[R] Calculating Percentile in R
Hi all, I have a problem on how R calculates Percentiles : Suppose I have following data set: data1 [1] -16648185 -14463457 -14449400 -13905309 -13443436 -13234755 -12956282 -11660896 [9] -10061040 -9805005 -9789583 -9754642 -9562164 -9391709 -9212182 -9151073 [17] -9092732 -9068214 -8978151 -8943912 -8761890 -8632106 -8541580 -8501249 [25] -8234466 -8219015 -8193543 -7488279 -7340768 -7236684 -7225458 -7159465 [33] -6819625 -6810858 -6755620 -6626439 -6610901 -6551762 -6207377 -6192583 [41] -6106783 -6077051 -6035300 -6035195 -6019017 -5954375 -5946285 -5886082 [49] -5880402 -5723368 -5668698 -5599168 -5548276 -5445734 -5412312 -5384707 [57] -5309365 -5303425 -5285274 -5204585 -5096301 -5092182 -5053349 -5041533 [65] -5021234 -5005402 -4984232 -4981990 -4964457 -4936653 -4920384 -4918021 [73] -4895351 -4843258 -4824730 -4774792 -4771018 -4616156 -4590430 -262 [81] -4443954 -4435397 -4415112 -4374465 -4341858 -4267891 -4252410 -4185021 [89] -4164458 -4158863 -4020436 -4006030 -3975819 -3959667 -3916414 -3876878 [97] -3765340 -3729338 -3713670 -3634991 Now the 5th percentile should be value corresponding to 0.05*(100+1) = 5.05 = 5 (rounded) hence : -13443436 But R give the value : quantile(data1, 0.05) 5% -13245189 Can anyone clarify me on this regards? Thanks Megh Comedy with an Edge to see what's on, when. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loading problem with R2HTML package
On Sun, 17 Jun 2007, spime wrote: I have downloaded latest version of R2HTML (v1.54) for 64-bit windows PC. The latest version is 1.58 from September 2006. See http://cran.r-project.org/src/contrib/Descriptions/R2HTML.html My R version 2.5.0. My problem arises when i want to install SciViews-R which need R2HTML package. library(R2HTML) Error in `parent.env-`(`*tmp*`, value = NULL) : use of NULL environment is defunct Error: package/namespace load failed for 'R2HTML' Any remedy ? Use the real 'latest version'. Have you perhaps used a broken 'mirror' of CRAN? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating Percentile in R
Quantiles aren't uniquely defined. Type ?quantile to learn more about the various possibilities built in to R. Ted. Megh Dal wrote on 06/18/2007 04:37 PM: Hi all, I have a problem on how R calculates Percentiles : Suppose I have following data set: data1 [1] -16648185 -14463457 -14449400 -13905309 -13443436 -13234755 -12956282 -11660896 [9] -10061040 -9805005 -9789583 -9754642 -9562164 -9391709 -9212182 -9151073 [17] -9092732 -9068214 -8978151 -8943912 -8761890 -8632106 -8541580 -8501249 [25] -8234466 -8219015 -8193543 -7488279 -7340768 -7236684 -7225458 -7159465 [33] -6819625 -6810858 -6755620 -6626439 -6610901 -6551762 -6207377 -6192583 [41] -6106783 -6077051 -6035300 -6035195 -6019017 -5954375 -5946285 -5886082 [49] -5880402 -5723368 -5668698 -5599168 -5548276 -5445734 -5412312 -5384707 [57] -5309365 -5303425 -5285274 -5204585 -5096301 -5092182 -5053349 -5041533 [65] -5021234 -5005402 -4984232 -4981990 -4964457 -4936653 -4920384 -4918021 [73] -4895351 -4843258 -4824730 -4774792 -4771018 -4616156 -4590430 -262 [81] -4443954 -4435397 -4415112 -4374465 -4341858 -4267891 -4252410 -4185021 [89] -4164458 -4158863 -4020436 -4006030 -3975819 -3959667 -3916414 -3876878 [97] -3765340 -3729338 -3713670 -3634991 Now the 5th percentile should be value corresponding to 0.05*(100+1) = 5.05 = 5 (rounded) hence : -13443436 But R give the value : quantile(data1, 0.05) 5% -13245189 Can anyone clarify me on this regards? Thanks Megh Comedy with an Edge to see what's on, when. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr E.A. Catchpole Visiting Fellow Univ of New South Wales at ADFA, Canberra, Australia _ and University of Kent, Canterbury, England 'v'- www.pems.adfa.edu.au/~ecatchpole / \ - fax: +61 2 6268 8786 m m- ph: +61 2 6268 8895 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating Percentile in R
On Sun, 17 Jun 2007, Megh Dal wrote: Hi all, I have a problem on how R calculates Percentiles : Suppose I have following data set: data1 [1] -16648185 -14463457 -14449400 -13905309 -13443436 -13234755 -12956282 -11660896 [9] -10061040 -9805005 -9789583 -9754642 -9562164 -9391709 -9212182 -9151073 [17] -9092732 -9068214 -8978151 -8943912 -8761890 -8632106 -8541580 -8501249 [25] -8234466 -8219015 -8193543 -7488279 -7340768 -7236684 -7225458 -7159465 [33] -6819625 -6810858 -6755620 -6626439 -6610901 -6551762 -6207377 -6192583 [41] -6106783 -6077051 -6035300 -6035195 -6019017 -5954375 -5946285 -5886082 [49] -5880402 -5723368 -5668698 -5599168 -5548276 -5445734 -5412312 -5384707 [57] -5309365 -5303425 -5285274 -5204585 -5096301 -5092182 -5053349 -5041533 [65] -5021234 -5005402 -4984232 -4981990 -4964457 -4936653 -4920384 -4918021 [73] -4895351 -4843258 -4824730 -4774792 -4771018 -4616156 -4590430 -262 [81] -4443954 -4435397 -4415112 -4374465 -4341858 -4267891 -4252410 -4185021 [89] -4164458 -4158863 -4020436 -4006030 -3975819 -3959667 -3916414 -3876878 [97] -3765340 -3729338 -3713670 -3634991 Now the 5th percentile should be value corresponding to 0.05*(100+1) = 5.05 = 5 (rounded) According to you, but not according to ?quantile, which gives you a choice of 9 definitions, yours not being the default. hence : -13443436 But R give the value : quantile(data1, 0.05) 5% -13245189 Can anyone clarify me on this regards? Thanks Megh [[alternative HTML version deleted]] PLEASE do read the posting guide http://www.R-project.org/posting-guide.html PLEASE do as we ask. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] to read table
Hello, I have a problem to read a csv table. To read it I used this syntax donParCara - read.table(C:/Documents and Settings/melyakhlifi/Mes documents/feuilles excel/calcul2.csv,header=TRUE,sep=;,quote=,dec=,) I don't understand my errors Erreur dans scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : la ligne 16 n'avait pas 21 éléments Can you help me please? _ Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo! Mail [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lines connecting the boxes in a boxplot
Dear Arne I' recommend to save the information of your boxplots a - boxplot(...) str(a) Then you have the information that you need about your boxplot (e.g. the value of the median) and can use segments() to draw the lines you want. Hope this helps Best regards, Christoph -- Credit and Surety PML study: visit our web page www.cs-pml.org -- Christoph Buser [EMAIL PROTECTED] Seminar fuer Statistik, LEO C13 ETH Zurich 8092 Zurich SWITZERLAND phone: x-41-44-632-4673 fax: 632-1228 http://stat.ethz.ch/~buser/ -- Arne Brutschy writes: Hello, I'm currently using a boxplot to visualize data for three different models. As I have three models, I'm plotting three parallel boxplots for each factor. This works fine - what I need now is a line connecting the medians of each boxplot of each model. I want to do this in order to visualize the trend that one of the models exhibit. Basically, I want to plot a curve for each model (slightly offset on the x axis), with a boxplot on each datapoint. It's only an idea, and I don't know if it's not too confusing after adding the lines... Is it possible? Has anyone done this before? Sorry if this has been asked before or is a standard feature, I simply have now clue how to name the feature I want. Ergo: I cannot search for it.. :\ Regards, Arne PS: this is my current code require(gplots) boxwex=0.15 data - read.table(all_runs_fitness.data); colnames(data)=c(model,matrix,fitness) boxplot(fitness ~ matrix, data=data, boxwex=boxwex, at=(1:7 - 0.2), main=Fitness for Matrix/Models, xlab=Matrixtype, ylab=Fitness, ylim=c(20,100), subset=(model==dyn), col=lightblue, xaxt=n, whisklty=1) boxplot(fitness ~ matrix, data=data, boxwex=boxwex, at = 1:7, add=TRUE, subset=(model==dl3), col=mistyrose, xaxt=n, whisklty=1) boxplot(fitness ~ matrix, data=data, boxwex=boxwex, at=(1:7 + 0.2), add=TRUE, subset=(model==dl4), col=lightcyan, xaxt=n, whisklty=1) axis(1, 1:8-0.5, labels=FALSE) axis(1, 1:7, tck=FALSE, labels=levels(data[,2])) smartlegend(x=left, y=bottom, inset = 0.01, c(dyn,dl3,dl4), fill = c(lightblue, mistyrose, lightcyan)) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] what about options in BATCH
hello, when I run a calcul in BATCH the screen displays all the code of my programs and even the introduction of R how can I do to don't display it? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] to read table
Why are you not using read.csv or read.csv2 when you are reading a csv file? Original Message Subject: [R] to read table From: elyakhlifi mustapha [EMAIL PROTECTED] To: R-help@stat.math.ethz.ch Date: 18.06.2007 09:23 Hello, I have a problem to read a csv table. To read it I used this syntax donParCara - read.table(C:/Documents and Settings/melyakhlifi/Mes documents/feuilles excel/calcul2.csv,header=TRUE,sep=;,quote=,dec=,) I don't understand my errors Erreur dans scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : la ligne 16 n'avait pas 21 éléments Can you help me please? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] to read table
That means that R doesn't see 21 elements in line 16 of your file. This can happen if one of your entries consists of two words, which are then read as two elements. Katharina elyakhlifi mustapha schrieb: Hello, I have a problem to read a csv table. To read it I used this syntax donParCara - read.table(C:/Documents and Settings/melyakhlifi/Mes documents/feuilles excel/calcul2.csv,header=TRUE,sep=;,quote=,dec=,) I don't understand my errors Erreur dans scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : la ligne 16 n'avait pas 21 éléments Can you help me please? _ Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo! Mail [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with binding data-frames
Hello, I'm having a problem concerning r-binding datasets. I have six datasets, from six different plates, and two different days. I want to combine these datasets together for analysis. Datasets from day 2, have all the same columns than datasets from day 1. However in addition, there are few columns more in day 2. Thus, using rbind for this, results a error, because the objects are not the same length. Error in paste(nmi[nii == 0L], collapse = , ) : object nii not found In addition: Warning message: longer object length is not a multiple of shorter object length in: clabs == nmi What I need, is to combine all the six together, and give for example NA-value in day 1, for those columns which can only be found in day 2. Is this somehow possible? I have several of these six-datasets groups, and only few of them are having this problem described above, and I cannot know in advance which. With most of the groups writing rbind(data1,data2,data3,data4,data5,data6) works easily, but these few problematic groups need also to be combined... Any help greatly appreciated! -Jouni __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lines connecting the boxes in a boxplot
Hi, C I' recommend to save the information of your boxplots C Then you have the information that you need about your boxplot C (e.g. the value of the median) and can use segments() to draw C the lines you want. Thanks, works like a charm! Regards, Arne C Arne Brutschy writes: Hello, I'm currently using a boxplot to visualize data for three different models. As I have three models, I'm plotting three parallel boxplots for each factor. This works fine - what I need now is a line connecting the medians of each boxplot of each model. I want to do this in order to visualize the trend that one of the models exhibit. Basically, I want to plot a curve for each model (slightly offset on the x axis), with a boxplot on each datapoint. It's only an idea, and I don't know if it's not too confusing after adding the lines... Is it possible? Has anyone done this before? Sorry if this has been asked before or is a standard feature, I simply have now clue how to name the feature I want. Ergo: I cannot search for it.. :\ Regards, Arne PS: this is my current code require(gplots) boxwex=0.15 data - read.table(all_runs_fitness.data); colnames(data)=c(model,matrix,fitness) boxplot(fitness ~ matrix, data=data, boxwex=boxwex, at=(1:7 - 0.2), main=Fitness for Matrix/Models, xlab=Matrixtype, ylab=Fitness, ylim=c(20,100), subset=(model==dyn), col=lightblue, xaxt=n, whisklty=1) boxplot(fitness ~ matrix, data=data, boxwex=boxwex, at = 1:7, add=TRUE, subset=(model==dl3), col=mistyrose, xaxt=n, whisklty=1) boxplot(fitness ~ matrix, data=data, boxwex=boxwex, at=(1:7 + 0.2), add=TRUE, subset=(model==dl4), col=lightcyan, xaxt=n, whisklty=1) axis(1, 1:8-0.5, labels=FALSE) axis(1, 1:7, tck=FALSE, labels=levels(data[,2])) smartlegend(x=left, y=bottom, inset = 0.01, c(dyn,dl3,dl4), fill = c(lightblue, mistyrose, lightcyan)) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] merging dataframes with diffent rownumbers
Dear R-Helpers, I have following problem: I do have two data frames dat1 and dat2 with a commen column BNUM (long integer). dat1 has a larger number of BNUM than dat2 and different rows of dat2 have equal BNUM. The numbers of rows in dat1 and dat2 is not equal. I applied the tapply-function to dat2 with BNUM as index. I would like to add the columns from dat1 to the results of b.sum - tapply(dat2, BNUM, sum). However the BNUM of b.sum are only a subset of the dat1. Does anybody knows a elegant way to solve the problem? Thanks in advance Thomas H. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help with panel.lda
I work with Windows, R version 2.4.1
I am trying to plot the results of a discriminant analysis done using the
lda function in the MASS library. The discriminant. analysis goes like this:
data.tb-read.table('C:\\Documents and
Settings\\silvia\\Desktop\\dicrim_test.txt', header=T) ## the actual made-up
test matrix is pasted below
train-sample (1:36, 15)
table (data.tb$group[train])
data.lda-lda(group~., data.tb, subset = train)
predict (data.lda, data.tb[-train,])$class
Then I want to obtain a plot by writing:
plot(data.lda, cex=0.7, 2, xlab='LD1', ylab='LD2')
but it's not working. It says that it could not find the function panel.
When I include panel=panel.lda, it tells me that the object 'panel.lda' was
not found. All I get is an empty plot in the graphics window. Was I
supposed to create an object called panel.lda? I cannot find in the help
what that object might be.
Any help would be appreciated.
TABLE USED:
group var1 var2 var3
1 13 556
2 14 667
3 15 558
4 14 667
5 13 446
6 13 555
7 13 444
8 14 443
9 14 447
10 14 666
11 25 889
12 24 998
13 28 889
14 29 768
15 28 669
16 29 99 10
17 2 10 1009
18 24 999
19 28 888
20 29 769
21 28 66 10
22 29 999
23 32 11 11
24 33 222
25 31 333
26 31 111
27 32 442
28 33 223
29 34 111
30 31 111
31 32 222
32 33 333
33 31 111
34 31 442
35 32 223
36 33 111
--
View this message in context:
http://www.nabble.com/help-with-panel.lda-tf3939027.html#a11172049
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Readline - wait for user input
Hello,
I also have problems to get to run the following lines. If I run the
block instead of every single line, it simply does not wait for the input.
Can anybody help me?
pos_name - readline(Please type: )
r - substr(pos_name, 1,1)
c - substr(pos_name, 2,nchar(pos_name))
Thank you!
Antje
Peter Dalgaard schrieb:
Forest Floor wrote:
Hi,
I've seen various posts on this question, but still can't get the code
right.
