2008, Keizer_71 [EMAIL PROTECTED] wrote: sorry...in step 4-i need the R code to output in this format when i export to excel. ProbeID Variance 1224588_at 21.58257457 thanks Keizer_71 wrote: Hello, Thanks everyone for helping me with the previous queries. step 1: Here is the orginal data: short sample ProbeID Sample_1_D Sample_1_CSample_2_D Sample_2_C 1 224588_at 2.425509867 11.34031409 11.46868531 11.75741478 step 2: i calculate the variance of the sample using this R code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) step 3: however, when i type in variableprobe, it gives me this. 1 21.58257457 step 4: I need the code to output this: ProbeID Variance 1 224588_at 21.58257457 What do i need to do to modify the code to give me better description like the one above? thank. Keizer -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15790768.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Variance-Calculation-in-R-tp15790621p15792364.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] discrete variable
Hello,
I am sorry for asking such a basic question. I could not find an answer to
it using google.
I have a discrete variable (a vector x) taking for example the following
values : 0, 3, 4, 3, 15, 5, 6, 5
Is it possible to know how many different values (modalities) it takes ?
Here it takes 6 different values but the length of the vector is 8.
I would like to know if there is a way to get the set of the
modalities {0,3,4,15,5,6} with the number of times each one is taken
{1,2,1,1,2,1}
Thank you very much
P.S. : is there some useful functions for discrete variables ?
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[R] emacs and R
At the suggestion of many people, I have installed emacs on my linux (Fedora
8.0) computer with the intention of using emacs as window interface to R
(2.6.0). I have gone though the emacs tutorial and don't see any information
about how I should use emacs to run R. Can anyone suggest a document that I
might read? In the past I have used R on a Windows XP system and used the
built-in windowing interface.
Thank you,
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
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Re: [R] emacs and R
At the suggestion of many people, I have installed emacs on my linux (Fedora 8.0) computer with the intention of using emacs as window interface to R (2.6.0). I have gone though the emacs tutorial and don't see any information about how I should use emacs to run R. Can anyone suggest a document that I might read? In the past I have used R on a Windows XP system and used the built-in windowing interface. Download Emacs Speak Statistics which is a LISP package for emacs. http://ess.r-project.org/ When installed you can e.g. M-x R Start an R process in Emacs % C-c C-c Sends a Control-C signal to the ESS process. This has the effect of aborting the current command. % C-c M-b Send the contents of the edit buffer to the ESS process and returns you to the ESS process buffer as well. % C-c M-r Send the text between point and mark to the ESS process and returns to the ESS process buffer afterwards. % C-c M-j Send the line containing point to the ESS process, and return to the ESS process buffer. % C-c C-n Sends the current line to the ESS process, echoing it in the process buffer, and moves point to the next line. % C-M-q Indents each line in the expression. % M-; Indents an existing comment line appropriately, or inserts an appropriate comment marker. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] discrete variable
See
length(unique(x))
[1] 6
table(x)
On 02/03/2008, Pete Dorothy [EMAIL PROTECTED] wrote:
Hello,
I am sorry for asking such a basic question. I could not find an answer to
it using google.
I have a discrete variable (a vector x) taking for example the following
values : 0, 3, 4, 3, 15, 5, 6, 5
Is it possible to know how many different values (modalities) it takes ?
Here it takes 6 different values but the length of the vector is 8.
I would like to know if there is a way to get the set of the
modalities {0,3,4,15,5,6} with the number of times each one is taken
{1,2,1,1,2,1}
Thank you very much
P.S. : is there some useful functions for discrete variables ?
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
--
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25° 25' 40 S 49° 16' 22 O
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Re: [R] summarizing matrix data
On 02/03/2008, Emilio Gagliardi [EMAIL PROTECTED] wrote: Henrique, Again, thank you very much for your code snippets, I am learning from them, but they are also creating as many questions as answers... On Sun, Mar 2, 2008 at 11:48 AM, Henrique Dallazuanna [EMAIL PROTECTED] wrote: Then you can try this: apply(array(as.vector(sapply(patt, get)), dim = c(dim(get(patt[1])), ngroups, nsubs)), 1:3, mean) For create the matrices in a loop one option is: lapply(seq(1, 51, by=10), function(x)matrix(seq(x, x+8), nrow=3, ncol=3)) if I use the above code, I get back a list of matrices, which is awesome! The problem is that your code to get the means doesn't work on it ( apply(array(as.vector(...))...additionally, I'm not creating the data, but reading it in, file by file and I don't know what group to assign a file until I open it (group2, group4, group4,group1,group3,group2,group1,etc) so I have 120 files to open and process and each file can be assigned to 1 of 4 groups(or lists as your code demonstrates)...but I don't know until I open the file which group hence my loop and meta structure approach. So I need to open the file, create a matrix from it (which I have working), store it along with other data from the same group, and then after I've read all the files and associated the data with a group, then I can generate the means per cell. So I can create 4 lists of matrices, which is fine by me but I don't see how I extend your code snippets to the case of reading the data from files...I currently have 1 master matrix which holds all the subject data as I mentioned in my previous post. data[3,3,1,nsubjects] represents all the subject data in group 1 data[3,3,2,nsubjects] represents all the subject data in group 2 data[3,3,3,nsubjects] represents all the subject data in group 3 how do i pullout just one group's worth of data and apply the mean function to each cell ? group1 - get(data[,,1,] ? Here, don't needed 'get' : group1 - data[,,1,] mean1 - apply(group1, 1:2, mean) group2 - get(data[,,2,] ? mean1 - apply(group1,1:3,mean) # mean per cell, so nine means (3 x 3) for each group... mean2 - apply(group2,1:3,mean) # mean per cell, so nine means (3 x 3) for each group... thanks so much for your patience! emilio [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] discrete variable
On 02-Mar-08 20:27:20, Pete Dorothy wrote:
Hello,
I am sorry for asking such a basic question. I could not find
an answer to it using google.
I have a discrete variable (a vector x) taking for example the
following values : 0, 3, 4, 3, 15, 5, 6, 5
Is it possible to know how many different values (modalities)
it takes?
Here it takes 6 different values but the length of the vector is 8.