If I run the following code one line at a time, it works fine. If I run
it together as a block, however, it doesn't wait for the input and gives
an error.
There must be a way to have are pause/wait for an answer, but I can't
seem to find it. Thanks! J
Code:
choosefunction - function(){readline(1. linear, 2. linear with lag, 3.
nonlinear )}
ans - as.integer(choosefunction())
if (ans==1){K2=x1}
if (ans==2){K2=x2 }
if (ans==3){K2=x3 }
ans
Error text:
ans - as.integer(choosefunction())
1. linear, 2. linear with lag, 3. nonlinear if (ans==1) {K2=x1}]}
Warning message:
NAs introduced by coercion
if (ans==2){K2=x2) }
Error in if (ans == 2) { : missing value where TRUE/FALSE needed
if (ans==3){K2=x3}
Error in if (ans == 3) { : missing value where TRUE/FALSE needed
ans
[1] NA
As you may have realized already, the issue is that choosefunction()
takes the next command as its input. Since if (ans==1){K2=x1} isn't an
integer ans becomes NA, and it just goes downhill from there.
An extra set of braces may help
choosefunction - function(){readline(1. linear, 2. linear with lag, 3.
+ nonlinear )}
{ans - as.integer(choosefunction())
+ if (ans==1){K2=x1}
+ if (ans==2){K2=x2 }
+ if (ans==3){K2=x3 }
+ ans}
1. linear, 2. linear with lag, 3.
nonlinear 3
Error: object x3 not found
It still doesn't quite work, but the reason(s) for that should be plain
to see.
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[R] change fold from packageDEDS
Hi all: packageDEDS can find out differentially expressed genes via computing changefold. But if there're 3(or more)groups(1\2\3 for instance),I wanna know the which group vs group the changefold is referd to(1 vs 2/1 vs 3/2 vs 3 for instance)? Thanks a lot! My best! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Prediction accuracy of poisson regression model
Dear all, I'd like to measure the prediction accuracy of a model I have derived from poisson regression. Can somebody help me with suggesting a good approach please? Thanks very much in advance. Best wishes, Des [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Inverse BoxCox transformation
Hi, I can't seem to find a function in R that will reverse a BoxCox transformation. Can somebody help me locate one please? Thanks in advance. Best wishes, Des [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Responding to a posting in the digest
Thanks a lot, Ted, for your comprehensive answer! [See one short note way below: ] TH == Ted Harding [EMAIL PROTECTED] on Thu, 14 Jun 2007 09:54:04 +0100 (BST) writes: TH On 14-Jun-07 07:26:26, Moshe Olshansky wrote: Is there a convenient way to respond to a particular posting which is a part of the digest? I mean something that will automatically quote the original message, subject, etc. Thank you! Moshe Olshansky [EMAIL PROTECTED] TH This will depend on two things. TH 1. Whether the mail software you use has the capability; TH 2. Whether the digest format would permit it anyway. TH Regarding (2), if you are receiving R-help in TH traditional digest format (all the messages, each with TH its principal headers, as one single long message-body), TH then the only way to respond to a particular message is TH to start to compose a new message and copy what you need TH from the digest. TH While I've never reveived R-help in digest format TH myself, according to Martin Maechler: TH http://finzi.psych.upenn.edu/R/Rhelp02a/archive/59429.html TH Please open the URL at the end of every message TH https://stat.ethz.ch/mailman/listinfo/r-help go to the TH bottom and log in -- clicking the [Unsubscribe or Edit TH Options] field. You need your mailing list password TH sooner or later. The one you get sent every 1st of the TH month; or you can have it sent to you again. TH Then you are in a page entitled R-help Membership TH Configuration for foo@bar Scroll down to the section TH Your R-help Subscription where the 3rd entry is TH entitled Get MIME or Plain Text Digests? and now you TH want MIME. TH In MIME digest format, each message with its own main TH headers is a separate MIME attachment, and suitable mail TH software can bring any message up on its own, You can TH then reply in the normal way. TH However (and here is where I'm ignorant as a result of TH never having received R-help as digest), your reply may TH not continue the thread -- since this depends on TH message-identifier headers being present which allow TH threading software to trace which messages are replies TH to which message. The JISCMAIL MIME digest for the TH AllStat mailing list only includes a Message-ID for the TH digest as a whole, i.e. the ID for the entire digest TH message. Message-IDs for the individual messages in the TH digest (as would be seen by people who received them TH singly) are absent: you only get the likes of TH Date: DoW, DD Mon HH:MM:SS TZ From: Sender TH (person who sent the message to the list) Subject: TH Subject of individual message MIME-Version: 1.0 TH Content-Type: text/plain; charset=iso-8859-1 TH Content-Transfer-Encoding: quoted-printable TH and no Message ID for the original message from TH Sender. So any reply to this component message is not TH identifiable as belonging to its thread. TH I don't know whether R-help's 'mailman' provides such TH headers (Martin??). Yes, it does (I've checked with a pseudo-user who receives r-help in digests in MIME format). So you can indeed do the following. In my limited experience, the main problem is the bad quality of people'e e-mail software which does not properly work with the (typically invisible) 'References:' and 'In-Reply-To:' headers which mailman indeed does preserve in its MIME-digests. TH If it does, then your reply could TH include an In-Reply-To: which identifies the TH thread. Otherwise it can't. TH As to (1), you will probably get several suggestions for TH suitable mail software. My own (see below) opens an TH AllStat digest in a window with attachment tags TH displayed, one for Tablf of Contents, one for each TH message. Clicking on one of these opens a new window TH with the message attached to that tag displayed, and now TH the usual reply/forward etc mail sunctions can be TH applied to that message. But it will reply only to the TH address given in the From: header (i.e. the original TH sender, as above), not to the AllStat list (so you have TH to enter that address by hand, if you want to reply to TH the list). TH In principle, mailer software could also identify the TH address of the list from which the digest has been sent, TH as well as the sender of the original message, so you TH could get the option to reply to either or both. But my TH XFMail does not, and only offers the original TH sender. Whether other mailer software can do this is for TH others to comment on! TH Hoping this helps, Ted. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
[R] Question about lmer
Hello,
We have a problem with function lmer. This is our code:
Get_values-function(ff_count, fixed_factors, rf_count, random_factors,
y_values)
{
SA-matrix(as.array(c(fixed_factors, random_factors)), ncol=3)
data-as.data.frame(SA)
y-as.array(y_values)
dd-data.frame(SA)
for(i in 1:(ff_count+rf_count)){
dd[,i]-as.factor(data[,i])
}
fit_full=lmer(y~dd[,1]+dd[,2]+(1|dd[,3]),method=ML)
fit_full
}
A-c(0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1)
B-c(0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1)
C-c(0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1)
Y-c(5,3,4,1,1,2,6,1,5,3,7,1,2,3,1,1,5,3,4,1,1,2,6,1,5,3,7,1,2,3,1,1)
r-Get_values(2, c(A,B),1,c(C),Y)
r
R output:
Error in inherits(x, factor) : object dd not found
Can this function work with random array? Because this code is
working:
D-as.factor(data[,3])
fit_full=lmer(y~dd[,1]+dd[,2]+(1|D),method=ML)
--
Truly yours,
Julia mailto:[EMAIL PROTECTED]
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[R] Unix-like permissions to allow a user to update recommended packages
I installed R from the tar.gz file (as root) in a directory under
/usr/local. The recommended packages are installed in a library in
that directory whereas additional packages I install in a directory
under the /home directory as a user.
Updating the additional packages is very easy with update.packages()
as a non-root user, but the recommended packages cannot be done so
readily because of file permissions.
My question is how do I set the permissions or ownerships in the
/usr/local/R-2.5.0 directory so that everything necessary can be
writable by a user? Should I make a group for R users (total of one
member) or is it simpler than that?
TIA
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] Inverse BoxCox transformation
to backtransform 'estimate':
if (lambda == 0 ) {
log(estimate)
} else {
estimate^(1/lambda)
}
Des Callaghan
[EMAIL PROTECTED]
er.co.uk To
Sent by: r-help@stat.math.ethz.ch
[EMAIL PROTECTED] cc
at.math.ethz.ch
Subject
[R] Inverse BoxCox transformation
18/06/2007 09:32
Hi,
I can't seem to find a function in R that will reverse a BoxCox
transformation. Can somebody help me locate one please? Thanks in advance.
Best wishes,
Des
[[alternative HTML version deleted]]
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Re: [R] Inverse BoxCox transformation
Look at the definition for the transform. For example in the car package, ?box.cox Then do the simple algebraic manipulations yourself. Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Des Callaghan Sent: Monday, June 18, 2007 3:33 AM To: r-help@stat.math.ethz.ch Subject: [R] Inverse BoxCox transformation Hi, I can't seem to find a function in R that will reverse a BoxCox transformation. Can somebody help me locate one please? Thanks in advance. Best wishes, Des [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging dataframes with diffent rownumbers
At 09:09 18/06/2007, Thomas Hoffmann wrote: Dear R-Helpers, I have following problem: I do have two data frames dat1 and dat2 with a commen column BNUM (long integer). dat1 has a larger number of BNUM than dat2 and different rows of dat2 have equal BNUM. The numbers of rows in dat1 and dat2 is not equal. I applied the tapply-function to dat2 with BNUM as index. I would like to add the columns from dat1 to the results of b.sum - tapply(dat2, BNUM, sum). However the BNUM of b.sum are only a subset of the dat1. Does anybody knows a elegant way to solve the problem? If I understand you correctly ?merge should help you here Thanks in advance Thomas H. Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging dataframes with diffent rownumbers
No easy to understand what exactly you mean, but try ?merge ?cbind ?rbind Petr Thomas Hoffmann napsal(a): Dear R-Helpers, I have following problem: I do have two data frames dat1 and dat2 with a commen column BNUM (long integer). dat1 has a larger number of BNUM than dat2 and different rows of dat2 have equal BNUM. The numbers of rows in dat1 and dat2 is not equal. I applied the tapply-function to dat2 with BNUM as index. I would like to add the columns from dat1 to the results of b.sum - tapply(dat2, BNUM, sum). However the BNUM of b.sum are only a subset of the dat1. Does anybody knows a elegant way to solve the problem? Thanks in advance Thomas H. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Petr Klasterecky Dept. of Probability and Statistics Charles University in Prague Czech Republic __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unix-like permissions to allow a user to update recommen
On 18-Jun-07 10:11:43, Patrick Connolly wrote: I installed R from the tar.gz file (as root) in a directory under /usr/local. The recommended packages are installed in a library in that directory whereas additional packages I install in a directory under the /home directory as a user. Updating the additional packages is very easy with update.packages() as a non-root user, but the recommended packages cannot be done so readily because of file permissions. My question is how do I set the permissions or ownerships in the /usr/local/R-2.5.0 directory so that everything necessary can be writable by a user? Should I make a group for R users (total of one member) or is it simpler than that? Since you have root access, do you need to segregate the additional packages to a particular user? Though I don't run R as root for general use, I always install/update by running R CMD as root. This makes all of R (recommended and also any extras) available system-wide, and no pemission problems arise. This of course does not stop you from setting up a special .Rprofile for each user, since this by definition lives in the user's home directory. Does this help? Or are there issues you haven't mentioned which make such an approach not feasible? Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 18-Jun-07 Time: 11:53:19 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with binding data-frames
Junnila, Jouni napsal(a): Hello, I'm having a problem concerning r-binding datasets. I have six datasets, from six different plates, and two different days. I want to combine these datasets together for analysis. Datasets from day 2, have all the same columns than datasets from day 1. However in addition, there are few columns more in day 2. Thus, using rbind for this, results a error, because the objects are not the same length. Error in paste(nmi[nii == 0L], collapse = , ) : object nii not found In addition: Warning message: longer object length is not a multiple of shorter object length in: clabs == nmi Hi, 1. the error has nothing to do with differing lengths of your objects - that's what the following warning is about. The error occured because your indexing object 'nii' does not exist where R is looking for it. 2. using rbind on dataframes is a bad practice, since the input is converted to marices if possible. Use merge() instead. Petr What I need, is to combine all the six together, and give for example NA-value in day 1, for those columns which can only be found in day 2. Is this somehow possible? I have several of these six-datasets groups, and only few of them are having this problem described above, and I cannot know in advance which. With most of the groups writing rbind(data1,data2,data3,data4,data5,data6) works easily, but these few problematic groups need also to be combined... Any help greatly appreciated! -Jouni __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Petr Klasterecky Dept. of Probability and Statistics Charles University in Prague Czech Republic __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] what about options in BATCH
?BATCH Read second paragraph of Details section. On 6/18/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote: hello, when I run a calcul in BATCH the screen displays all the code of my programs and even the introduction of R how can I do to don't display it? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Christophe Pallier (http://www.pallier.org) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Readline
Hello,
I also have problems to get to run the following lines. If I run the
block instead of every single line, it simply does not wait for the input.
Can anybody help me?
pos_name - readline(Please type: )
r - substr(pos_name, 1,1)
c - substr(pos_name, 2,nchar(pos_name))
Thank you!
Antje
Peter Dalgaard schrieb:
Forest Floor wrote:
Hi,
I've seen various posts on this question, but still can't get the
code
right.
If I run the following code one line at a time, it works fine.
If I run
it together as a block, however, it doesn't wait for the input
and gives
an error.
There must be a way to have are pause/wait for an answer, but I
can't
seem to find it. Thanks! J
Code:
choosefunction - function(){readline(1. linear, 2. linear with
lag, 3.
nonlinear )}
ans - as.integer(choosefunction())
if (ans==1){K2=x1}
if (ans==2){K2=x2 }
if (ans==3){K2=x3 }
ans
Error text:
ans - as.integer(choosefunction())
1. linear, 2. linear with lag, 3. nonlinear if (ans==1) {K2=x1}]}
Warning message:
NAs introduced by coercion
if (ans==2){K2=x2) }
Error in if (ans == 2) { : missing value where TRUE/FALSE needed
if (ans==3){K2=x3}
Error in if (ans == 3) { : missing value where TRUE/FALSE needed
ans
[1] NA
As you may have realized already, the issue is that choosefunction()
takes the next command as its input. Since if (ans==1){K2=x1}
isn't an
integer ans becomes NA, and it just goes downhill from there.
An extra set of braces may help
choosefunction - function(){readline(1. linear, 2. linear with
lag, 3.
+ nonlinear )}
{ans - as.integer(choosefunction())
+ if (ans==1){K2=x1}
+ if (ans==2){K2=x2 }
+ if (ans==3){K2=x3 }
+ ans}
1. linear, 2. linear with lag, 3.
nonlinear 3
Error: object x3 not found
It still doesn't quite work, but the reason(s) for that should be plain
to see.
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Re: [R] Problem with binding data-frames
On Mon, 18 Jun 2007, Petr Klasterecky wrote: Junnila, Jouni napsal(a): Hello, I'm having a problem concerning r-binding datasets. I have six datasets, from six different plates, and two different days. I want to combine these datasets together for analysis. Datasets from day 2, have all the same columns than datasets from day 1. However in addition, there are few columns more in day 2. Thus, using rbind for this, results a error, because the objects are not the same length. Error in paste(nmi[nii == 0L], collapse = , ) : object nii not found In addition: Warning message: longer object length is not a multiple of shorter object length in: clabs == nmi Hi, 1. the error has nothing to do with differing lengths of your objects - that's what the following warning is about. The error occured because your indexing object 'nii' does not exist where R is looking for it. It's because the dataframes have differing number of columns, and that has not been allowed for in the error message in that version of R. 2. using rbind on dataframes is a bad practice, since the input is converted to marices if possible. Use merge() instead. Not so: rbind on data frames does no such conversion, and is not problematic provided they have the same column names (and hence the same number of columns). You may have missed in ?rbind The functions 'cbind' and 'rbind' are S3 generic, with methods for data frames. The data frame method will be used if at least one argument is a data frame and the rest are vectors or matrices. ... and a later description of the data frame method for 'rbind'. Petr What I need, is to combine all the six together, and give for example NA-value in day 1, for those columns which can only be found in day 2. Is this somehow possible? I have several of these six-datasets groups, and only few of them are having this problem described above, and I cannot know in advance which. With most of the groups writing rbind(data1,data2,data3,data4,data5,data6) works easily, but these few problematic groups need also to be combined... Any help greatly appreciated! -Jouni __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about lmer
On 6/18/07, Julia Proudnikova [EMAIL PROTECTED] wrote:
Hello,
We have a problem with function lmer. This is our code:
Get_values-function(ff_count, fixed_factors, rf_count, random_factors,
y_values)
{
SA-matrix(as.array(c(fixed_factors, random_factors)), ncol=3)
data-as.data.frame(SA)
y-as.array(y_values)
dd-data.frame(SA)
for(i in 1:(ff_count+rf_count)){
dd[,i]-as.factor(data[,i])
}
fit_full=lmer(y~dd[,1]+dd[,2]+(1|dd[,3]),method=ML)
fit_full
}
A-c(0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1)
B-c(0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1)
C-c(0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1)
Y-c(5,3,4,1,1,2,6,1,5,3,7,1,2,3,1,1,5,3,4,1,1,2,6,1,5,3,7,1,2,3,1,1)
r-Get_values(2, c(A,B),1,c(C),Y)
r
R output:
Error in inherits(x, factor) : object dd not found
Can this function work with random array? Because this code is
working:
The full explanation of why lmer fails to find dd has to do with the
way names are resolved in a call to model.frame. However, there may
be a way to solve your problem by redesigning your function so you
don't need to worry about what model.frame does.
Why not pass the data as a data frame and pass the names of the fixed
factors, random factors and response variable as character strings?
Your current design of creating a matrix, then converting it to a data
frame then converting numeric variables back to factors is a bit
convoluted.
If you knew that you were only going to have one random factor you
could generate the formula as
substitute(y ~ ff + (1|rf), list(y = as.name(y_name), ff =
parse(paste(ff_names, collapse = +)), rf = as.name(rf_name))
It gets a bit trickier with multiple random factors.
Having said all this, it does appear that the call to model.frame
inside lmer is getting the wrong environment from the formula and I
will correct that.
If you need more detail about the redesign I am suggesting, feel free
to contact me off-list.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Automatic paren/bracket closing in 2.5.0?
On 18/06/2007 12:30 AM, Adam D. I. Kramer wrote: Hello, Just upgraded to 2.5.0, and found that R now includes an rparen (right parentheses) or rbracket whenever I enter in an lparen. While I can see the use of this function, it doesn't mesh well with my personal style of using R (e.g., using the up arrow, adding an rparen, jumping to the beginning of the line, and then wrapping a summary, for instance). Some 10 minutes of google searching has failed to come up with a solution for turning this feature off--any suggestions from the list? You don't say your OS. If it's MacOSX (which I think is the only platform with this feature), then see the R-sig-mac list, and in particular Simon Urbanek's posting on May 23: On May 23, 2007, at 3:55 PM, Roberto Osorio wrote: I can't find a preference to disable brace completion in the console in R 2.5.0 GUI 1.19. Unfortunately it didn't make it to the Preferences UI, so you have to paste this in Terminal: defaults write org.R-project.R auto.close.parens NO If you want to revert back to the default you can use: defaults delete org.R-project.R auto.close.parens Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Readline
Antje wrote: Hello, I also have problems to get to run the following lines. If I run the block instead of every single line, it simply does not wait for the input. Can anybody help me? pos_name - readline(Please type: ) r - substr(pos_name, 1,1) c - substr(pos_name, 2,nchar(pos_name)) Hi Antje, What you seem to be doing is pasting the lines into the R window. The second line looks like the input line that the first line is expecting, and if you remove the empty line, you will see that pos_name has been assigned the text of the second line. If you put these lines into a text file and then use source to read it, it works as I think you expect. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] triangle contour plots
Suppose I have three numbers p1, p2, p3 with 0 = p1,p2,p3 = 1 and p1+p2+p3=1, and a function f=f(p1,p2,p3) = f(p1,p2,1-p1-p2). How to draw a contour plot of f() on the p1+p2+p3=1 plane, that is, an equilateral triangle? Functions triplot(), triangle.plot(), and ternaryplot() give only scatterplots, AFAICS -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] BATCH
Hello, I run some programs R from BATCH using this syntax R CMD BATCH options(echo = FALSE) C:\R\copie.r C:\PHP\sortie.r but the options doesn't work do you know why? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Inverse BoxCox transformation
Dear Des, The following should do the trick: invBoxCox - function(x, lambda) if (lambda == 0) exp(x) else (lambda*x + 1)^(1/lambda) I hope this helps, John John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Des Callaghan Sent: Monday, June 18, 2007 3:33 AM To: r-help@stat.math.ethz.ch Subject: [R] Inverse BoxCox transformation Hi, I can't seem to find a function in R that will reverse a BoxCox transformation. Can somebody help me locate one please? Thanks in advance. Best wishes, Des [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] BATCH
?BATCH By default, the input commands are printed along with the output. To suppress this behavior, add 'options(echo = FALSE)' at the beginning of 'infile'. at the begining of 'infile' means that 'options(echo = FALSE)' must be included inside 'infile'( in your case copie.r), as a first line. Not before 'infile' on the command line. Btw, the help does not mention that you can put several script files on the command line. I am not sure if it works. Christophe On 6/18/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote: Hello, I run some programs R from BATCH using this syntax R CMD BATCH options(echo = FALSE) C:\R\copie.r C:\PHP\sortie.r but the options doesn't work do you know why? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Christophe Pallier (http://www.pallier.org) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data.frame
hello, are there functions giving the columns number and the rows number of a matrix? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame
See help(dim) and please read the manuals before asking basic questions like this. Thank you. elyakhlifi mustapha wrote: hello, are there functions giving the columns number and the rows number of a matrix? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame
On 6/18/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote: hello, are there functions giving the columns number and the rows number of a matrix? Yes, there are. Are you trying to use R without reading *any* documentation??? The mailing list is not a substitute for the manuals. See the Posting guide. Christophe thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Christophe Pallier (http://www.pallier.org) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame
On 18 Jun 2007, at 14:16, Adaikalavan Ramasamy wrote: See help(dim) and please read the manuals before asking basic questions like this. Thank you. I think the questioner was looking for row() and col(), which (IMO) are difficult to find if you don't know of their existence. [as indeed are slice.index() or arow() for the array case] Robin elyakhlifi mustapha wrote: hello, are there functions giving the columns number and the rows number of a matrix? thanks. _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data.frame
hello,
I'm trying to write a function which take a matrix and give a dataframe with
column names and row names but the problem I meet it's that the column number
is changing and the vector containing the column names is also changing how can
I do to write a good progam for the moment I tryied like follow:
dm - ncol(M)
v - vector()
t - 1
while (dm 0) {
v - c(v,paste(Rép,t,sep=))
t - t + 1
dm - dm - 1
}
nv - noquote(v)
df - function (M,x) {
return(data.frame(nv[1] = M[,1], nv[2] = M[,2],nv[3] = M[,3], row.names =
var[[1]], check.rows = TRUE, check.names = TRUE))
}
I know that there are errors but the important is that R doesn't recognize nv.
For more precision the martix M is like follow:
M
[,1] [,2] [,3]
[1,] 6.52 NA 6.59
[2,] 6.99 6.85 6.38
[3,] 6.92 6.72 6.99
[4,] 6.59 5.51 6.45
[5,] 6.65 7.12 6.99
[6,] 6.18 5.71 5.78
[7,] 6.65 6.52 6.72
[8,] 6.65 6.79 6.12
[9,] 6.59 6.65 6.32
[10,] 5.85 6.05 6.38
[11,] 6.38 6.79 6.65
[12,] 6.79 6.52 6.72
[13,] 6.12 6.25 6.38
[14,] 6.99 6.72 6.38
[15,] 6.59 6.65 6.99
[16,] 6.45 6.18 6.59
[17,] 5.65 6.05 6.52
[18,] 6.52 6.85 6.65
[19,] 6.18 6.32 6.32
[20,] 6.99 6.65 6.72
[21,] 6.52 6.99 6.32
Can you help me?
thanks.
_
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[[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.
[R] how to obtain the OR and 95%CI with 1 SD change of a continue variable
Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with 1 SD change of waist circumference.how? Any default package(s) or options in glm available now? if not, how to calculate them by hand? many thanks. yours,sincerely, Xingwang Ye __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] triangle contour plots
On 6/18/2007 8:12 AM, Robin Hankin wrote: Suppose I have three numbers p1, p2, p3 with 0 = p1,p2,p3 = 1 and p1+p2+p3=1, and a function f=f(p1,p2,p3) = f(p1,p2,1-p1-p2). How to draw a contour plot of f() on the p1+p2+p3=1 plane, that is, an equilateral triangle? The usual contour function leaves blanks where you give it NA values, so you could put the f values into a rectangular array with NA outside the triangle and use that. I don't know how you're thinking of displaying things, but one possible transformation from (x,y) to (p1, p2, p3) would be f - function(p1, p2, p3) p3 # just to illustrate maxy - sin(pi/3) x - seq(0,1,len=100) y - seq(0, maxy, len=100) p1 - outer(x,y, function(x,y) x - y/maxy/2) p2 - outer(x,y, function(x,y) y/maxy) p3 - 1-p1-p2 z - ifelse(0 p1 0 p3, f(p1,p2,p3), NA) contour(x,y,z) This puts p1==1 at the bottom right, p2==1 at the top, and p3==1 at the bottom left. Duncan Murdoch Functions triplot(), triangle.plot(), and ternaryplot() give only scatterplots, AFAICS -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame
On 6/18/07, Robin Hankin [EMAIL PROTECTED] wrote: I think the questioner was looking for row() and col(), which (IMO) are difficult to find if you don't know of their existence. Searching for R matrix number of rows columns in google returns, in fourth position, the manual page for 'nrow'. -- Christophe Pallier (http://www.pallier.org) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame
If M is the original matrix,
k - as.data.frame(M)
names(k) - paste(Rép,1:ncol(k),sep=)
rownames(k) - paste(Col,1:nrow(k),sep=) # replace by what you want
k
On 6/18/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote:
hello,
I'm trying to write a function which take a matrix and give a dataframe
with column names and row names but the problem I meet it's that the column
number is changing and the vector containing the column names is also
changing how can I do to write a good progam for the moment I tryied like
follow:
dm - ncol(M)
v - vector()
t - 1
while (dm 0) {
v - c(v,paste(Rép,t,sep=))
t - t + 1
dm - dm - 1
}
nv - noquote(v)
df - function (M,x) {
return(data.frame(nv[1] = M[,1], nv[2] = M[,2],nv[3] = M[,3], row.names =
var[[1]], check.rows = TRUE, check.names = TRUE))
}
I know that there are errors but the important is that R doesn't recognize
nv.
For more precision the martix M is like follow:
M
[,1] [,2] [,3]
[1,] 6.52 NA 6.59
[2,] 6.99 6.85 6.38
[3,] 6.92 6.72 6.99
[4,] 6.59 5.51 6.45
[5,] 6.65 7.12 6.99
[6,] 6.18 5.71 5.78
[7,] 6.65 6.52 6.72
[8,] 6.65 6.79 6.12
[9,] 6.59 6.65 6.32
[10,] 5.85 6.05 6.38
[11,] 6.38 6.79 6.65
[12,] 6.79 6.52 6.72
[13,] 6.12 6.25 6.38
[14,] 6.99 6.72 6.38
[15,] 6.59 6.65 6.99
[16,] 6.45 6.18 6.59
[17,] 5.65 6.05 6.52
[18,] 6.52 6.85 6.65
[19,] 6.18 6.32 6.32
[20,] 6.99 6.65 6.72
[21,] 6.52 6.99 6.32
Can you help me?
thanks.
_
Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo!
Mail
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Christophe Pallier (http://www.pallier.org)
[[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.
[R] source a specific function
Dear Listers: For example, if I have a .R source file which has more than one function, and I want to just load only one of the functions, how could I do that? (removing the rest after sourcing is not what I intend b/c in my workspace, I might have some of the rest and I don't want to change my workspace: i.e., I only change my workspace by adding one function from a R source file). Thanks, -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: data.frame
Hi
Why scratching your left ear with your right hand?
If M is numeric matrix
d.m - data.frame(M)
names(d.m) - paste(Rep, 1:dim(M)[2], sep=)
not sure what you want as row names as var is not defined anywhere in your
code, but you can use the same principle for changing row names. Jus use
row.names(d.m) - whatever vector of names you can elaborate
Regards
Petr Pikal
[EMAIL PROTECTED]
BTW. R is not C and you shall use strong features of R not to try to avoid
them.
[EMAIL PROTECTED] napsal dne 18.06.2007 15:40:38:
hello,
I'm trying to write a function which take a matrix and give a dataframe
with
column names and row names but the problem I meet it's that the column
number
is changing and the vector containing the column names is also changing
how
can I do to write a good progam for the moment I tryied like follow:
dm - ncol(M)
v - vector()
t - 1
while (dm 0) {
v - c(v,paste(Rép,t,sep=))
t - t + 1
dm - dm - 1
}
nv - noquote(v)
df - function (M,x) {
return(data.frame(nv[1] = M[,1], nv[2] = M[,2],nv[3] = M[,3], row.names
=
var[[1]], check.rows = TRUE, check.names = TRUE))
}
I know that there are errors but the important is that R doesn't
recognize nv.
For more precision the martix M is like follow:
M
[,1] [,2] [,3]
[1,] 6.52 NA 6.59
[2,] 6.99 6.85 6.38
[3,] 6.92 6.72 6.99
[4,] 6.59 5.51 6.45
[5,] 6.65 7.12 6.99
[6,] 6.18 5.71 5.78
[7,] 6.65 6.52 6.72
[8,] 6.65 6.79 6.12
[9,] 6.59 6.65 6.32
[10,] 5.85 6.05 6.38
[11,] 6.38 6.79 6.65
[12,] 6.79 6.52 6.72
[13,] 6.12 6.25 6.38
[14,] 6.99 6.72 6.38
[15,] 6.59 6.65 6.99
[16,] 6.45 6.18 6.59
[17,] 5.65 6.05 6.52
[18,] 6.52 6.85 6.65
[19,] 6.18 6.32 6.32
[20,] 6.99 6.65 6.72
[21,] 6.52 6.99 6.32
Can you help me?
thanks.
_
Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo!
Mail
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame
Is this close to what you want? 'var' was not defined for row.names.
M - matrix(scan(textConnection(x)), ncol=3, byrow=TRUE)
Read 63 items
dm - ncol(M)
v - vector()
t - 1
while (dm 0) {
+ v - c(v,paste(Rép,t,sep=))
+ t - t + 1
+ dm - dm - 1
+ }
df - as.data.frame(M)
colnames(df) - v
df
Rép1 Rép2 Rép3
1 6.52 NA 6.59
2 6.99 6.85 6.38
3 6.92 6.72 6.99
4 6.59 5.51 6.45
5 6.65 7.12 6.99
6 6.18 5.71 5.78
7 6.65 6.52 6.72
8 6.65 6.79 6.12
9 6.59 6.65 6.32
10 5.85 6.05 6.38
11 6.38 6.79 6.65
12 6.79 6.52 6.72
13 6.12 6.25 6.38
14 6.99 6.72 6.38
15 6.59 6.65 6.99
16 6.45 6.18 6.59
17 5.65 6.05 6.52
18 6.52 6.85 6.65
19 6.18 6.32 6.32
20 6.99 6.65 6.72
21 6.52 6.99 6.32
On 6/18/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote:
hello,
I'm trying to write a function which take a matrix and give a dataframe
with column names and row names but the problem I meet it's that the column
number is changing and the vector containing the column names is also
changing how can I do to write a good progam for the moment I tryied like
follow:
dm - ncol(M)
v - vector()
t - 1
while (dm 0) {
v - c(v,paste(Rép,t,sep=))
t - t + 1
dm - dm - 1
}
nv - noquote(v)
df - function (M,x) {
return(data.frame(nv[1] = M[,1], nv[2] = M[,2],nv[3] = M[,3], row.names =
var[[1]], check.rows = TRUE, check.names = TRUE))
}
I know that there are errors but the important is that R doesn't recognize
nv.