I would like to know if there is a way to get the set of the
modalities {0,3,4,15,5,6} with the number of times each one is
taken {1,2,1,1,2,1}
This particulat kind of question is very simple:
table(c(0, 3, 4, 3, 15, 5, 6, 5))
# 0 3 4 5 6 15
# 1 2 1 2 1 1
Note that the first row of the output is simply the column
names, and is not part of the value returned by table() in
this case, which is the vector c(1,2,1,2,1,1) printed in
the second row.
table(c(0, 3, 4, 3, 15, 5, 6, 5))[4]
# 5
# 2
table(c(0, 3, 4, 3, 15, 5, 6, 5))[4] + 1.1
# 5
# 3.1
sum(table(c(0, 3, 4, 3, 15, 5, 6, 5)))
# [1] 8
Hoping this helps.
Thank you very much
P.S. : is there some useful functions for discrete variables ?
Many!
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 02-Mar-08 Time: 20:52:40
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Re: [R] discrete variable
Pete,
try
x - c(0, 3, 4, 3, 15, 5, 6, 5)
table(x)
or
length(table(x))
Cheers
Andrew
On Sun, Mar 02, 2008 at 09:27:20PM +0100, Pete Dorothy wrote:
Hello,
I am sorry for asking such a basic question. I could not find an answer to
it using google.
I have a discrete variable (a vector x) taking for example the following
values : 0, 3, 4, 3, 15, 5, 6, 5
Is it possible to know how many different values (modalities) it takes ?
Here it takes 6 different values but the length of the vector is 8.
I would like to know if there is a way to get the set of the
modalities {0,3,4,15,5,6} with the number of times each one is taken
{1,2,1,1,2,1}
Thank you very much
P.S. : is there some useful functions for discrete variables ?
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Andrew Robinson
Department of Mathematics and StatisticsTel: +61-3-8344-6410
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/
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Re: [R] Need help to locate my mistake
On 3/03/2008, at 9:18 AM, Louise Hoffman wrote:
Dear readers
I would like to make General Linear Model (GLM) for the following
data set
http://louise.hoffman.googlepages.com/fuel.csv
The code I have written is
fuelData-read.table('fuel.csv',header=TRUE, sep=',')
n-dim(fuelData)[1]
xOnes- matrix(1,nrow=n,ncol=1)
x-cbind(xOnes,fuelData[,3])
y-fuelData[,4]
theta-((t(x)%*%x)^(-1))%*%t(x)%*%y
which gives
theta
[,1]
[1,] 215.8374077
[2,] 0.1083491
When I do it in Matlab I get theta to be
79.69
0.18
which is correct. ~79 is the crossing of the y-axis.
This is certainly ***NOT*** correct. (If you really got those numbers
from Matlab, then Matlab is up to Puttee.)
Have you plotted your data?
(1) Fitting a straight line is ridiculous.
(2) If you are so foolish as to fit a straight line, you get
theta to have entries -4197.96 (intercept) and 2.16 (slope).
The line y = 79.69 + 0.18*x is off the edge of the graph and
does not even appear.
Have I perhaps written theta wrong?
Yes. The expression (t(x)%*%x)^(-1) is the matrix of entry
by entry reciprocals of the entries of t(x)%*%x.
You want:
theta - solve(t(x)%*%x))%*%t(x)%*%y
Anyhow, if you're going to use R, why not ***use R***?
fit - lm(fpi ~ rtime,data=fuelData)
theta - coef(fit)
This gives an answer identical to that from the corrected version of
your ``from scratch'' expression. (That expression, while
theoretically
correct, is numerically ill-advised. The cognoscenti use either the
Choleski or the ``qr'' decomposition of t(x)%*%x to effect the
calculations.
One of these is what is going on in the bowels of lm(). But here I
speak
of that of which I know little.)
cheers,
Rolf Turner
##
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Re: [R] Need help to locate my mistake
On 02-Mar-08 20:18:29, Louise Hoffman wrote:
Dear readers
I would like to make General Linear Model (GLM) for the
following data set http://louise.hoffman.googlepages.com/fuel.csv
The code I have written is
fuelData-read.table('fuel.csv',header=TRUE, sep=',')
n-dim(fuelData)[1]
xOnes- matrix(1,nrow=n,ncol=1)
x-cbind(xOnes,fuelData[,3])
y-fuelData[,4]
theta-((t(x)%*%x)^(-1))%*%t(x)%*%y
which gives
theta
[,1]
[1,] 215.8374077
[2,] 0.1083491
When I do it in Matlab I get theta to be
79.69
0.18
which is correct. ~79 is the crossing of the y-axis.
Have I perhaps written theta wrong? The formula for theta is
(alpha,beta)^T = (x^T * x)^(-1) * x^T * Y
where ^T means transposed.
Unfortunately, x^(-1) is not the inverse of x:
x-matrix(c(2,4,4,5),nrow=2)
x
# [,1] [,2]
# [1,]24
# [2,]45
x^(-1)
# [,1] [,2]
# [1,] 0.50 0.25
# [2,] 0.25 0.20
i.e. it is the matrix which you get by applying the operation
(...)^(-1) to each element of x.
In R, the inverse of a non-singular matrix is (somewhat obscurely)
denoted by solve(x):
solve(x)
#[,1] [,2]
# [1,] -0.833 0.667
# [2,] 0.667 -0.333
solve(x)%*%x
# [,1] [,2]
# [1,]1 1.110223e-16
# [2,]0 1.00e+00
(Note the slight rounding error); whereas
(x^(-1))%*%x
# [,1] [,2]
# [1,] 2.0 3.25
# [2,] 1.3 2.00
Best wishes,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 02-Mar-08 Time: 21:05:50
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[R] listing components of an object
Is there a method to list the components of an object, instead of looking at the help for that method? Let me be more clear with an example data(iris) ## tune `svm' for classification with RBF-kernel (default in svm), ## using one split for training/validation set obj - tune(svm, Species~., data = iris, ranges = list(gamma = 2^(-1:1), cost = 2^(2:4)), tunecontrol = tune.control(sampling = fix) ) ## alternatively: ## obj - tune.svm(Species~., data = iris, gamma = 2^(-1:1), cost = 2^(2:4)) summary(obj) plot(obj) - For tune, an object of class tune, including the components: best.parameters a 1 x k data frame, k number of parameters. best.performance best achieved performance. performances if requested, a data frame of all parameter combinations along with the corresponding performance results. train.ind list of index vectors used for splits into training and validation sets. best.model if requested, the model trained on the complete training data using the best parameter combination. I got the above by doing ?tune. Is there a function that helps be do this? Thanks ../Murli __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [OT] normal (as in Guassian)
Johannes Hüsing johannes at huesing.name writes: Am 02.03.2008 um 17:44 schrieb Gabor Csardi: I'm not a statistician, but do i remember well that among all distributions with a given mean and variance, the normal distribution has the highest entropy? This is good enough for me to call it normal Also, the formula for the standard normal distribution is the only one that is its own Fourier transform. So, if we assume the same distribution for a momentum and a location of a physical object, according to Heisenberg's Law it has to be the normal. It's not the only one. There is also the comb function, an infinite train of evenly spaced impulse functions that is its own transform, and then there is abs(x)^-0.5 and sech(x), but I'm just reading out of the appendix of Bracewell, 1978, The Fourier Transformation and Its Applications, McGraw-Hill. best, Ken __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [PS] discrete variable
Try table(), with the name of your vector inside the parentheses.