For more precision the martix M is like follow:
M
[,1] [,2] [,3]
[1,] 6.52 NA 6.59
[2,] 6.99 6.85 6.38
[3,] 6.92 6.72 6.99
[4,] 6.59 5.51 6.45
[5,] 6.65 7.12 6.99
[6,] 6.18 5.71 5.78
[7,] 6.65 6.52 6.72
[8,] 6.65 6.79 6.12
[9,] 6.59 6.65 6.32
[10,] 5.85 6.05 6.38
[11,] 6.38 6.79 6.65
[12,] 6.79 6.52 6.72
[13,] 6.12 6.25 6.38
[14,] 6.99 6.72 6.38
[15,] 6.59 6.65 6.99
[16,] 6.45 6.18 6.59
[17,] 5.65 6.05 6.52
[18,] 6.52 6.85 6.65
[19,] 6.18 6.32 6.32
[20,] 6.99 6.65 6.72
[21,] 6.52 6.99 6.32
Can you help me?
thanks.
_
Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo!
Mail
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
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?
[[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.
Re: [R] source a specific function
One way to do it would be to surround the function(s) you don't want
sourced, like this:
if (FALSE) {
## function definition here
}
But you might find it easier to just put each function in its own file.
At 9:54 AM -0400 6/18/07, Weiwei Shi wrote:
Dear Listers:
For example, if I have a .R source file which has more than one
function, and I want to just load only one of the functions, how could
I do that? (removing the rest after sourcing is not what I intend b/c
in my workspace, I might have some of the rest and I don't want to
change my workspace: i.e., I only change my workspace by adding one
function from a R source file).
Thanks,
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] source a specific function
This loads all the functions into an anonymous environment defined
by local and then exports f to the global environment.
f - local({
source(/a.R, local = TRUE)
environment(f) - .GlobalEnv
f
})
On 6/18/07, Weiwei Shi [EMAIL PROTECTED] wrote:
Dear Listers:
For example, if I have a .R source file which has more than one
function, and I want to just load only one of the functions, how could
I do that? (removing the rest after sourcing is not what I intend b/c
in my workspace, I might have some of the rest and I don't want to
change my workspace: i.e., I only change my workspace by adding one
function from a R source file).
Thanks,
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] source a specific function
On 6/18/2007 9:54 AM, Weiwei Shi wrote: Dear Listers: For example, if I have a .R source file which has more than one function, and I want to just load only one of the functions, how could I do that? (removing the rest after sourcing is not what I intend b/c in my workspace, I might have some of the rest and I don't want to change my workspace: i.e., I only change my workspace by adding one function from a R source file). In Windows, open the file in an editor, copy (e.g. by highlighting it and hitting Ctrl-C) the part you want to source to the clipboard, and then in R enter source(clipboard), or just paste the selected text. I think source(clipboard) is Windows-specific, but other platforms support copy and paste in their own ways. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] triangle contour plots
The triplot function in the TeachingDemos package (I don't know about the one in klaR, or the others mentioned) honors the type='l' argument and passes it on to points. So if you know where you want the contours drawn, you can use triplot to draw the lines (it also has an add argument that could be used to add labels after plotting the lines). You can also look at the source code to see how the plotting is done and modify it to do the plot you are interested in. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Robin Hankin Sent: Monday, June 18, 2007 6:12 AM To: R program Subject: [R] triangle contour plots Suppose I have three numbers p1, p2, p3 with 0 = p1,p2,p3 = 1 and p1+p2+p3=1, and a function f=f(p1,p2,p3) = f(p1,p2,1-p1-p2). How to draw a contour plot of f() on the p1+p2+p3=1 plane, that is, an equilateral triangle? Functions triplot(), triangle.plot(), and ternaryplot() give only scatterplots, AFAICS -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to obtain the OR and 95%CI with 1 SD change of a continue variable
felix wrote: Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with 1 SD change of waist circumference.how? Any default package(s) or options in glm available now? if not, how to calculate them by hand? Unless you want to do something advanced like factoring in the sampling error of the SD (I don't think anyone bothers with that), probably the easiest way is to scale() the predictor and look at the relevant line of exp(confint(glm(.))). As in (library(MASS); example(confint.glm)) budworm.lg0 - glm(SF ~ sex + scale(ldose), family = binomial) exp(confint(budworm.lg0)) Waiting for profiling to be done... 2.5 % 97.5 % (Intercept) 0.2652665 0.7203169 sexM 1.5208018 6.1747207 scale(ldose) 4.3399952 10.8983903 Or, if you insist on getting asymptotic Wald-statistic based intervals: exp(confint.default(budworm.lg0)) 2.5 % 97.5 % (Intercept) 0.269864 0.7294944 sexM 1.496808 6.0384756 scale(ldose) 4.220890 10.5546837 You can also get it from the coefficients of the unscaled analysis, like in budworm.lg0 - glm(SF ~ sex + ldose, family = binomial) confint(budworm.lg0) Waiting for profiling to be done... 2.5 %97.5 % (Intercept) -4.4582430 -2.613736 sexM 0.4192377 1.820464 ldose0.8229072 1.339086 exp(confint(budworm.lg0)[3,]*sd(ldose)) Waiting for profiling to be done... 2.5 %97.5 % 4.339995 10.898390 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] discriminant analysis with lda(MASS)
I use Widows, R version 2.4.1
I have 4 questions on lda (MASS) (code is pasted below):
1st. How can I obtain the statistics and p-value associated with
discriminant analysis? Am I supposed to calculate that manually by squaring
the svd value and looking the p value up in a table? I am writing the
following code:
training.mx-read.table('C:\\Documents and Settings\\silvia\\My
Documents\\silvia\\paper Martin\\trainingAndvalidation.txt', header=T)
train - sample (1:148) ##in a file with 399 cases, I am using the first 148
as a training set
table(training.mx$disperser[train])
training.df - lda (disperser~., training.mx, subset=train)
predict (training.df, training.mx[-train,])$class
2nd. How can I get the scores for each species on the discriminant
functions? I only get the scores for the group means, but I need the values
for all species.
3rd. Is it possible to obtain confidence intervals for my groups?
4th. (this is part of a previous posting but it's related to all my previous
questions so here it goes again) When I try to plot the resulting
discriminant functions following the example I found in the help, I get an
error saying that 'panel.lda' doesn't exist. Am I supposed to create it?
Here is the code for the plot:
plot(x, panel = panel.lda, cex = 0.7, dimen=1,
xlab = LD1, ylab = LD2)
Help on any or all of these questions will be greatly appreciated!
Silvia.
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[R] Optimization
Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01,
x1 is the quantile of normal distribution (0.0032,x) with probability of
0.7, and the changing value should be x. Initial value for x is 0.0207. I am
using the following codes, but it does not work.
fr - function(x) {
x1-qnorm(0.7,0.0032,x)
x2=0.01
x1-x2
}
xsd - optim(0.0207, fr, NULL,method=BFGS)
It is the first time I am trying to use optimization. Could anyone give me
some advice?
--
View this message in context:
http://www.nabble.com/Optimization-tf3941212.html#a11178663
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Re: [R] source a specific function
On 18-Jun-07 14:28:35, Gabor Grothendieck wrote:
This loads all the functions into an anonymous environment defined
by local and then exports f to the global environment.
f - local({
source(/a.R, local = TRUE)
environment(f) - .GlobalEnv
f
})
That looks neat! Two questions:
1. Would something similar work for extracting selected functions
from a library (assuming that you know about interdependencies)?
E.g. something like
f - local({
library(f.etc.lib)
environment(f) - .GlobalEnv
f
})
2. Having done what you describe to extract just f from a source
file, can one then delete the local environment used to load
the source? I think what I'm basically asking is whether the
exporting is done by value (local environment deletion OK)
or by reference (deletion would destroy the exported object).
Apologies, but for instance ?local is a bit too deep for me!
The underlying agenda behind these queries is the saving of
memory space.
With theanks,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 18-Jun-07 Time: 17:11:15
-- XFMail --
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[R] Large Binary file reader for Simple minds
I'm more like a caveman when it comes to programming tools. So, with that in mind, is there a way to use readBin in a batch format to read in pieces of a large binary file? Thank you for the consideration of my question. Todd Remund __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] source a specific function
1. You can do this:
library(plotrix)
environment(draw.arc) - .GlobalEnv # implicitly copies it
detach()
plot(1,1)
draw.arc(1, 1, .1) # its there
2. Since the local environment we created in the prior post was anonymous
and since there are no other references to it either I assume it gets deleted
on the next garbage collection automatically.
On 6/18/07, Ted Harding [EMAIL PROTECTED] wrote:
On 18-Jun-07 14:28:35, Gabor Grothendieck wrote:
This loads all the functions into an anonymous environment defined
by local and then exports f to the global environment.
f - local({
source(/a.R, local = TRUE)
environment(f) - .GlobalEnv
f
})
That looks neat! Two questions:
1. Would something similar work for extracting selected functions
from a library (assuming that you know about interdependencies)?
E.g. something like
f - local({
library(f.etc.lib)
environment(f) - .GlobalEnv
f
})
2. Having done what you describe to extract just f from a source
file, can one then delete the local environment used to load
the source? I think what I'm basically asking is whether the
exporting is done by value (local environment deletion OK)
or by reference (deletion would destroy the exported object).
Apologies, but for instance ?local is a bit too deep for me!
The underlying agenda behind these queries is the saving of
memory space.
With theanks,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 18-Jun-07 Time: 17:11:15
-- XFMail --
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Re: [R] Optimization
You don't need optimization for the solution to your problem. You
just need an understanding of the meaning of qnorm() and some simple algebra.
Try: x- (0.01-0.0032)/qnorm(0.7,0,1)
At 12:01 PM 6/18/2007, you wrote:
Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01,
x1 is the quantile of normal distribution (0.0032,x) with probability of
0.7, and the changing value should be x. Initial value for x is 0.0207. I am
using the following codes, but it does not work.
fr - function(x) {
x1-qnorm(0.7,0.0032,x)
x2=0.01
x1-x2
}
xsd - optim(0.0207, fr, NULL,method=BFGS)
It is the first time I am trying to use optimization. Could anyone give me
some advice?
--
View this message in context:
http://www.nabble.com/Optimization-tf3941212.html#a11178663
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and provide commented, minimal, self-contained, reproducible code.
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED]
Least Cost Formulations, Ltd.URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239Fax: 757-467-2947
Vere scire est per causas scire
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Re: [R] Optimization
livia wrote:
Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01,
x1 is the quantile of normal distribution (0.0032,x) with probability of
0.7, and the changing value should be x. Initial value for x is 0.0207. I am
using the following codes, but it does not work.
fr - function(x) {
x1-qnorm(0.7,0.0032,x)
x2=0.01
x1-x2
}
xsd - optim(0.0207, fr, NULL,method=BFGS)
I guess you want to use optimize() and change the last line of fr to
(x1-x2)^2 as in:
fr - function(x) {
x1 - qnorm(0.7, 0.0032, x)
x2 - 0.01
(x1-x2)^2
}
optimize(fr, c(-5, 5))
Uwe Ligges
It is the first time I am trying to use optimization. Could anyone give me
some advice?
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Re: [R] Optimization
On 18-Jun-07 16:01:03, livia wrote: Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01, x1 is the quantile of normal distribution (0.0032,x) with probability of 0.7, and the changing value should be x. Initial value for x is 0.0207. I'm a bit puzzled by the question. If I understand it right, we can ignore x2 (since it is a fixed value) and simply consider minimising x1 (instead of x1-x2). Then, denoting by P(u) the cumulative normal distribution function for mean=0 and variance=1 (i.e. in R: pnorm(u,0,1)), and by Q(p) its inverse, corresponding to qnorm(p,0,1), we have (again if I have understood right): P((x1 - 0.0032)/x) = 0.7 so x1 = 0.0032 + x*Q(0.7) and therefore, since Q(0.7) 0 and x must be positive, the value of x1 can be made as close to 0.032 as you please (but greater than 0.032) by taking x small enough. Hence there is no strictly minimising value of x, but the greatest lower bound of all possible values of x1 is 0.032. Then you can subtract x2. The fact that there is no positive value of x which gives this bound as the value probably explains the failure of your optim() attempt. Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 18-Jun-07 Time: 17:46:01 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with RSVGTipsDevice
The new version of RSVGTipsDevice (0.7.1) that is now available on CRAN
should fix this problem. Please let me know if it doesn't, or if there
are other problems.
-- Tony Plate
mister_bluesman wrote:
Hi there.
I am still trying to get the RSVGTipsDevice to work, yet I can not.
I have copied the first example from RSVGTipsDevice documentation:
library(RSVGTipsDevice)
devSVGTips(C:\\svgplot1.svg, toolTipMode=1,
title=SVG example plot 1: shapes and points, tooltips are title + 1 line)
plot(c(0,10),c(0,10), type=n, xlab=x, ylab=y,
main=Example SVG plot with title + 1 line tips (mode=1))
setSVGShapeToolTip(title=A rectangle, desc=that is yellow)
rect(1,1,4,6, col='yellow')
setSVGShapeToolTip(title=1st circle with title only)
points(5.5,7.5,cex=20,pch=19,col='red')
setSVGShapeToolTip(title=A triangle, desc=big and green)
polygon(c(3,6,8), c(3,6,3), col='green')
# no tooltips on these points
points(2:8, 8:2, cex=3, pch=19, col='black')
# tooltips on each these points
invisible(sapply(1:7, function(x)
{setSVGShapeToolTip(title=paste(point, x))
points(x+1, 8-x, cex=3, pch=1, col='black')}))
dev.off()
This results in the following output:
http://www.nabble.com/file/p11064573/svgplot1.svg svgplot1.svg
It opens but when I try and hover over the triangle, for example, I do not
get a topptip box appear. I have tried opening the file though firefox, and
XP IE - and on more than one computer yet it does not work. Do I need to
install something else as well?
Many thanks
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Re: [R] Automatic paren/bracket closing in 2.5.0?
Many thanks, Duncan. I did not expect this to be an OS-specific issue, and figured it must be the new default and thus configurable from within the program, though indeed I am using MacOS. --Adam On Mon, 18 Jun 2007, Duncan Murdoch wrote: On 18/06/2007 12:30 AM, Adam D. I. Kramer wrote: Hello, Just upgraded to 2.5.0, and found that R now includes an rparen (right parentheses) or rbracket whenever I enter in an lparen. While I can see the use of this function, it doesn't mesh well with my personal style of using R (e.g., using the up arrow, adding an rparen, jumping to the beginning of the line, and then wrapping a summary, for instance). Some 10 minutes of google searching has failed to come up with a solution for turning this feature off--any suggestions from the list? You don't say your OS. If it's MacOSX (which I think is the only platform with this feature), then see the R-sig-mac list, and in particular Simon Urbanek's posting on May 23: On May 23, 2007, at 3:55 PM, Roberto Osorio wrote: I can't find a preference to disable brace completion in the console in R 2.5.0 GUI 1.19. Unfortunately it didn't make it to the Preferences UI, so you have to paste this in Terminal: defaults write org.R-project.R auto.close.parens NO If you want to revert back to the default you can use: defaults delete org.R-project.R auto.close.parens Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Optimization
From the help page:
Note:
'optim' will work with one-dimensional 'par's, but the default
method does not work well (and will warn). Use 'optimize'
instead.
Next, there is a constraint of x=0 that you are not imposing.
Finally, it is easy to see that qnorm(0.7, 0.0032, x) is monotome in x, so
the solution is x=0. In fact, x1 = 0.0032 + sqrt(x) * qnorm(0.7).
optim(0.0207, fr) does a good enough job, as does
optimize(fr, low=0, up=0.05)
Advice: numerical optimization is not a black box, and has to be used with
some analysis of the problem to hand. See e.g. MASS4, chapter 16.
On Mon, 18 Jun 2007, livia wrote:
Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01,
x1 is the quantile of normal distribution (0.0032,x) with probability of
0.7, and the changing value should be x. Initial value for x is 0.0207. I am
using the following codes, but it does not work.
fr - function(x) {
x1-qnorm(0.7,0.0032,x)
x2=0.01
x1-x2
}
xsd - optim(0.0207, fr, NULL,method=BFGS)
It is the first time I am trying to use optimization. Could anyone give me
some advice?
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Optimization
Hi,
my first guess is that the algorithm returns a negative value in some
step - recall that you start from 0.0207!! This negative value is then
passed as standard error to qnorm and that cannot work...
My guess is based on a small experiment where I tried a different
starting point (.02 is so close to 0 that one cannot see anything):
xsd - optim(20, fr, NULL,method=BFGS,control=list(trace=6))
The warnings which you didn't include also tell you about NaNs in
qnorm() - another strong indication of wrong arguments to qnorm().
Try constrained optimization to resctrict to positive values.
See ?constrOptim or use optim() with a method allowing for box
constraints - see ?optim, arguments lower, upper.
Petr
livia napsal(a):
Hi, I would like to minimize the value of x1-x2, x2 is a fixed value of 0.01,
x1 is the quantile of normal distribution (0.0032,x) with probability of
0.7, and the changing value should be x. Initial value for x is 0.0207. I am
using the following codes, but it does not work.
fr - function(x) {
x1-qnorm(0.7,0.0032,x)
x2=0.01
x1-x2
}
xsd - optim(0.0207, fr, NULL,method=BFGS)
It is the first time I am trying to use optimization. Could anyone give me
some advice?
--
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic
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Re: [R] Large Binary file reader for Simple minds
On 6/18/2007 12:17 PM, Todd Remund wrote: I'm more like a caveman when it comes to programming tools. So, with that in mind, is there a way to use readBin in a batch format to read in pieces of a large binary file? Thank you for the consideration of my question. I'm not sure what you mean by batch format, but you can use readBin to read bits and pieces of a file, by opening a connection to the file and reading from there. For example, to read a single unsigned byte value at offset 1, do something like this: con - file(myfile.dat, open=rb) # open for binary reading seek(con, 1) result - readBin(con, integer, size=1, signed=FALSE) close(con) Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] chron() question
Hi all, I'm using chron and it seems to me that there is a strange behaviour when constructing chronological objects. An extract of my source data is: tdr.hhmm[4860:4870] [1] 22:22:00 22:42:00 23:02:00 23:22:00 23:42:00 00:02:00 [7] 00:22:00 00:42:00 01:02:00 01:22:00 01:42:00 tdr.dat$year[4860:4870] [1] 2005 2005 2005 2005 2005 2006 2006 2006 2006 2006 2006 tdr.dat$day[4860:4870] [1] 365 365 365 365 365 1 1 1 1 1 1 And if I use: tdr.chron - chron(dates.=tdr.dat$day,times.=tdr.hhmm,origin.=c(month=1,day=0,year=tdr.dat$year),format=c(dates=d/m/y,times=h:m:s)) The result is: tdr.chron[4860:4870] [1] (31/12/05 22:22:00) (31/12/05 22:42:00) (31/12/05 23:02:00) [4] (31/12/05 23:22:00) (31/12/05 23:42:00) (01/01/05 00:02:00) [7] (01/01/05 00:22:00) (01/01/05 00:42:00) (01/01/05 01:02:00) [10] (01/01/05 01:22:00) (01/01/05 01:42:00) While it seems to me that, through the R recycling rule, it should consider the year 2006 in the corresponding results. Isn't it so? Wishes, Javier -- Javier García-Pintado Institute of Earth Sciences Jaume Almera (CSIC) Lluis Sole Sabaris s/n, 08028 Barcelona Phone: +34 934095410 Fax: +34 934110012 e-mail:[EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] source a specific function
On 6/18/2007 12:11 PM, (Ted Harding) wrote:
On 18-Jun-07 14:28:35, Gabor Grothendieck wrote:
This loads all the functions into an anonymous environment defined
by local and then exports f to the global environment.
f - local({
source(/a.R, local = TRUE)
environment(f) - .GlobalEnv
f
})
That looks neat! Two questions:
1. Would something similar work for extracting selected functions
from a library (assuming that you know about interdependencies)?
E.g. something like
f - local({
library(f.etc.lib)
environment(f) - .GlobalEnv
f
})
The exact syntax you list there won't work, but in any case, changing
the environment of a function in a package is a bad idea -- it may need
to reference things from the namespace of the package.
2. Having done what you describe to extract just f from a source
file, can one then delete the local environment used to load
the source? I think what I'm basically asking is whether the
exporting is done by value (local environment deletion OK)
or by reference (deletion would destroy the exported object).
Gabor answered this: it will go away automatically.
Duncan Murdoch
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Re: [R] Large Binary file reader for Simple minds
Hi Todd Remund wrote: I'm more like a caveman when it comes to programming tools. So, with that in mind, is there a way to use readBin in a batch format to read in pieces of a large binary file? Thank you for the consideration of my question. The 'hexView' package might be useful to you. See Viewing Binary Files with the hexView Package in R News 7/1. Paul Todd Remund __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Stacked barchart color
Hi Hadley, Great, I am starting to get it. It's working for me, but there is one more thing I am having trouble with. The ordering of the stacked bars seems to be dictated by the name of the color, I guess because of the fill=color argument in aes(). In other words, if I set up my colors like this: y$color = c(gray1,gray35,gray45,gray65) the bars get stacked in the opposite order than if I set up the colors like this: y$color = c(gray65,gray45,gray35,gray1). How can I control the order of the bars independent of the name of the colors? Thanks so much in advance! Really neat package you've made. FYI, my plot command now looks like this: p = ggplot(y, aes(x=locus, y=Freq, fill=color)) p = p + geom_bar(position=fill) p = p + scale_fill_identity(labels=levels(y$Fnd), grob=tile, name=Fnd Results) p = p + coord_flip() And the data table is similar as before: y Fnd locusFreq color 1 signeg DPB1 0.013071895 gray1 2 neg DPB1 0.581699346 gray35 3 pos DPB1 0.379084967 gray45 4 sigpos DPB1 0.026143791 gray65 5 signeg DPA1 0.068181818 gray1 6 neg DPA1 0.659090909 gray35 7 pos DPA1 0.25000 gray45 8 sigpos DPA1 0.022727273 gray65 hadley wrote: Hi Owen, The identity scale won't create a legend, unless you tell it what labels it should use - there's an example at http://had.co.nz/ggplot2/scale_identity.html. Otherwise, if you have a continuous scale and you want something that works in black and white, p + scale_fill_gradient(low=white, high=black) might be easier. Hadley y$color = factor(y$Fnd) y$color = c(black,darkgray,lightgray,white) y Fnd locusFreq color 1 signeg A 0.087248322 black 2 neg A 0.711409396 darkgray 3 pos A 0.201342282 lightgray 4 sigpos A 0.0 white 5 signeg C 0.320754717 black 6 neg C 0.603773585 darkgray 7 pos C 0.075471698 lightgray 8 sigpos C 0.0 white 9 signeg B 0.157534247 black 10neg B 0.732876712 darkgray 11pos B 0.109589041 lightgray 12 sigpos B 0.0 white p = ggplot(y, aes(x=locus, y=Freq, fill=color)) + geom_bar(position=fill) + scale_fill_identity() p hadley wrote: Hi Dieter, You can do this with ggplot2 (http://had.co.nz/ggplot2) as follows: library(ggplot2) barley1 - subset(barley, site==Grand Rapids variety %in% c(Velvet,Peatland)) barley1[] - lapply(barley1, [, drop=TRUE) qplot(variety, yield, data=barley1, geom=bar, stat=identity, fill=factor(year)) barley1$fill - c(red,green,blue,gray) qplot(variety, yield, data=barley1, geom=bar, stat=identity, fill=fill) + scale_fill_identity() See http://had.co.nz/ggplot2/scale_identity.html and http://had.co.nz/ggplot2/position_stack.html for more details. Hadley -- View this message in context: http://www.nabble.com/Stacked-barchart-color-tf3909162.html#a11149419 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Stacked-barchart-color-tf3909162.html#a11182581 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] psm/survreg coefficient values ?
I am using psm to model some parametric survival data, the data is for length of stay in an emergency department. There are several ways a patient's stay in the emergency department can end (discharge, admit, etc..) so I am looking at modeling the effects of several covariates on the various outcomes. Initially I am trying to fit a survival model for each type of outcome using the psm function in the design package, i.e., all patients who's visits come to an end due to any event other than the event of interest are considered to be censored. Being new to the psm and survreg packages (and to parametric survival modeling) I am not entirely sure how to interpret the coefficient values that psm returns. I have included the following code to illustrate code similar to what I am using on my data. I suppose that the coefficients are somehow rescaled , but I am not sure how to return them to the original scale and make sense out of the coefficients, e.g., estimate the the effect of higher acuity on time to event in minutes. Any explanation or direction on how to interpret the coefficient values would be greatly appreciated. this is from the documentation for survreg.object. coefficientsthe coefficients of the linear.predictors, which multiply the columns of the model matrix. It does not include the estimate of error (sigma). The names of the coefficients are the names of the single-degree-of-freedom effects (the columns of the model matrix). If the model is over-determined there will be missing values in the coefficients corresponding to non-estimable coefficients. code: LOS - sort(rweibull(1000,1.4,108)) AGE - sort(rnorm(1000,41,12)) ACUITY - sort(rep(1:5,200)) EVENT - sample(x=c(0,1),replace=TRUE,1000) psm(Surv(LOS,EVENT)~AGE+as.factor(ACUITY),dist='weibull') output: psm(formula = Surv(LOS, CENS) ~ AGE + as.factor(ACUITY), dist = weibull) Obs Events Model L.R. d.f. P R2 10005132387.62 5 0 0.91 Value Std. Error z p (Intercept) 1.10550.04425 24.98 8.92e-138 AGE 0.07720.00152 50.93 0.00e+00 ACUITY=2 0.09440.01357 6.96 3.39e-12 ACUITY=3 0.17520.02111 8.30 1.03e-16 ACUITY=4 0.13910.02722 5.11 3.18e-07 ACUITY=5-0.05440.03789 -1.43 1.51e-01 Log(scale)-2.72870.03780 -72.18 0.00e+00 Scale= 0.0653 best, Spencer [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unix-like permissions to allow a user to update recommen
On Mon, 18-Jun-2007 at 11:53AM +0100, Ted Harding wrote:
| On 18-Jun-07 10:11:43, Patrick Connolly wrote:
| I installed R from the tar.gz file (as root) in a directory under
| /usr/local. The recommended packages are installed in a library in
| that directory whereas additional packages I install in a directory
| under the /home directory as a user.
|
| Updating the additional packages is very easy with update.packages()
| as a non-root user, but the recommended packages cannot be done so
| readily because of file permissions.
|
| My question is how do I set the permissions or ownerships in the
| /usr/local/R-2.5.0 directory so that everything necessary can be
| writable by a user? Should I make a group for R users (total of one
| member) or is it simpler than that?
|
| Since you have root access, do you need to segregate the additional
| packages to a particular user?
It's handy to not have to reload packages that don't change between
versions of the basic installation.
|
| Though I don't run R as root for general use, I always install/update
| by running R CMD as root. This makes all of R (recommended and also
| any extras) available system-wide, and no pemission problems arise.
|
| This of course does not stop you from setting up a special .Rprofile
| for each user, since this by definition lives in the user's home
| directory.
|
| Does this help? Or are there issues you haven't mentioned which make
| such an approach not feasible?
I don't exactly have issues. It's not a huge problem I'm dealing
with. It's simple enough for me to use update.packages() as a user
which will download the appropriate packages. Though they won't be
installed, they are all in the one place in the /tmp/ directory from
where I can install them as root. I just thought there must be a more
elegant way to set permissions so that users could write to the
subdirectories under /usr/local/R-2.xxx/. So much of the installation
process of R and its packages is so elegant, I'd like to retain some
of that elegance.
best
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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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] Second y-axis in xyplot (lattice) where y1 and y2 have different ranges
Hi all,
I realize this is asking a lot of lattice, but I want to add a second y
axis inside a xyplot and have y1 and y2 have different ranges. Given dat
below, I can add a second y axis by overlaying a new plot with
par(new=T) and label axis 4 with standard graphics. I've seen an example
for doing something similar in xyplot even though Deepayan has indicated
that lattice isn't the right tool for the job.
However, is there a way to gracefully add a second y-axis to a xyplot
where y1 and y2 have different scales as in the example below? I've seen
the experimental tools to focus and modify lattice graphics but do not
know if these are applicable.