Ben
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Pete Dorothy
Sent: Sunday, March 02, 2008 2:27 PM
To: r-help@r-project.org
Subject: [PS] [R] discrete variable
Hello,
I am sorry for asking such a basic question. I could not find an answer
to
it using google.
I have a discrete variable (a vector x) taking for example the following
values : 0, 3, 4, 3, 15, 5, 6, 5
Is it possible to know how many different values (modalities) it takes ?
Here it takes 6 different values but the length of the vector is 8.
I would like to know if there is a way to get the set of the
modalities {0,3,4,15,5,6} with the number of times each one is taken
{1,2,1,1,2,1}
Thank you very much
P.S. : is there some useful functions for discrete variables ?
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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Re: [R] emacs and R
Louise Hoffman wrote: At the suggestion of many people, I have installed emacs on my linux (Fedora 8.0) computer with the intention of using emacs as window interface to R (2.6.0). I have gone though the emacs tutorial and don't see any information about how I should use emacs to run R. Can anyone suggest a document that I might read? In the past I have used R on a Windows XP system and used the built-in windowing interface. Download Emacs Speak Statistics which is a LISP package for emacs. http://ess.r-project.org/ It's easier still (on F8): yum install emacs-ess and you may also notice the user-friendly menus supplementing the key combinations below. These appear automagically if you are in an R interaction buffer, or a .R or .Rout file. -p When installed you can e.g. M-x R Start an R process in Emacs % C-c C-c Sends a Control-C signal to the ESS process. This has the effect of aborting the current command. % C-c M-b Send the contents of the edit buffer to the ESS process and returns you to the ESS process buffer as well. % C-c M-r Send the text between point and mark to the ESS process and returns to the ESS process buffer afterwards. % C-c M-j Send the line containing point to the ESS process, and return to the ESS process buffer. % C-c C-n Sends the current line to the ESS process, echoing it in the process buffer, and moves point to the next line. % C-M-q Indents each line in the expression. % M-; Indents an existing comment line appropriately, or inserts an appropriate comment marker. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] listing components of an object
On Mar 2, 2008, at 4:12 PM, Nair, Murlidharan T wrote: Is there a method to list the components of an object, instead of looking at the help for that method? Let me be more clear with an example data(iris) ## tune `svm' for classification with RBF-kernel (default in svm), ## using one split for training/validation set obj - tune(svm, Species~., data = iris, ranges = list(gamma = 2^(-1:1), cost = 2^(2:4)), tunecontrol = tune.control(sampling = fix) ) ## alternatively: ## obj - tune.svm(Species~., data = iris, gamma = 2^(-1:1), cost = 2^(2:4)) summary(obj) plot(obj) - For tune, an object of class tune, including the components: best.parameters a 1 x k data frame, k number of parameters. best.performance best achieved performance. performances if requested, a data frame of all parameter combinations along with the corresponding performance results. train.ind list of index vectors used for splits into training and validation sets. best.model if requested, the model trained on the complete training data using the best parameter combination. I got the above by doing ?tune. Is there a function that helps be do this? Usually str(obj), or possibly as.list(obj) for many objects, will give you the sort of information that I think you want, but what's wrong with ?tune ? In other words, why do you want this list? Typically, it is advisable that one uses methods to access the components of an object, like for instance coef(obj) for the coefficients in a model object. This is so that the internal representation of an object can change without breaking your code, and a host of other reasons. Thanks ../Murli Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [OT] normal (as in Guassian)
There is some information and references regarding the name 'normal' in the internet article 'Earliest Known Uses of Some of the Words of Mathematics (N)', http://members.aol.com/jeff570/n.html, by John Aldrich. It contains the comment, Galton does not explain why he uses the term normal but the sense of conforming to a norm ( = 'A standard, model, pattern, type.' (OED)) seems implied. On Sun, 2 Mar 2008 [EMAIL PROTECTED] wrote: Hi Folks, Apologies to anyone who'd prefer not to see this query on this list; but I'm asking because it is probably the forum where I'm most likely to get a good answer! I'm interested in the provenance of the name normal distribution (for what I'd really prefer to call the Gaussian distribution). According to Wikipedia, The name normal distribution was coined independently by Charles S. Peirce, Francis Galton and Wilhelm Lexis around 1875. So be it, if that was the case -- but I would like to know why they chose the name normal: what did they intend to convey? As background: I'm reflecting a bit on the usage in statistics of everyday language as techincal terms, as in significantly different. This, for instance, is likely to be misunderstood by the general publidc when they encounter statements in the media. Likewise, normally distributed would probably be interpreted as distributed in the way one would normally expect or, perhaps, there was nothing unusual about the distribution. Comments welcome! With thanks, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 02-Mar-08 Time: 13:04:17 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] emacs and R