I have unreasonable faith that lattice can do anything. Since my
eventual goal is to make use of a grouping variable as with dat2 below,
lattice will be preferable to complex layouts. Thanks, Andy
dat - data.frame(Year = 1751:2000,
Stuff = rnorm(250),
Samples = floor(seq(5,30,length.out=250)
+rnorm(250,5)),
Grp = rep('SiteOne',250))
par(mar=c(5,4,4,4) + 0.1)
plot(Stuff~Year, data=dat, type='l')
par(new=T)
plot(Samples~Year, data=dat, type=l, axes=F, bty=n,
xlab=, ylab=)
axis(4, at=pretty(range(dat$Samples)))
mtext(Number of Samples, 4, 3)
xyplot(Stuff + Samples ~ Year | Grp, data=dat,
layout = c(1, 1),
panel = panel.superpose.2,
ylab = Stuff,
legend = list(right =
list(fun = grid::textGrob(Samples, rot = 90))),
type = c('l', 'l'))
dat2 - data.frame(Year = rep(1751:2000,2),
Stuff = rep(rnorm(250),2),
Samples = rep(floor(seq(5,30,length.out=250)+
rnorm(250,5)),2),
Grp = c(rep('SiteOne',250),
rep('SiteTwo',250)))
xyplot(Stuff + Samples ~ Year | Grp, data=dat2,
panel = panel.superpose.2,
ylab = Stuff,
legend = list(right =
list(fun = grid::textGrob(Samples, rot = 90))),
type = c('l', 'l'))
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[R] Help: Upgrading to R2.5 on Ubuntu (Feisty)
Thank you in advance for reading this help request. I am pretty new to R. I am experiencing some issues getting 2.5 installed on my Ubuntu Fiesty system and seek your advice. To the best of my ability I followed the instructions here: http://cran.r-project.org/bin/linux/ubuntu/README Setting this as the last line in my sources.list: deb http://cran.fhcrc.org/bin/linux/ubuntu feisty/ When I typed in: [EMAIL PROTECTED]:/usr/local/lib/R/site-library$ sudo apt-get install r-base Reading package lists... Done Building dependency tree Reading state information... Done r-base is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. [EMAIL PROTECTED]:/usr/local/lib/R/site-library$ But when I go to R and check my version: version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) My version is still 2.4.1. I must be missing something. What do I need to do to get R version 2.5 installed on my ubuntu feisty (7.04) system? Let me know if there is any additional information I need to give to be helped out with this. Thank you for taking a look at this, Sincerely, Matt __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] source a specific function
On 18-Jun-07 18:53:50, Duncan Murdoch wrote: The exact syntax you list there won't work, but in any case, changing the environment of a function in a package is a bad idea -- it may need to reference things from the namespace of the package. Well, as I said before, (assuming that you know about interdependencies)! I tried Gabor's suggested syntax as follows, bearing in mind that mvrnorm in MASS is pure R code calling only base functions: library(MASS) environment(mvrnorm) - .GlobalEnv mu=c(0,0) V-matrix(c(1.0,0.5,0.5,1.0),ncol=2) detach() ls() [1] %.+%%+.%mu mvrnorm V mvrnorm(10,mu,V) [,1] [,2] [1,] -1.80466069 -1.8229928 [2,] 0.05565147 -1.6279434 [3,] -0.28505572 -0.8927696 [4,] -0.48919795 0.0750501 [5,] -0.08437832 0.1349296 [6,] 2.17399713 1.2881640 [7,] 1.59934824 1.3784665 [8,] 0.30555420 0.3835743 [9,] 0.11120527 -0.7287910 [10,] -0.77281783 -1.2265502 so that one seems to have worked. However, I'm not sure about the space used by MASS going away after detach(): Starting R from scratch: Type 'q()' to quit R. gc() used (Mb) gc trigger (Mb) Ncells 412589 11.1 597831 16 Vcells 102417 0.8 7864326 library(MASS) gc() used (Mb) gc trigger (Mb) Ncells 459471 12.3 667722 17.9 Vcells 87 0.9 786432 6.0 environment(mvrnorm) - .GlobalEnv detach() gc() used (Mb) gc trigger (Mb) Ncells 459297 12.3 741108 19.8 Vcells 05 0.9 786432 6.0 gc() used (Mb) gc trigger (Mb) Ncells 459304 12.3 818163 21.9 Vcells 18 0.9 786432 6.0 so only about 170KB of the 2.2MB used by MASS has been recovered after detach(). Or am I looking at the wrong indicator of space used? On the other hand, if I first extract the code 9f mvrnorm() which is a few lines of pure R, I get: 1. Start R from scratch: gc() used (Mb) gc trigger (Mb) Ncells 412589 11.1 597831 16 Vcells 102417 0.8 7864326 2. Paste in the code for mvrnorm, and then: gc() used (Mb) gc trigger (Mb) Ncells 412844 11.1 667722 17.9 Vcells 102591 0.8 786432 6.0 so mrvnorm on its own is only taking up about (412844-412589) + (102591-102417) = 429KB Hence I wonder whether the space first occupied by MASS (2.2MB) apart from mvrnorm gets freed up at all? Thanks for the comments. Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 18-Jun-07 Time: 22:00:33 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unix-like permissions to allow a user to update recommen
On 18-Jun-07 20:27:56, Patrick Connolly wrote: On Mon, 18-Jun-2007 at 11:53AM +0100, Ted Harding wrote: | On 18-Jun-07 10:11:43, Patrick Connolly wrote: | I installed R from the tar.gz file (as root) in a directory under | /usr/local. The recommended packages are installed in a library in | that directory whereas additional packages I install in a directory | under the /home directory as a user. | | Updating the additional packages is very easy with | update.packages() | as a non-root user, but the recommended packages cannot be done so | readily because of file permissions. | | My question is how do I set the permissions or ownerships in the | /usr/local/R-2.5.0 directory so that everything necessary can be | writable by a user? Should I make a group for R users (total of | one | member) or is it simpler than that? | | Since you have root access, do you need to segregate the additional | packages to a particular user? It's handy to not have to reload packages that don't change between versions of the basic installation. | | Though I don't run R as root for general use, I always install/update | by running R CMD as root. This makes all of R (recommended and also | any extras) available system-wide, and no pemission problems arise. | | This of course does not stop you from setting up a special .Rprofile | for each user, since this by definition lives in the user's home | directory. | | Does this help? Or are there issues you haven't mentioned which make | such an approach not feasible? I don't exactly have issues. It's not a huge problem I'm dealing with. It's simple enough for me to use update.packages() as a user which will download the appropriate packages. Though they won't be installed, they are all in the one place in the /tmp/ directory from where I can install them as root. I just thought there must be a more elegant way to set permissions so that users could write to the subdirectories under /usr/local/R-2.xxx/. So much of the installation process of R and its packages is so elegant, I'd like to retain some of that elegance. On my Linux, all the places where components of R might normally be installed (/usr/lib or /usr/local/lib) are user=root, group=root, and when I look under them practically everything is writeable only by user=root. So you'd have to change a lot of permissions before any other user got rights to write to these directories. Even adding a user to group=root wouldn't change things, since group does not have write permissions (unless user=root too). I'm still wondering, though, why you don't just run the command update.packages() as root. You have root access, and you said (in the adding user to group context) that only one user is involved (presumably yourself?). In that case, why not start R as root and run update.packages()? Or am I missing something? Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 18-Jun-07 Time: 22:25:18 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] psm/survreg coefficient values ?
sj wrote: I am using psm to model some parametric survival data, the data is for length of stay in an emergency department. There are several ways a patient's stay in the emergency department can end (discharge, admit, etc..) so I am looking at modeling the effects of several covariates on the various outcomes. Initially I am trying to fit a survival model for each type of outcome using the psm function in the design package, i.e., all patients who's visits come to an end due to any event other than the event of interest are considered to be censored. Being new to the psm and survreg packages (and to parametric survival modeling) I am not entirely sure how to interpret the coefficient values that psm returns. I have included the following code to illustrate code similar to what I am using on my data. I suppose that the coefficients are somehow rescaled , but I am not sure how to return them to the original scale and make sense out of the coefficients, e.g., estimate the the effect of higher acuity on time to event in minutes. Any explanation or direction on how to interpret the coefficient values would be greatly appreciated. this is from the documentation for survreg.object. coefficientsthe coefficients of the linear.predictors, which multiply the columns of the model matrix. It does not include the estimate of error (sigma). The names of the coefficients are the names of the single-degree-of-freedom effects (the columns of the model matrix). If the model is over-determined there will be missing values in the coefficients corresponding to non-estimable coefficients. code: LOS - sort(rweibull(1000,1.4,108)) AGE - sort(rnorm(1000,41,12)) ACUITY - sort(rep(1:5,200)) EVENT - sample(x=c(0,1),replace=TRUE,1000) psm(Surv(LOS,EVENT)~AGE+as.factor(ACUITY),dist='weibull') output: psm(formula = Surv(LOS, CENS) ~ AGE + as.factor(ACUITY), dist = weibull) Obs Events Model L.R. d.f. P R2 10005132387.62 5 0 0.91 Value Std. Error z p (Intercept) 1.10550.04425 24.98 8.92e-138 AGE 0.07720.00152 50.93 0.00e+00 ACUITY=2 0.09440.01357 6.96 3.39e-12 ACUITY=3 0.17520.02111 8.30 1.03e-16 ACUITY=4 0.13910.02722 5.11 3.18e-07 ACUITY=5-0.05440.03789 -1.43 1.51e-01 Log(scale)-2.72870.03780 -72.18 0.00e+00 Scale= 0.0653 best, Spencer I have a case study using psm (survreg wrapper) in my book. Briefly, coefficients are on the log median survival time scale. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data type for block data?
Dear All, I have a matrix with data that is not organised. I would like to go through this and extract it. Each feature has 2 vectors which express the data. I also have an index of the places where the data should be cut. eg. class(cc) matrix cc [,1] [,2] [1,]1 26 [2,]2 27 [3,]3 28 [4,]4 29 [5,]5 30 [6,]6 31 [7,]7 32 [8,]8 33 [9,]9 34 [10,]1 27 [11,]1 28 [12,]2 30 [13,]3 34 ect.. index [1] 10 40 Is there a way to take cc[i:index[i-1],] to another format as to where each block could be worked on separately. ie so in one block would be rows1:10 the next block would be rows11:40 and so on. Thanks, Paul -- Research Technician Mass Spectrometry o The / o Scripps \ o Research / o Institute __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Second y-axis in xyplot (lattice) where y1 and y2 have different ranges
On 6/18/07, Andy Bunn [EMAIL PROTECTED] wrote:
Hi all,
I realize this is asking a lot of lattice, but I want to add a second y
axis inside a xyplot and have y1 and y2 have different ranges. Given dat
below, I can add a second y axis by overlaying a new plot with
par(new=T) and label axis 4 with standard graphics. I've seen an example
for doing something similar in xyplot even though Deepayan has indicated
that lattice isn't the right tool for the job.
However, is there a way to gracefully add a second y-axis to a xyplot
where y1 and y2 have different scales as in the example below? I've seen
the experimental tools to focus and modify lattice graphics but do not
know if these are applicable.
You could use those, but one drawback there is that you don't get the
usual benefit of automatic allocation of space. Here is a ``better''
solution (as long as you realize that this is still a hack):
[Note: this won't work if scales=free or sliced]
[...]
dat2 - data.frame(Year = rep(1751:2000,2),
Stuff = rep(rnorm(250),2),
Samples = rep(floor(seq(5,30,length.out=250)+
rnorm(250,5)),2),
Grp = c(rep('SiteOne',250),
rep('SiteTwo',250)))
scale.pars - function(x)
{
c(mx = min(x), dx = diff(range(x)))
}
rescale - function(x, pars = scale.pars(x))
{
(x - pars[mx]) / pars[dx]
}
pars.Stuff - scale.pars(dat2$Stuff)
pars.Samples - scale.pars(dat2$Samples)
rng.Stuff - range(dat2$Stuff)
rng.Samples - range(dat2$Samples)
my.yscale.components - function(lim, ...)
{
## template we will modify
ans - yscale.components.default(lim, ...)
## labels for Stuff in original scale
Stuff - yscale.components.default(rng.Stuff, ...)
Stuff$left$ticks$at -
rescale(Stuff$left$ticks$at, pars.Stuff)
Stuff$left$labels$at -
rescale(Stuff$left$labels$at, pars.Stuff)
## labels for Samples in original scale
Samples - yscale.components.default(rng.Samples, ...)
Samples$left$ticks$at -
rescale(Samples$left$ticks$at, pars.Samples)
Samples$left$labels$at -
rescale(Samples$left$labels$at, pars.Samples)
## modified 'components'
ans$left - Stuff$left
ans$right - Samples$left
ans
}
xyplot(rescale(Stuff, pars.Stuff) +
rescale(Samples, pars.Samples) ~ Year | Grp,
data=dat2,
panel = panel.superpose.2,
## newlay added:
yscale.components = my.yscale.components,
scales = list(alternating = 3),
ylab = Stuff,
legend = list(right =
list(fun = grid::textGrob(Samples, rot = 90))),
type = c('l', 'l'))
-Deepayan
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[R] viewing source code
Hi, could somebody give me a quick hint how to view the source code of a function if sole entering of the function name does not work? In particular, I am trying to look at cd_plot from the vcd package. Many thanks in advance, Werner __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] viewing source code
Werner: could somebody give me a quick hint how to view the source code of a function if sole entering of the function name does not work? In particular, I am trying to look at cd_plot from the vcd package. Strategy 1: Typing cd_plot tells you that it is a generic function and methods(cd_plot) shows you which methods exist (default and formula) which are both non-visible. You can still directly access them via vcd:::cd_plot (which is the main work horse). Strategy 2 (preferred, especially if you want to take a closer look): Obtain the source package from CRAN. Unpack the tar.gz file and look into the vcd/R folder where you will find cd_plot.R containing the sources of both methods. grx, Z __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to compare GLM and GAM models
Dear Listers, I want to compare two negative binomial models fitted using glm.nb and gam(mgcv) based on the same data. What would be the most appropriate criteria to compare these two models? Can someone point me to some references? Thank you very much. Yuanchang Xie __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Histogram using frequency data
Hello, I wanted to know how to plot a histogram using a vector of frequencies rather than the data vector as a whole. So I have two vectors: a vector of labels V1= c(A,B,C,D) and vector B which is a vector of frequencies of A, B, C and D respectively V2=c(20,50,60,30). I wanted to plot a histogram of the labels using the frequencies. I could not figure out a way to do this using the 'hist' function which takes only the full data vector as input. Could you please help me with this? Thank you, Suman [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help: Upgrading to R2.5 on Ubuntu (Feisty)
Dear Matt, Did you issue: $ sudo apt-get update before running: $ sudo apt-get install r-base Now, let me tell you one thing about Linux and particularly Debian/Ubuntu. We are spoiled to the point that we love the official repositories. Because the official packages go through some testing, we tend to sacrifice a little bit of cutting edge for stability/reliability. If you don't think you need anything specific from version 2.5.0, I would recommend you to stick with the current version, 2.4.1. You'll also have several packages already compiled for you if you do that. I hope it helps. Paulo On 6/18/07, M. Jankowski [EMAIL PROTECTED] wrote: Thank you in advance for reading this help request. I am pretty new to R. I am experiencing some issues getting 2.5 installed on my Ubuntu Fiesty system and seek your advice. To the best of my ability I followed the instructions here: http://cran.r-project.org/bin/linux/ubuntu/README Setting this as the last line in my sources.list: deb http://cran.fhcrc.org/bin/linux/ubuntu feisty/ When I typed in: [EMAIL PROTECTED]:/usr/local/lib/R/site-library$ sudo apt-get install r-base Reading package lists... Done Building dependency tree Reading state information... Done r-base is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. [EMAIL PROTECTED]:/usr/local/lib/R/site-library$ But when I go to R and check my version: version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) My version is still 2.4.1. I must be missing something. What do I need to do to get R version 2.5 installed on my ubuntu feisty (7.04) system? Let me know if there is any additional information I need to give to be helped out with this. Thank you for taking a look at this, Sincerely, Matt __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Retain names in conversion of matrix to vector
Fantastic. All of those methods worked, though I did have to first convert my matrices using the data.matrix command. Thank you for the assistance. Is there any equally simple way to re-convert the resulting table/matrix to its original NxN form? I do not see any obvious opposites to the cbind or as.table commands. Also, Jeff Laake, is there a way to have your routine output the vectors as numeric values (i.e., without the surrounding quote marks)? Best regards, Jeff Laake [EMAIL PROTECTED] wrote: x=matrix(1:9,nrow=3) colnames(x)=c(a,b,c) row.names(x)=c(1,2,3) x a b c 1 1 4 7 2 2 5 8 3 3 6 9 cbind(as.vector(x),colnames(x)[as.vector(col(x))],row.names(x)[as.vector(row(x))]) [,1] [,2] [,3] [1,] 1 a 1 [2,] 2 a 2 [3,] 3 a 3 [4,] 4 b 1 [5,] 5 b 2 [6,] 6 b 3 [7,] 7 c 1 [8,] 8 c 2 [9,] 9 c 3 philozine wrote: Hi R-listers, I'm using R only for a few basic functions but am having difficulty doing something that *should* be simple. I have an nxn matrix, Q, where Q[i,j] is a directed value (in this case, oil exports from i to j). Note that Q[i,j]~=Q[j,i]. I imported column names along with the matrix then copied them to the rows using rownames(Q) - colnames(Q). Simple so far. What I'd like to do now is convert Q for export into a vector of values with the original row and column names intact. Having one vector each for row, column, and cell would be ideal, e.g., [1,1] = i's name, [1,2] = j's name, and [1,3] = Q[i, j]. But just being able to export my matrix data in vector form with the correct row/col names for each observation would be sufficient. Thus far I've tried c(), vector(), and a few others, but can't get the correct results. They do generate the correct vector of matrix values, but they do not appear to retain both row and column names. (Or, rather, I have not discovered how to make them do so.) To illustrate, my data currently look something like this: ABCD A | 0 |.1 |.4 |.6 | B |.2 | 0 |.2 |.1 | C |.5 |.9 | 0 |.9 | D |.7 | 0 |.3 | 0 | I would like them to look like this (at least when exported as a .txt file, if not necessary when displayed within R): i j Q | A | A | 0 | | A | B |.1 | | A | C |.4 | | A | D |.6 | | B | A |.2 | | B | B | 0 | | B | C |.2 | [...] and so on If anybody knows how to do this, I will be extremely appreciative! Best regards, - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - The fish are biting. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] String manipulation, insert delim
Hello All, I've been using R for two years now and I am happy to say this is the first time I could not find the answer to my problem in the R-help archives. Here is the pending problem: I want to be able to insert delimiters, say commas, into a string of characters at uneven intervals such that: foo-c(haveaniceday)#my string of character bar-c(4,1,4,3) # my vector of uneven intervals my.fun(foo,bar) # some function that places delimiters appropriately have,a,nice,day # what the function would ideally return I've tried multiple for-loops using cut and paste but have not had success. Thanks! Chris Marcum UCI Sociology __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histogram using frequency data
On Mon, 2007-06-18 at 19:40 -0400, suman Duvvuru wrote: Hello, I wanted to know how to plot a histogram using a vector of frequencies rather than the data vector as a whole. So I have two vectors: a vector of labels V1= c(A,B,C,D) and vector B which is a vector of frequencies of A, B, C and D respectively V2=c(20,50,60,30). I wanted to plot a histogram of the labels using the frequencies. I could not figure out a way to do this using the 'hist' function which takes only the full data vector as input. Could you please help me with this? Thank you, Suman See ?barplot To wit: V1 - c(A, B, C, D) V2 - c(20, 50, 60, 30) # Do the barplot, saving the bar midpoints in 'mp' mp - barplot(V2, names.arg = V1, ylim = c(0, 80)) # Now add the bar values above the bars text(mp, V2, V2, pos = 3) See ?text and ?mtext for adding annotation HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] classical plant hybrid analysis
Hi, I am looking to perform some classical hybrid analysis on hybrid plants including GCA and SCA components through R. I was wondering if anybody knew of existing packages that can perform these analyses (I cannot find any) or any examples of these analyses being worked through. Cameron Beeck __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] genetics package not working
Has something changed in R that requires an update in the genetics package by Gregory Warnes? I am using R version 2.5.0 This used to work summary(founders[,59]) to prove that it is a genotype class class(founders[,59]) [1] genotype factor Now when I issue the command: summary(founders[,59]) I get: Error in attr(retval, which) - which : attempt to set an attribute on NULL In addition: Warning message: $ operator is deprecated for atomic vectors, returning NULL in: x$allele.names Clearly, I am missing something. What am I missing? -- Farrel Buchinsky [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] String manipulation, insert delim
This should work for you.
foo-c(haveaniceday)#my string of character
bar-c(4,1,4,3) # my vector of uneven intervals
my.function - function(foo, bar){
+ # construct a matrix with start/end character positions
+ start - head(cumsum(c(1, bar)), -1) # delete last one
+ sel - cbind(start=start,
+ end=start + bar -1)
+ strings - apply(sel, 1, function(x) substr(foo, x[1], x[2]))
+ paste(strings, collapse=',')
+ }
my.function(foo, bar)
[1] have,a,nice,day
On 6/18/07, Christopher Marcum [EMAIL PROTECTED] wrote:
Hello All,
I've been using R for two years now and I am happy to say this is the
first time I could not find the answer to my problem in the R-help
archives. Here is the pending problem:
I want to be able to insert delimiters, say commas, into a string of
characters at uneven intervals such that:
foo-c(haveaniceday)#my string of character
bar-c(4,1,4,3) # my vector of uneven intervals
my.fun(foo,bar) # some function that places delimiters appropriately
have,a,nice,day # what the function would ideally return
I've tried multiple for-loops using cut and paste but have not had
success.
Thanks!
Chris Marcum
UCI Sociology
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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?
[[alternative HTML version deleted]]
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Re: [R] String manipulation, insert delim
Try this: paste(read.fwf(textConnection(foo), bar, as.is = TRUE), collapse = ,) [1] have,a,nice,day On 6/18/07, Christopher Marcum [EMAIL PROTECTED] wrote: Hello All, I've been using R for two years now and I am happy to say this is the first time I could not find the answer to my problem in the R-help archives. Here is the pending problem: I want to be able to insert delimiters, say commas, into a string of characters at uneven intervals such that: foo-c(haveaniceday)#my string of character bar-c(4,1,4,3) # my vector of uneven intervals my.fun(foo,bar) # some function that places delimiters appropriately have,a,nice,day # what the function would ideally return I've tried multiple for-loops using cut and paste but have not had success. Thanks! Chris Marcum UCI Sociology __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] String manipulation, insert delim
On Mon, 2007-06-18 at 16:54 -0700, Christopher Marcum wrote:
Hello All,
I've been using R for two years now and I am happy to say this is the
first time I could not find the answer to my problem in the R-help
archives. Here is the pending problem:
I want to be able to insert delimiters, say commas, into a string of
characters at uneven intervals such that:
foo-c(haveaniceday)#my string of character
bar-c(4,1,4,3) # my vector of uneven intervals
my.fun(foo,bar) # some function that places delimiters appropriately
have,a,nice,day # what the function would ideally return
I've tried multiple for-loops using cut and paste but have not had success.
Thanks!
Chris Marcum
UCI Sociology
One more variation on the replies already provided:
foo - c(haveaniceday)
bar - c(4, 1, 4, 3)
insert.char - function(x, at, char = ,)
{
cs.at - cumsum(at)
vec - unlist(strsplit(x, ))
for (i in seq(length(cs.at) - 1))
vec - append(vec, char, cs.at[i] + i - 1)
paste(vec, collapse = )
}
insert.char(foo, bar)
[1] have,a,nice,day
See ?append
HTH,
Marc Schwartz
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[R] Histograms with strings, grouped by repeat count (w/ data)
Hello R gurus, I just spent my first weekend wrestling with R, but so far have come up empty handed. I have a dataset that represents file downloads; it has 4 dimensions: date, filename, email, and country. (sample data below) My first goal is to get an idea of the frequency of repeated downloads. Let me explain that. Some people tend to download multiple times, e.g. if the download fails they keep trying over and over. I'm trying to build a histogram that shows the repeat count along the x-axis, that is, how many people downloaded once, twice, three times, etc. I plan to compare the median of that before and after we switched ISPs. To accomplish this, I'm assuming that I'll first need to combine the email and filename columns so as to represent a single download attempt by an individual. Does that sound right? Later, it would be nice to limit the histogram to a single filename, country, or company. I can probably figure that out myself after I understand how to write this funky histogram expression. With the help of Verzani's introductory text, I've learned how to read in the CSV data and do some simple tables, like this: hist(table(d$filename)) hist(table(d$filename[substring(d$filename, 1, 5)==file1])) hist(sort(table(d$filename[substring(d$filename, 1, 5)==file1]))) Obviously, these commands count the frequency of the files. What I'd like to see are the repeats grouped along the x-axis; I'd like to find, for all files, the distribution of retries. I hope that makes sense. :) Can someone point me in the right direction? I'm very new to R and to statistics, but I write code for a living. At this point I'd almost be better off writing a program do this kind of simple counting... but I have a feeling R would be so useful if I could just get past the initial learning curve. Thank you in advance, Matt Here's some real data, with the private info replaced :) d-read.table(file=C:\\users\\trunnellm\\downloads\\statistics\\downloads.csv, sep=,, quote=\, header=TRUE) filename,last_modified,email_addr,country_residence file1,3/4/2006 13:54,email1,Korea (South) file2,3/4/2006 14:33,email2,United States file2,3/4/2006 16:03,email2,United States file2,3/4/2006 16:17,email3,United States file2,3/4/2006 16:28,email3,United States file3,3/4/2006 19:13,email4,United States file2,3/4/2006 21:22,email5,India file4,3/4/2006 21:46,email6,United States file1,3/4/2006 22:04,email7,Japan file2,3/4/2006 22:09,email8,Croatia file1,3/4/2006 22:22,email7,Japan file1,3/4/2006 22:29,email9,India file1,3/4/2006 23:06,email6,United States file1,3/4/2006 23:33,email6,United States file5,3/4/2006 23:44,email10,China file1,3/5/2006 0:13,email9,India file2,3/5/2006 0:52,email8,Croatia file2,3/5/2006 0:54,email8,Croatia file2,3/5/2006 1:10,email5,India file6,3/5/2006 2:17,email9,India file2,3/5/2006 2:24,email11,Italy file7,3/5/2006 2:36,email12,Italy file8,3/5/2006 2:52,email12,Italy file2,3/5/2006 3:09,email13,United Kingdom file2,3/5/2006 4:02,email14,India file2,3/5/2006 4:07,email14,India file2,3/5/2006 4:14,email14,India file2,3/5/2006 4:37,email5,India file2,3/5/2006 4:44,email15,Belgium file1,3/5/2006 5:02,email9,India file1,3/5/2006 5:24,email16,Taiwan file2,3/5/2006 6:06,email17,Saudi Arabia file2,3/5/2006 7:32,email17,Saudi Arabia file2,3/5/2006 8:12,email18,Brazil file2,3/5/2006 8:26,email18,Brazil file2,3/5/2006 9:49,email19,United Kingdom file1,3/5/2006 10:49,email11,Italy file1,3/5/2006 11:16,email13,United Kingdom file1,3/5/2006 11:16,email13,United Kingdom file1,3/5/2006 11:45,email13,United Kingdom file1,3/5/2006 14:34,email20,Australia file9,3/5/2006 14:56,email20,Australia file9,3/5/2006 14:56,email20,Australia file5,3/5/2006 16:43,email21,United States file1,3/5/2006 17:17,email7,Japan file2,3/5/2006 17:26,email22,Japan file2,3/5/2006 17:27,email22,Japan file2,3/5/2006 17:33,email23,China file1,3/5/2006 17:45,email22,Japan file2,3/5/2006 17:45,email22,Japan file2,3/5/2006 17:59,email23,China file1,3/5/2006 18:27,email24,Japan file1,3/5/2006 18:47,email25,Taiwan file2,3/5/2006 18:48,email26,New Zealand file2,3/5/2006 19:15,email27,Canada file2,3/5/2006 19:23,email28,Canada file2,3/5/2006 19:24,email28,Canada file10,3/5/2006 19:49,email29,Japan file10,3/5/2006 19:52,email29,Japan file10,3/5/2006 19:57,email29,Japan file2,3/5/2006 20:01,email29,Japan file2,3/5/2006 20:02,email29,Japan file2,3/5/2006 20:06,email29,Japan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] categorical time series
Hello, Can anyone tell me how to simulate multinomial time series in R? Thanks, Jaydip Jaydip Mukhopadhyay Graduate Student Dept of Statistics University Of Connecticut - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histograms with strings, grouped by repeat count (w/ data)
You should be using barplot and not hist. I think this produces what you
want:
x - filename,last_modified,email_addr,country_residence
file1,3/4/2006 13:54,email1,Korea (South)
file2,3/4/2006 14:33,email2,United States
file2,3/4/2006 16:03,email2,United States
file2,3/4/2006 16:17,email3,United States
file2,3/4/2006 16:28,email3,United States
file3,3/4/2006 19:13,email4,United States
file2,3/4/2006 21:22,email5,India
file4,3/4/2006 21:46,email6,United States
file1,3/4/2006 22:04,email7,Japan
file2,3/4/2006 22:09,email8,Croatia
file1,3/4/2006 22:22,email7,Japan
file1,3/4/2006 22:29,email9,India
file1,3/4/2006 23:06,email6,United States
file1,3/4/2006 23:33,email6,United States
file5,3/4/2006 23:44,email10,China
file1,3/5/2006 0:13,email9,India
file2,3/5/2006 0:52,email8,Croatia
file2,3/5/2006 0:54,email8,Croatia
file2,3/5/2006 1:10,email5,India
file6,3/5/2006 2:17,email9,India
file2,3/5/2006 2:24,email11,Italy
file7,3/5/2006 2:36,email12,Italy
file8,3/5/2006 2:52,email12,Italy
file2,3/5/2006 3:09,email13,United Kingdom
file2,3/5/2006 4:02,email14,India
file2,3/5/2006 4:07,email14,India
file2,3/5/2006 4:14,email14,India
file2,3/5/2006 4:37,email5,India
file2,3/5/2006 4:44,email15,Belgium
file1,3/5/2006 5:02,email9,India
file1,3/5/2006 5:24,email16,Taiwan
file2,3/5/2006 6:06,email17,Saudi Arabia
file2,3/5/2006 7:32,email17,Saudi Arabia
file2,3/5/2006 8:12,email18,Brazil
file2,3/5/2006 8:26,email18,Brazil
file2,3/5/2006 9:49,email19,United Kingdom
file1,3/5/2006 10:49,email11,Italy
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:45,email13,United Kingdom
file1,3/5/2006 14:34,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file5,3/5/2006 16:43,email21,United States
file1,3/5/2006 17:17,email7,Japan
file2,3/5/2006 17:26,email22,Japan
file2,3/5/2006 17:27,email22,Japan
file2,3/5/2006 17:33,email23,China
file1,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:59,email23,China
file1,3/5/2006 18:27,email24,Japan
file1,3/5/2006 18:47,email25,Taiwan
file2,3/5/2006 18:48,email26,New Zealand
file2,3/5/2006 19:15,email27,Canada
file2,3/5/2006 19:23,email28,Canada
file2,3/5/2006 19:24,email28,Canada
file10,3/5/2006 19:49,email29,Japan
file10,3/5/2006 19:52,email29,Japan
file10,3/5/2006 19:57,email29,Japan
file2,3/5/2006 20:01,email29,Japan
file2,3/5/2006 20:02,email29,Japan
file2,3/5/2006 20:06,email29,Japan
d - read.csv(textConnection(x))
barplot(table(d$filename), main=All Files, las=2) # plot counts for all
the files
# generate plots for each file name showing which emails used them
counts - table(d$filename, d$email_addr)
for (i in seq(nrow(counts))){
.index - which(counts[i,] 0)
barplot(counts[i, .index], las=2,
names.arg=colnames(counts)[.index], main=rownames(counts)[i])
}
On 6/18/07, Matthew Trunnell [EMAIL PROTECTED] wrote:
Hello R gurus,
I just spent my first weekend wrestling with R, but so far have come
up empty handed.
I have a dataset that represents file downloads; it has 4 dimensions:
date, filename, email, and country. (sample data below)
My first goal is to get an idea of the frequency of repeated
downloads. Let me explain that. Some people tend to download
multiple times, e.g. if the download fails they keep trying over and
over. I'm trying to build a histogram that shows the repeat count
along the x-axis, that is, how many people downloaded once, twice,
three times, etc. I plan to compare the median of that before and
after we switched ISPs.
To accomplish this, I'm assuming that I'll first need to combine the
email and filename columns so as to represent a single download
attempt by an individual. Does that sound right? Later, it would be
nice to limit the histogram to a single filename, country, or company.
I can probably figure that out myself after I understand how to write
this funky histogram expression.
With the help of Verzani's introductory text, I've learned how to read
in the CSV data and do some simple tables, like this:
hist(table(d$filename))
hist(table(d$filename[substring(d$filename, 1, 5)==file1]))
hist(sort(table(d$filename[substring(d$filename, 1, 5)==file1])))
Obviously, these commands count the frequency of the files. What I'd
like to see are the repeats grouped along the x-axis; I'd like to
find, for all files, the distribution of retries. I hope that makes
sense. :)
Can someone point me in the right direction? I'm very new to R and to
statistics, but I write code for a living. At this point I'd almost
be better off writing a program do this kind of simple counting... but
I have a feeling R would be so useful if I could just get past the
initial learning curve.
Thank you in advance,
Matt
Here's some real data, with the private info replaced :)
d-read.table
(file=C:\\users\\trunnellm\\downloads\\statistics\\downloads.csv,
sep=,, quote=\, header=TRUE)
Re: [R] data type for block data?