On Sun, 2008-03-02 at 21:45 +0100, Louise Hoffman wrote: At the suggestion of many people, I have installed emacs on my linux (Fedora 8.0) computer with the intention of using emacs as window interface to R (2.6.0). I have gone though the emacs tutorial and don't see any information about how I should use emacs to run R. Can anyone suggest a document that I might read? In the past I have used R on a Windows XP system and used the built-in windowing interface. Download Emacs Speak Statistics which is a LISP package for emacs. http://ess.r-project.org/ On a Fedora 8 box, one can install ess from exiting Fedora package repositories. yum install emacs-ess as root in a terminal (or use Pirut via Applications Menu Add/Remove software) to install ESS. Then follow Louise's crib sheet to start R within Emacs etc. There is nothing wrong with installing from the tar ball at the ess homepage, but with the availability of a Fedora package I keep ESS more up-to-date than I ever did when I had to manually upgrade. G When installed you can e.g. M-x R Start an R process in Emacs % C-c C-c Sends a Control-C signal to the ESS process. This has the effect of aborting the current command. % C-c M-b Send the contents of the edit buffer to the ESS process and returns you to the ESS process buffer as well. % C-c M-r Send the text between point and mark to the ESS process and returns to the ESS process buffer afterwards. % C-c M-j Send the line containing point to the ESS process, and return to the ESS process buffer. % C-c C-n Sends the current line to the ESS process, echoing it in the process buffer, and moves point to the next line. % C-M-q Indents each line in the expression. % M-; Indents an existing comment line appropriately, or inserts an appropriate comment marker. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Recommended Packages
Having just update to R 2.6.2 on my old Windows laptop I notice that the number of packages is growing exponentially and my usual approach of get-em-all may not be viable much longer. Has any thought been given to dividing contributed binaries into a recommended set, perhaps a couple of hundred, and the remained. That way one could install the recommended ones routinely and add in the others as required. Any comments? Murray Jorgensen -- Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html Department of Statistics, University of Waikato, Hamilton, New Zealand Email: [EMAIL PROTECTED]Fax 7 838 4155 Phone +64 7 838 4773 wkHome +64 7 825 0441Mobile 021 1395 862 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [OT] normal (as in Guassian)
Thanks, Katherine! Now I wonder what, in particular, Peirce might have had in mind (he was a particularly sharp philosophical thinker, and might be expected to pay attention to the semantic baggage of what he said). I'm also enjoying the other delightful OT (= On Tangent) responses that my query has prompted! Best wishes to all, Ted. On 02-Mar-08 21:19:24, Katharine Mullen wrote: There is some information and references regarding the name 'normal' in the internet article 'Earliest Known Uses of Some of the Words of Mathematics (N)', http://members.aol.com/jeff570/n.html, by John Aldrich. It contains the comment, Galton does not explain why he uses the term normal but the sense of conforming to a norm ( = 'A standard, model, pattern, type.' (OED)) seems implied. On Sun, 2 Mar 2008 [EMAIL PROTECTED] wrote: Hi Folks, Apologies to anyone who'd prefer not to see this query on this list; but I'm asking because it is probably the forum where I'm most likely to get a good answer! I'm interested in the provenance of the name normal distribution (for what I'd really prefer to call the Gaussian distribution). According to Wikipedia, The name normal distribution was coined independently by Charles S. Peirce, Francis Galton and Wilhelm Lexis around 1875. So be it, if that was the case -- but I would like to know why they chose the name normal: what did they intend to convey? As background: I'm reflecting a bit on the usage in statistics of everyday language as techincal terms, as in significantly different. This, for instance, is likely to be misunderstood by the general publidc when they encounter statements in the media. Likewise, normally distributed would probably be interpreted as distributed in the way one would normally expect or, perhaps, there was nothing unusual about the distribution. Comments welcome! With thanks, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 02-Mar-08 Time: 13:04:17 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 02-Mar-08 Time: 21:52:20 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] discrete variable
Pete Dorothy [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Hello,
I am sorry for asking such a basic question. I could not find an
answer to it using google.
I have a discrete variable (a vector x) taking for example the
following values : 0, 3, 4, 3, 15, 5, 6, 5
Is it possible to know how many different values (modalities) it
takes ? Here it takes 6 different values but the length of the
vector is 8.
Try:
test - c(0, 3, 4, 3, 15, 5, 6, 5)
length(table(test))
#[1] 6
I would like to know if there is a way to get the set of the
modalities {0,3,4,15,5,6} with the number of times each one is taken
{1,2,1,1,2,1}
table(test)
#test
# 0 3 4 5 6 15
# 1 2 1 2 1 1
Thank you very much
P.S. : is there some useful functions for discrete variables ?
Yes...
https://home.comcast.net/~lthompson221/Splusdiscrete2.pdf
--
David Winsemius
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[R] help.start() on linux (fedora 8) when firefox alreading running - a way to open a new tab?
Using linux fedora 8 (x86_64) I get the following when firefox is already running. Is there a way to adjust settings in either R or firefox to open a new tab when help.start() is invoked? Thanks. --Dale help.start() Making links in per-session dir ... If '/usr/bin/firefox' is already running, it is *not* restarted, and you must switch to its window. Otherwise, be patient ... __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Runtime error
Hi everybody!
I try to implements Self-Organizing Maps with R and the Kohonen package.
I have made an algorithm that try differents width and height. For each
step, Silhouette index, Davies-Bouldin index and Dunn index are computed.
However, there is always after severals minutes a *Runtime error* appears on
the window:
Microsoft visual c++ runtime library
this application has requested the runtime to terminate
or
R for Windows GUI front-end has encountered a problem and needs to close.
We are sorry for the inconvenience.
it happens also that the window disppeared without message error
Please, if you have already see this problem or if you understand it and
know the solution, please contact me...
code:
results.som - matrix(NA, 7, 7)
results.sil - matrix(NA, 7, 7)
results.dbi - matrix(NA, 7, 7)
results.dun - matrix(NA, 7, 7)
for (i in 8:14){
for (j in 8:14){
# SOM with Kohonen package
som - som(data.scaled, grid = somgrid(i, j, hexagonal),
radius=j, rlen=1000)
# Summary SOM
results.som[i-7,j-7] - mean(som$distances)
# Indexes
results.sil[i-7,j-7] - summary(silhouette(som$unit.classif,
dist))$avg.width
results.dbi[i-7,j-7] - index.DB(species.data,som$unit.classif)$DB
results.dun[i-7,j-7] - dunn(dist, som$unit.classif)
}
}
--
Gwénaël Leday
Bio-Protection and Ecology Division
Lincoln University
New Zealand
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
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Re: [R] plotting to stdout (while reading from stdin?)
Hi Dirk, I didn't at first pay attention to your comment about littler, as my original problem of plotting to stdout was solved. But it was just part of the larger problem: I actually need to be piping the data with the code for making the picture in, and getting the picture out without opening any files. This is to be run in a server environment, where the use of temp files is not really acceptable. Having solved half of the problem, I feel cornered again, because I haven't found a nice way of mixing code and data in R. I am aware that this issue was brought up before; I am not sure I like the solutions suggested, because they involve the R language parsing and interpreting for each data row -- if I understand correctly. http://tolstoy.newcastle.edu.au/R/help/03a/6750.html Are these solutions the best currently available? That thread is almost 5 years old now. Let me give you an example of what I would like to be possible: In postgres, I can read this from stdin: CREATE TABLE remark ( case smallint, text text ); COPY remark (case, text) FROM stdin; 877 lymph node biopsy 909 Unresectable mass in the body of the pancreas \. ... more SQL It allows the code and input chunks to be mixed, because the input for stdin is always terminated by a special token, '\.', and so the parser can skip the data chunks without interpreting them. Almost similarly, perl has the __DATA__ token, that allows a portion of text within a program to be treated as stdin. I wonder whether anything like this is possible in R or littler. Thanks, --Gene On Fri, 7 Sep 2007, Dirk Eddelbuettel wrote: On Fri, Sep 07, 2007 at 02:27:00PM -0400, Duncan Murdoch wrote: On 9/7/2007 2:15 PM, Gene Selkov wrote: Thanks a ton, Duncan! So I have verified that this line works: echo postscript(file=\\, command=\cat\); plot(0) | r --vanilla --slave Wonderful! (albeit a little unobvious) I would include an explicit dev.off() after the plotting; I'm not sure all devices guarantee a clean shutdown when R quits. And for the record, both littler and Rscript can do that without the need for double quotes, at least under Linux. E.g. both $ r -e 'postscript(file=, command=cat); plot(0)' | head $ Rscript -e 'postscript(file=, command=cat); plot(0)' | head provide the same output (of the beginning of the postscript output). Our r is as usual somewhat faster, not that this matters in this non-repeat context. Dirk -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Recommended Packages
On Mon, 2008-03-03 at 10:52 +1300, Murray Jorgensen wrote: Having just update to R 2.6.2 on my old Windows laptop I notice that the number of packages is growing exponentially and my usual approach of get-em-all may not be viable much longer. Has any thought been given to dividing contributed binaries into a recommended set, perhaps a couple of hundred, and the remained. That way one could install the recommended ones routinely and add in the others as required. Any comments? Murray Jorgensen Hi Murray, What is one person's recommended set might not correspond with the recommended set of another nor that of a third party. Grouping contributed packages into topics or themes might be a more appropriate classification, and to that end we have CRAN Task Views (thanks to Achim Zeileis and Kurt Hornik and their band of merry CTV-maintainers). There are some 15 of these views now available. Using the tools in the ctv package a user can install these groupings of packages. Do these Task Views meet, or go some way towards, your goal? See: http://cran.r-project.org/web/views/ HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need help to locate my mistake
This is certainly ***NOT*** correct. (If you really got those numbers from Matlab, then Matlab is up to Puttee.) It was my mistake =) I had calculated the straight line so the edge of the plot was the y-axis =) Have you plotted your data? (1) Fitting a straight line is ridiculous. Yes, I guess we have to realise that in the project, and explain why it is not good enough =) Next step is to use exponential smoothing =) (2) If you are so foolish as to fit a straight line, you get theta to have entries -4197.96 (intercept) and 2.16 (slope). The line y = 79.69 + 0.18*x is off the edge of the graph and does not even appear. It was clearly a bad mistake of mine! =( Yes. The expression (t(x)%*%x)^(-1) is the matrix of entry by entry reciprocals of the entries of t(x)%*%x. You want: theta - solve(t(x)%*%x))%*%t(x)%*%y Thanks =) Anyhow, if you're going to use R, why not ***use R***? fit - lm(fpi ~ rtime,data=fuelData) theta - coef(fit) This gives an answer identical to that from the corrected version of your ``from scratch'' expression. (That expression, while theoretically correct, is numerically ill-advised. The cognoscenti use either the Choleski or the ``qr'' decomposition of t(x)%*%x to effect the calculations. This is great stuff! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] About R-Project
I'm not familiar with the Schuster system, but RSiteSearch(Schuster) produced 34 hits and RSiteSearch(river, fun) produced 133 for me just now. If this does not lead you to an answer to your question, PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Spencer Ion Andronache wrote: Hello, My name is Ion Andronache and I am a candidate for a doctor's degree on geomorphology in Braila, Romania. I would like to ask for some advice. I would like to make an analysis on the Danube's flow capacity and to determine the circles and periodicity using the Schuster system. I understood that your soft can do these operations but I don't know how. Would you be so kind to guide me? Best wishes, Ion Andronache. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] insert vector image into a plot
How can I insert a vector image in svg or pdf format into a plot. Basically i need the equivalent of what the 'pixmap' package does for bitmap images. Cheers, Will Morris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] emacs and R
Hi, John,
you don't have to switch to linux in order to use ess + emacs with R.
just follow the installation instruction of ess and it will take you 5
minutes at most. i also feel that xemacs seems more friendly than
gnuemacs for windows user.
On Sun, Mar 2, 2008 at 3:40 PM, John Sorkin [EMAIL PROTECTED] wrote:
At the suggestion of many people, I have installed emacs on my linux (Fedora
8.0) computer with the intention of using emacs as window interface to R
(2.6.0). I have gone though the emacs tutorial and don't see any information
about how I should use emacs to run R. Can anyone suggest a document that I
might read? In the past I have used R on a Windows XP system and used the
built-in windowing interface.
Thank you,
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
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--
===
WenSui Liu
ChoicePoint Precision Marketing
Phone: 678-893-9457
Email : [EMAIL PROTECTED]
Blog : statcompute.spaces.live.com
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Re: [R] Need help to locate my mistake
On 3/03/2008, at 1:58 PM, Louise Hoffman wrote:
What if I need to calculate the variance for the fuel data?
Are there a 'R' way to do that?
I have derived
variance = (Y-x*theta)^T * (Y - x*theta) / (n-p)
This should be (something like)
tr((Y - X%*%theta))%*%(Y-X%*%theta)/(n-p)
where X is your design matrix.
To get the information in R:
fit - lm(fpi ~ rtime,data=fuelData)
summary(fit)
This tells you the residual standard error (standard deviation) is
7.681. Typing
summary(fit)$sigma
gives the answer to more decimal places --- 7.681288.
The residual variance can of course be obtained by
summary(fit)$sigma^2
Note that the name ``sigma'' is bad --- this terminology should
be reserved for population quantities. What we have is ``sigma-hat'',
an *estimate* of sigma.
Note also that for your fuel data the residual variance is pretty
much meaningless since your model is highly inappropriate for these
data.
cheers,
Rolf Turner
##
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Re: [R] insert vector image into a plot
Thanks Paul. As always you spend an hour searching and find nothing, if i'd used the search term import instead of insert I would have found it in 2 secs. Lesson learnt, again. Cheers, Will Morris On 03/03/2008, at 12:50 PM, Paul Murrell wrote: Hi William Morris wrote: How can I insert a vector image in svg or pdf format into a plot. Basically i need the equivalent of what the 'pixmap' package does for bitmap images. 'grImport' See ... http://cran.r-project.org/web/packages/grImport/index.html http://www.stat.auckland.ac.nz/~paul/Talks/import.pdf Paul Cheers, Will Morris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R data Export to Excel
Here is my R Code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) variableprobe #output variance across probesets hist(variableprobe) #displaying histogram of variableprobe write.table(cbind(data[1], Variance=apply(data[,y],1,var)),file='c://variance.csv') #export as a .csv file. Output in Excel all in 1 column. ProbeID Variance 1 224588_at 21.5825745738848 How do i separate them so that i can have three columns ProbeID Variance 1 224588_at 21.582. thanks, Kei -- View this message in context: http://www.nabble.com/R-data-Export-to-Excel-tp15796903p15796903.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] glm: offset
R 2.6.0
Windows XP
A question about running a generalized linear model.
I am running a glm with
(1) a poisson distribution and a log link:
family=poisson(link = log)
and an offset.
I would like to know if I should express the offset as the log of the offset
value, i.e.
offset=log(NumUniqPt)
or as:
offset=NumUniqPt
I suspect I need to use the log, bu t I can't find any discussion of this in
MASS 1994 or on the man page for glm.
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] R data Export to Excel
If you are asking how to convert to multiple columns in Excel, look at the text to column option in I think the data tab. On Sun, Mar 2, 2008 at 9:59 PM, Keizer_71 [EMAIL PROTECTED] wrote: Here is my R Code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) variableprobe #output variance across probesets hist(variableprobe) #displaying histogram of variableprobe write.table(cbind(data[1], Variance=apply(data[,y],1,var)),file='c://variance.csv') #export as a .csv file. Output in Excel all in 1 column. ProbeID Variance 1 224588_at 21.5825745738848 How do i separate them so that i can have three columns ProbeID Variance 1 224588_at 21.582. thanks, Kei -- View this message in context: http://www.nabble.com/R-data-Export-to-Excel-tp15796903p15796903.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] unable to start device PNG and unable to open connection to X11 display
Hi, I have installed R on a computational cluster, and am using putty to access R. Please help on how to solve the problem of saving png files. png(file=myplot.png, bg=transparent) Error in X11(paste(png::, filename, sep = ), width, height, pointsize, : unable to start device PNG In addition: Warning message: unable to open connection to X11 display '' [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] glm: offset
Yes, use the log. I've had the same problem in the past, too. Try it on
a toy example to confirm it for yourself.
Cheers,
Simon.
On Sun, 2008-03-02 at 22:01 -0500, John Sorkin wrote:
R 2.6.0
Windows XP
A question about running a generalized linear model.
I am running a glm with
(1) a poisson distribution and a log link:
family=poisson(link = log)
and an offset.
I would like to know if I should express the offset as the log of the offset
value, i.e.
offset=log(NumUniqPt)
or as:
offset=NumUniqPt
I suspect I need to use the log, bu t I can't find any discussion of this in
MASS 1994 or on the man page for glm.
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
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GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
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--
Simon Blomberg, BSc (Hons), PhD, MAppStat.
Lecturer and Consultant Statistician
Faculty of Biological and Chemical Sciences
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Re: [R] R data Export to Excel
hi, did you try write.xls in xlsReadWrite package? On Sun, Mar 2, 2008 at 9:59 PM, Keizer_71 [EMAIL PROTECTED] wrote: Here is my R Code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) variableprobe #output variance across probesets hist(variableprobe) #displaying histogram of variableprobe write.table(cbind(data[1], Variance=apply(data[,y],1,var)),file='c://variance.csv') #export as a .csv file. Output in Excel all in 1 column. ProbeID Variance 1 224588_at 21.5825745738848 How do i separate them so that i can have three columns ProbeID Variance 1 224588_at 21.582. thanks, Kei -- View this message in context: http://www.nabble.com/R-data-Export-to-Excel-tp15796903p15796903.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === WenSui Liu ChoicePoint Precision Marketing Phone: 678-893-9457 Email : [EMAIL PROTECTED] Blog : statcompute.spaces.live.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R data Export to Excel
thanks for your response. How do i install it? I try looking at the manual it doesn't seem indicate any installation instruction. I also download a windows version but it doesn't have an exe file. http://cran.r-project.org/web/packages/xlsReadWrite/index.html Newbie, Kei On 3/3/08, Wensui Liu [EMAIL PROTECTED] wrote: hi, did you try write.xls in xlsReadWrite package? On Sun, Mar 2, 2008 at 9:59 PM, Keizer_71 [EMAIL PROTECTED] wrote: Here is my R Code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) variableprobe #output variance across probesets hist(variableprobe) #displaying histogram of variableprobe write.table(cbind(data[1], Variance=apply(data[,y],1,var)),file='c://variance.csv') #export as a .csv file. Output in Excel all in 1 column. ProbeID Variance 1 224588_at 21.5825745738848 How do i separate them so that i can have three columns ProbeID Variance 1 224588_at 21.582. thanks, Kei -- View this message in context: http://www.nabble.com/R-data-Export-to-Excel-tp15796903p15796903.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === WenSui Liu ChoicePoint Precision Marketing Phone: 678-893-9457 Email : [EMAIL PROTECTED] Blog : statcompute.spaces.live.com === -- Christophe Lo (078) 8275 7029 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R data Export to Excel
i think you simply install it in the way you install other R packages. On Sun, Mar 2, 2008 at 10:40 PM, Christophe Lo [EMAIL PROTECTED] wrote: thanks for your response. How do i install it? I try looking at the manual it doesn't seem indicate any installation instruction. I also download a windows version but it doesn't have an exe file. http://cran.r-project.org/web/packages/xlsReadWrite/index.html Newbie, Kei On 3/3/08, Wensui Liu [EMAIL PROTECTED] wrote: hi, did you try write.xls in xlsReadWrite package? On Sun, Mar 2, 2008 at 9:59 PM, Keizer_71 [EMAIL PROTECTED] wrote: Here is my R Code x-1:2 y-2:141 data.matrix-data.matrix(data[,y])#create data.matrix variableprobe-apply(data.matrix[x,],1,var) variableprobe #output variance across probesets hist(variableprobe) #displaying histogram of variableprobe write.table(cbind(data[1], Variance=apply(data[,y],1,var)),file='c://variance.csv') #export as a .csv file. Output in Excel all in 1 column. ProbeID Variance 1 224588_at 21.5825745738848 How do i separate them so that i can have three columns ProbeID Variance 1 224588_at 21.582. thanks, Kei -- View this message in context: http://www.nabble.com/R-data-Export-to-Excel-tp15796903p15796903.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === WenSui Liu ChoicePoint Precision Marketing Phone: 678-893-9457 Email : [EMAIL PROTECTED] Blog : statcompute.spaces.live.com === -- Christophe Lo (078) 8275 7029 [EMAIL PROTECTED] -- === WenSui Liu ChoicePoint Precision Marketing Phone: 678-893-9457 Email : [EMAIL PROTECTED] Blog : statcompute.spaces.live.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting to stdout (while reading from stdin?)
On Sun, 2 Mar 2008, Gabor Grothendieck wrote: You can do this: Lines - A,B 1,2 3,4 DF - read.csv(textConnection(Lines)) which is slightly simpler than the examples there. Thank you Gabor, I made it even simpler by replacing the Lines object with a string contstant. It works perfectly well for me. Cheers, --Gene On Sun, Mar 2, 2008 at 5:59 PM, Gene Selkov [EMAIL PROTECTED] wrote: Hi Dirk, I didn't at first pay attention to your comment about littler, as my original problem of plotting to stdout was solved. But it was just part of the larger problem: I actually need to be piping the data with the code for making the picture in, and getting the picture out without opening any files. This is to be run in a server environment, where the use of temp files is not really acceptable. Having solved half of the problem, I feel cornered again, because I haven't found a nice way of mixing code and data in R. I am aware that this issue was brought up before; I am not sure I like the solutions suggested, because they involve the R language parsing and interpreting for each data row -- if I understand correctly. http://tolstoy.newcastle.edu.au/R/help/03a/6750.html Are these solutions the best currently available? That thread is almost 5 years old now. Let me give you an example of what I would like to be possible: In postgres, I can read this from stdin: CREATE TABLE remark ( case smallint, text text ); COPY remark (case, text) FROM stdin; 877 lymph node biopsy 909 Unresectable mass in the body of the pancreas \. ... more SQL It allows the code and input chunks to be mixed, because the input for stdin is always terminated by a special token, '\.', and so the parser can skip the data chunks without interpreting them. Almost similarly, perl has the __DATA__ token, that allows a portion of text within a program to be treated as stdin. I wonder whether anything like this is possible in R or littler. Thanks, --Gene On Fri, 7 Sep 2007, Dirk Eddelbuettel wrote: On Fri, Sep 07, 2007 at 02:27:00PM -0400, Duncan Murdoch wrote: On 9/7/2007 2:15 PM, Gene Selkov wrote: Thanks a ton, Duncan! So I have verified that this line works: echo postscript(file=\\, command=\cat\); plot(0) | r --vanilla --slave Wonderful! (albeit a little unobvious) I would include an explicit dev.off() after the plotting; I'm not sure all devices guarantee a clean shutdown when R quits. And for the record, both littler and Rscript can do that without the need for double quotes, at least under Linux. E.g. both $ r -e 'postscript(file=, command=cat); plot(0)' | head $ Rscript -e 'postscript(file=, command=cat); plot(0)' | head provide the same output (of the beginning of the postscript output). Our r is as usual somewhat faster, not that this matters in this non-repeat context. Dirk -- Three out of two people have difficulties with fractions. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R function to convert a number to text
hi, Dear R users - I wonder is there a written R function to convert a number to a text, say convert 1 to one , 100 to one hundred. I know in xls. has such a function BAHTTEXT, does anybody know is there a similar function in R ? Thanks. Lin - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] insert vector image into a plot
Hi William Morris wrote: How can I insert a vector image in svg or pdf format into a plot. Basically i need the equivalent of what the 'pixmap' package does for bitmap images. 'grImport' See ... http://cran.r-project.org/web/packages/grImport/index.html http://www.stat.auckland.ac.nz/~paul/Talks/import.pdf Paul Cheers, Will Morris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Constrained regression
Dear list members, I am trying to get information on how to fit a linear regression with constrained parameters. Specifically, I have 8 predictors , their coeffiecients should all be non-negative and add up to 1. I understand it is a quadratic programming problem but I have no experience in the subject. I searched the archives but the results were inconclusive. Could someone provide suggestions and references to the literature, please? Thank you very much. Carlos Carlos Alzola [EMAIL PROTECTED] (703) 242-6747 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] handling big data set in R
Hello R users,
I'm wondering whether it is possible to manage big data set in R? I
have a data set with 3 million rows and 3 columns (X,Y,Z), where X is
the group id. For each X, I need to run 2 regression on the submatrix.
I used the function split:
datamatrix-read.csv(datas.csv, header=F, sep=,)
dim(datamatrix)
# [1] 2980523 3
names(datamatrix)-c(X,Y,Z)
attach(datamatrix)
subX-split(X, X)
subY-split(Y,X)
subZ-split(Z,X)
n-length(subdata) ### number of groups
s1-s2-rep(NA, n) ### vector to store the regression slope
for (i in 1:n){
a-table(Y[[i]])
table.x-as.numeric(names(a))
table.y-as.numeric(a)
fit1-lm(table.y~table.x)# find the slope of the histogram of y
s1[i]-fit$coefficients[2]
fit2-lm(subY[[i]]~subZ[[i]]) ### regress y on z
s2[i]-fit$coefficients[2]
}
But my R died before completing the loop... (I've thought about doing
it in SAS, but I don't know how to write a loop combined with a PROC
REG...)
One thing that might be helpful is that my data set has already been
sorted based on X. I don't know whether this can be any helpful for
managing the dataset.
Any suggestion would be appreciated!
Thanks!
-Shu
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] handling big data set in R
Hello R users,
I'm wondering whether it is possible to manage big data set in R? I
have a data set with 3 million rows and 3 columns (X,Y,Z), where X is
the group id. For each X, I need to run 2 regression on the submatrix.
I used the function split:
datamatrix-read.csv(datas.csv, header=F, sep=,)
dim(datamatrix)
# [1] 2980523 3
names(datamatrix)-c(X,Y,Z)
attach(datamatrix)
subX-split(X, X)
subY-split(Y,X)
subZ-split(Z,X)
n-length(subdata) ### number of groups
s1-s2-rep(NA, n) ### vector to store the regression slope
for (i in 1:n){
a-table(Y[[i]])
table.x-as.numeric(names(a))
table.y-as.numeric(a)
fit1-lm(table.y~table.x)# find the slope of the histogram of y
s1[i]-fit$coefficients[2]
fit2-lm(subY[[i]]~subZ[[i]]) ### regress y on z
s2[i]-fit$coefficients[2]
}
But my R died before completing the loop... (I've thought about doing
it in SAS, but I don't know how to write a loop combined with a PROC
REG...)
One thing that might be helpful is that my data set has already been
sorted based on X. I don't know whether this can be any helpful for
managing the dataset.
Any suggestion would be appreciated!
Thanks!
-Shu
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Re: [R] Problem with PCA
This is a homework problem. I know how to do a PCA, you need to learn. I suggest you visit your textbook, then check the documentation for R's various PCA implementations to work out how to effect the analysis. phthao05 wrote: I have an exercise. With 3 kinds of yohourt a,b,c. There are 25 participatients estimate 3 norms: taste (va,vb,vc), structure (ca,cb,cc) and price (ga,gb,gc) and give the mark from 1 to 5. I don't know how to PCA this data. Please help me! I attached the data file follow: VaVb Vc Ca Cb Cc Ga Gb Gc 4 2 4 5 5 5 4 4 2 2 2 4 3 2 5 4 5 1 2 2 1 2 3 3 3 1 4 1 1 2 2 3 3 4 3 2 3 4 4 4 3 1 2 1 2 1 1 1 1 2 4 3 2 4 4 4 2 2 2 1 2 1 3 2 3 3 3 4 3 1 1 1 4 5 1 3 3 2 4 2 4 2 2 5 1 1 3 2 3 3 4 2 4 5 3 3 4 4 4 3 4 2 1 2 2 1 2 4 1 2 1 2 3 3 3 1 4 3 4 1 1 2 1 2 3 3 5 4 3 4 3 1 1 1 2 4 4 2 2 2 1 4 2 4 2 2 1 1 2 4 3 2 4 3 1 3 4 2 5 3 4 1 1 1 2 2 3 3 4 3 2 2 3 4 3 4 4 4 3 1 2 3 3 3 4 3 1 1 1 4 3 1 1 1 2 2 3 3 1 1 1 1 2 4 3 2 4 3 4 2 1 2 2 3 1 3 3 4 2 1 2 2 1 2 4 -- Dr Richard Rowe Zoology Tropical Ecology School of Marine Tropical Biology James Cook University Townsville 4811 AUSTRALIA ph +61 7 47 81 4851 fax +61 7 47 25 1570 JCU has CRICOS Provider Code 00117J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] unable to start device PNG and unable to open connection to X11 display
Please consult the help page for png. On Mon, 3 Mar 2008, Ng Stanley wrote: Hi, I have installed R on a computational cluster, and am using putty to access R. Please help on how to solve the problem of saving png files. png(file=myplot.png, bg=transparent) Error in X11(paste(png::, filename, sep = ), width, height, pointsize, : unable to start device PNG In addition: Warning message: unable to open connection to X11 display '' [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R data Export to Excel
it doesn't seem indicate any installation instruction. http://cran.r-project.org/web/packages/xlsReadWrite/index.html As Wensui Liu already mentioned, installation is the same as with other packages: In the RGui menu select Packages-Install Packages...-(Select a CRAN mirror)-(Select the package)-done. Best regards, Hans-Peter [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Attempting to connect to an Empress RDBMS via RODBC 1.2-3 causes R 2.6.2.pat to segfault
Hi I've experienced some unpleasant behaviour while attempting to connect to an Empress 8.6.2 RDBMS via RODBC 1.2-3 and a freshly minted R 2.6.2.pat on a SLED linux system where a call to odbcConnect() to initiate the odbc connection causes R 2.6.2.pat to segfault. The odbcConnect call is CHANNEL - odbcConnect(myEmpressDB, uid=myusername, pwd=mysecurepassword, believeNRows=FALSE) where myEmpressDB is an apparently correctly configured ODBC datasource pointing at the Empress RDBMS I wish to access on my system. The R version data for my system are platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status Patched major 2 minor 6.2 year 2008 month 03 day02 svn rev44661 language R version.string R version 2.6.2 Patched (2008-03-02 r44661) All a bit surprising (to me, at least) as: (i) testing connections to the Empress datasource with isql on my SLED linux system is successful, suggesting to me that the Empress driver and the datasource are correctly configured; (ii) I am able to successfully initiate a connection to a non-Empress odbc datasource (postgres and mysql databases) on my SLED linux system within R and to send and retrieve data from these; and (iii) an analogous setup (R, Empress odbc driver, datasource,...) on a windows XP system that I have access to works flawlessly. Any suggestions about how to proceed? Thanks muchly, Michael _ begin:vcard fn:Michael J. Manning n:Manning;Michael org:National Institute of Water and Atmospheric Research (NIWA);Inshore and Pelagic Fisheries adr:295-301 Evans Bay Parade, Kilbirnie;;Private Bag 14901;Wellington;;;New Zealand email;internet:[EMAIL PROTECTED] title:Scientist tel;work:+64 (0)4 386 0851 tel;fax:+64 (0)4 386 0574 x-mozilla-html:FALSE url:http://www.niwascience.co.nz version:2.1 end:vcard __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Formating a zoo dataset .
Suppose I have following dataset : head(data1) Date Return 1 03/31/00 0.14230650 2 04/28/00 -0.03276228 3 05/31/00 -0.06527890 4 06/30/00 -0.04999873 5 07/31/00 -0.01447902 6 08/31/00 0.22265729 Now I convert it to zoo object : data11 = zoo(data1[,2], as.Date(data1[,1], format=%m/%d/%y)) head(data11) 2000-03-31 2000-04-28 2000-05-31 2000-06-30 2000-07-31 2000-08-31 0.14230650 -0.03276228 -0.06527890 -0.04999873 -0.01447902 0.22265729 Clearly those are monthly data. Therefore I want to convert it to mm-yy format. I used following code : data111 = zoo(coredata(data11), format(index(data11), %m/%y)) However what I got is that : head(data111) 01/0101/0201/0301/0401/0501/06 -0.00139 -0.016274826 -0.047707664 0.001104362 -0.077961541 0.017637141 tail(data111) 12/0212/0312/0412/0512/0612/07 0.058660676 -0.018067833 -0.055569851 0.007142888 0.051162052 0.052643733 It is evident that month order has been changed. Can anyone here tell me how to get correct order like : 01/01, 02/01, 03/01.. Your help is highly appreciated Regards, - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