This will create a list of the matrix subsets:
# create a matrix
x - cbind(1:40, runif(40))
index - c(10,15,33,40) # cut points
# create a matrix with start and end points
slices - cbind(start=head(c(1,index + 1), -1), end=index)
# create a list with the matrices
matrix.subset - lapply(seq(nrow(slices)), function(.row){
x[slices[.row, 1]:slices[.row, 2], ]
})
matrix.subset
On 6/18/07, H. Paul Benton [EMAIL PROTECTED] wrote:
Dear All,
I have a matrix with data that is not organised. I would like to go
through this and extract it. Each feature has 2 vectors which express
the data. I also have an index of the places where the data should be cut.
eg.
class(cc)
matrix
cc
[,1] [,2]
[1,]1 26
[2,]2 27
[3,]3 28
[4,]4 29
[5,]5 30
[6,]6 31
[7,]7 32
[8,]8 33
[9,]9 34
[10,]1 27
[11,]1 28
[12,]2 30
[13,]3 34
ect..
index
[1] 10 40
Is there a way to take cc[i:index[i-1],] to another format as to where
each block could be worked on separately. ie so in one block would be
rows1:10 the next block would be rows11:40 and so on.
Thanks,
Paul
--
Research Technician
Mass Spectrometry
o The
/
o Scripps
\
o Research
/
o Institute
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
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?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histograms with strings, grouped by repeat count (w/ data)
Jim,
Thanks for the quick reply! When I run your code, I end up with a
single barplot of one datapoint, file9 vs email20 == 2.0. I see the
call to barplot is inside a for loop... maybe it's zooming through the
display of many barplots, but all I see is the last one?
In any case, I need to figure out the distribution of the retries, such as
No. Retries Count
1 6
2 13
3 5
4 3
5 2
6 1
That is, 6 people retried the download once; 13 people retried the
download twice, etc. So it would be counting the frequency of the
email-filename combination, and grouping those together by the number
of retries. Does that make sense?
When I look at the counts object from your code, I can see that it's
close to what I need. How do I access the properties of the counts
object-- it's a table, right? If I look at counts[1,1], that returns
1. But how do I get at the row/col name of that cell? Is that cell
an object? rownames(counts[1,1]) returns null.
Thanks,
Matt
On 6/18/07, jim holtman [EMAIL PROTECTED] wrote:
You should be using barplot and not hist. I think this produces what you
want:
x - filename,last_modified,email_addr,country_residence
file1,3/4/2006 13:54,email1,Korea (South)
file2,3/4/2006 14:33,email2,United States
file2,3/4/2006 16:03,email2,United States
file2,3/4/2006 16:17,email3,United States
file2,3/4/2006 16:28,email3,United States
file3,3/4/2006 19:13,email4,United States
file2,3/4/2006 21:22,email5,India
file4,3/4/2006 21:46,email6,United States
file1,3/4/2006 22:04,email7,Japan
file2,3/4/2006 22:09,email8,Croatia
file1,3/4/2006 22:22,email7,Japan
file1,3/4/2006 22:29,email9,India
file1,3/4/2006 23:06,email6,United States
file1,3/4/2006 23:33,email6,United States
file5,3/4/2006 23:44,email10,China
file1,3/5/2006 0:13,email9,India
file2,3/5/2006 0:52,email8,Croatia
file2,3/5/2006 0:54,email8,Croatia
file2,3/5/2006 1:10,email5,India
file6,3/5/2006 2:17,email9,India
file2,3/5/2006 2:24,email11,Italy
file7,3/5/2006 2:36,email12,Italy
file8,3/5/2006 2:52,email12,Italy
file2,3/5/2006 3:09,email13,United Kingdom
file2,3/5/2006 4:02,email14,India
file2,3/5/2006 4:07,email14,India
file2,3/5/2006 4:14,email14,India
file2,3/5/2006 4:37,email5,India
file2,3/5/2006 4:44,email15,Belgium
file1,3/5/2006 5:02,email9,India
file1,3/5/2006 5:24,email16,Taiwan
file2,3/5/2006 6:06,email17,Saudi Arabia
file2,3/5/2006 7:32,email17,Saudi Arabia
file2,3/5/2006 8:12,email18,Brazil
file2,3/5/2006 8:26,email18,Brazil
file2,3/5/2006 9:49,email19,United Kingdom
file1,3/5/2006 10:49,email11,Italy
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:45,email13,United Kingdom
file1,3/5/2006 14:34,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file5,3/5/2006 16:43,email21,United States
file1,3/5/2006 17:17,email7,Japan
file2,3/5/2006 17:26,email22,Japan
file2,3/5/2006 17:27,email22,Japan
file2,3/5/2006 17:33,email23,China
file1,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:59,email23,China
file1,3/5/2006 18:27,email24,Japan
file1,3/5/2006 18:47,email25,Taiwan
file2,3/5/2006 18:48,email26,New Zealand
file2,3/5/2006 19:15,email27,Canada
file2,3/5/2006 19:23,email28,Canada
file2,3/5/2006 19:24,email28,Canada
file10,3/5/2006 19:49,email29,Japan
file10,3/5/2006 19:52,email29,Japan
file10,3/5/2006 19:57,email29,Japan
file2,3/5/2006 20:01,email29,Japan
file2,3/5/2006 20:02,email29,Japan
file2,3/5/2006 20:06,email29,Japan
d - read.csv(textConnection(x))
barplot(table(d$filename), main=All Files, las=2) # plot counts for all
the files
# generate plots for each file name showing which emails used them
counts - table(d$filename, d$email_addr)
for (i in seq(nrow(counts))){
.index - which(counts[i,] 0)
barplot(counts[i, .index], las=2,
names.arg=colnames(counts)[.index], main=rownames(counts)[i])
}
On 6/18/07, Matthew Trunnell [EMAIL PROTECTED] wrote:
Hello R gurus,
I just spent my first weekend wrestling with R, but so far have come
up empty handed.
I have a dataset that represents file downloads; it has 4 dimensions:
date, filename, email, and country. (sample data below)
My first goal is to get an idea of the frequency of repeated
downloads. Let me explain that. Some people tend to download
multiple times, e.g. if the download fails they keep trying over and
over. I'm trying to build a histogram that shows the repeat count
along the x-axis, that is, how many people downloaded once, twice,
three times, etc. I plan to compare the median of that before and
after we switched ISPs.
To accomplish this, I'm assuming that I'll first need to combine the
email and filename columns so as to represent a single download
attempt by an individual. Does that sound right? Later,
Re: [R] Histograms with strings, grouped by repeat count (w/ data)
If you are running on windows, make sure you have 'recording' checked in the
history window of the graphics. You can also put the output to a pdf file
and view it later.
If you use table on the counts matrix:
table(counts)
counts
0 1 2 3
253 20 8 9
this shows that there were 20 single tries, 8 files downloaded twice and 9
three times. Is this what you want?
You can also get the indices of the non-zero entries by:
which(counts != 0, arr.ind=TRUE)
row col
file11 1
file56 2
file11 3
file23 3
file78 4
file89 4
file11 5
file23 5
file23 6
.
On 6/18/07, Matthew Trunnell [EMAIL PROTECTED] wrote:
Jim,
Thanks for the quick reply! When I run your code, I end up with a
single barplot of one datapoint, file9 vs email20 == 2.0. I see the
call to barplot is inside a for loop... maybe it's zooming through the
display of many barplots, but all I see is the last one?
In any case, I need to figure out the distribution of the retries, such as
No. Retries Count
1 6
2 13
3 5
4 3
5 2
6 1
That is, 6 people retried the download once; 13 people retried the
download twice, etc. So it would be counting the frequency of the
email-filename combination, and grouping those together by the number
of retries. Does that make sense?
When I look at the counts object from your code, I can see that it's
close to what I need. How do I access the properties of the counts
object-- it's a table, right? If I look at counts[1,1], that returns
1. But how do I get at the row/col name of that cell? Is that cell
an object? rownames(counts[1,1]) returns null.
Thanks,
Matt
On 6/18/07, jim holtman [EMAIL PROTECTED] wrote:
You should be using barplot and not hist. I think this produces what
you
want:
x - filename,last_modified,email_addr,country_residence
file1,3/4/2006 13:54,email1,Korea (South)
file2,3/4/2006 14:33,email2,United States
file2,3/4/2006 16:03,email2,United States
file2,3/4/2006 16:17,email3,United States
file2,3/4/2006 16:28,email3,United States
file3,3/4/2006 19:13,email4,United States
file2,3/4/2006 21:22,email5,India
file4,3/4/2006 21:46,email6,United States
file1,3/4/2006 22:04,email7,Japan
file2,3/4/2006 22:09,email8,Croatia
file1,3/4/2006 22:22,email7,Japan
file1,3/4/2006 22:29,email9,India
file1,3/4/2006 23:06,email6,United States
file1,3/4/2006 23:33,email6,United States
file5,3/4/2006 23:44,email10,China
file1,3/5/2006 0:13,email9,India
file2,3/5/2006 0:52,email8,Croatia
file2,3/5/2006 0:54,email8,Croatia
file2,3/5/2006 1:10,email5,India
file6,3/5/2006 2:17,email9,India
file2,3/5/2006 2:24,email11,Italy
file7,3/5/2006 2:36,email12,Italy
file8,3/5/2006 2:52,email12,Italy
file2,3/5/2006 3:09,email13,United Kingdom
file2,3/5/2006 4:02,email14,India
file2,3/5/2006 4:07,email14,India
file2,3/5/2006 4:14,email14,India
file2,3/5/2006 4:37,email5,India
file2,3/5/2006 4:44,email15,Belgium
file1,3/5/2006 5:02,email9,India
file1,3/5/2006 5:24,email16,Taiwan
file2,3/5/2006 6:06,email17,Saudi Arabia
file2,3/5/2006 7:32,email17,Saudi Arabia
file2,3/5/2006 8:12,email18,Brazil
file2,3/5/2006 8:26,email18,Brazil
file2,3/5/2006 9:49,email19,United Kingdom
file1,3/5/2006 10:49,email11,Italy
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:16,email13,United Kingdom
file1,3/5/2006 11:45,email13,United Kingdom
file1,3/5/2006 14:34,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file9,3/5/2006 14:56,email20,Australia
file5,3/5/2006 16:43,email21,United States
file1,3/5/2006 17:17,email7,Japan
file2,3/5/2006 17:26,email22,Japan
file2,3/5/2006 17:27,email22,Japan
file2,3/5/2006 17:33,email23,China
file1,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:45,email22,Japan
file2,3/5/2006 17:59,email23,China
file1,3/5/2006 18:27,email24,Japan
file1,3/5/2006 18:47,email25,Taiwan
file2,3/5/2006 18:48,email26,New Zealand
file2,3/5/2006 19:15,email27,Canada
file2,3/5/2006 19:23,email28,Canada
file2,3/5/2006 19:24,email28,Canada
file10,3/5/2006 19:49,email29,Japan
file10,3/5/2006 19:52,email29,Japan
file10,3/5/2006 19:57,email29,Japan
file2,3/5/2006 20:01,email29,Japan
file2,3/5/2006 20:02,email29,Japan
file2,3/5/2006 20:06,email29,Japan
d - read.csv(textConnection(x))
barplot(table(d$filename), main=All Files, las=2) # plot counts for
all
the files
# generate plots for each file name showing which emails used them
counts - table(d$filename, d$email_addr)
for (i in seq(nrow(counts))){
.index - which(counts[i,] 0)
barplot(counts[i, .index], las=2,
names.arg=colnames(counts)[.index], main=rownames(counts)[i])
}
On 6/18/07, Matthew Trunnell [EMAIL PROTECTED] wrote:
Hello R gurus,
I just spent my first weekend wrestling with R,
[R] Fwd: Help: Upgrading to R2.5 on Ubuntu (Feisty)
This fixed my problem: Thanks! Did you run 'sudo apt-get update' as well so that it actually reads the listing at CRAN / FHCRC ? What does 'apt-cache policy r-base' show? [ It should display the different vertsions it knows about; if you only see 2.4.1 then you have a problem which may just be the missing 'apt-get update' ] Hth, Dirk -- Forwarded message -- From: M. Jankowski Date: Jun 18, 2007 3:34 PM Subject: Help: Upgrading to R2.5 on Ubuntu (Feisty) To: r-help@stat.math.ethz.ch Thank you in advance for reading this help request. I am pretty new to R. I am experiencing some issues getting 2.5 installed on my Ubuntu Fiesty system and seek your advice. To the best of my ability I followed the instructions here: http://cran.r-project.org/bin/linux/ubuntu/README Setting this as the last line in my sources.list: deb http://cran.fhcrc.org/bin/linux/ubuntu feisty/ When I typed in: [EMAIL PROTECTED]:/usr/local/lib/R/site-library$ sudo apt-get install r-base Reading package lists... Done Building dependency tree Reading state information... Done r-base is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. [EMAIL PROTECTED]:/usr/local/lib/R/site-library$ But when I go to R and check my version: version _ platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) My version is still 2.4.1. I must be missing something. What do I need to do to get R version 2.5 installed on my ubuntu feisty (7.04) system? Let me know if there is any additional information I need to give to be helped out with this. Thank you for taking a look at this, Sincerely, Matt __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histograms with strings, grouped by repeat count (w/ data)
Aha! So to expand that from the original expression, table(table(d$filename, d$email_addr)) 0 1 2 3 253 20 8 9 I think that is exactly what I'm looking for. I knew it must be simple!!! What does the 0 column represent? Also, does this tell me the same thing, filtered by Japan? table(table(d$filename, d$email_addr, d$country_residence)[d$country_residence==Japan]) 0 1 2 3 958 5 2 1 How does that differ logically from this? table(table(d$filename, d$email_addr)[d$country_residence==Japan]) 0 1 2 3 51 4 2 1 I don't understand why that produces different results. The first one adds a third dimension to the table, but limits that third dimension to a single element, Japan. Shouldn't it be the same? And again, what's that zero column? Thank you, Matt On 6/18/07, jim holtman [EMAIL PROTECTED] wrote: If you are running on windows, make sure you have 'recording' checked in the history window of the graphics. You can also put the output to a pdf file and view it later. If you use table on the counts matrix: table(counts) counts 0 1 2 3 253 20 8 9 this shows that there were 20 single tries, 8 files downloaded twice and 9 three times. Is this what you want? You can also get the indices of the non-zero entries by: which(counts != 0, arr.ind=TRUE) row col file11 1 file56 2 file11 3 file23 3 file78 4 file89 4 file11 5 file23 5 file23 6 . On 6/18/07, Matthew Trunnell [EMAIL PROTECTED] wrote: Jim, Thanks for the quick reply! When I run your code, I end up with a single barplot of one datapoint, file9 vs email20 == 2.0. I see the call to barplot is inside a for loop... maybe it's zooming through the display of many barplots, but all I see is the last one? In any case, I need to figure out the distribution of the retries, such as No. Retries Count 1 6 2 13 3 5 4 3 5 2 6 1 That is, 6 people retried the download once; 13 people retried the download twice, etc. So it would be counting the frequency of the email-filename combination, and grouping those together by the number of retries. Does that make sense? When I look at the counts object from your code, I can see that it's close to what I need. How do I access the properties of the counts object-- it's a table, right? If I look at counts[1,1], that returns 1. But how do I get at the row/col name of that cell? Is that cell an object? rownames(counts[1,1]) returns null. Thanks, Matt On 6/18/07, jim holtman [EMAIL PROTECTED] wrote: You should be using barplot and not hist. I think this produces what you want: x - filename,last_modified,email_addr,country_residence file1,3/4/2006 13:54,email1,Korea (South) file2,3/4/2006 14:33,email2,United States file2,3/4/2006 16:03,email2,United States file2,3/4/2006 16:17,email3,United States file2,3/4/2006 16:28,email3,United States file3,3/4/2006 19:13,email4,United States file2,3/4/2006 21:22,email5,India file4,3/4/2006 21:46,email6,United States file1,3/4/2006 22:04,email7,Japan file2,3/4/2006 22:09,email8,Croatia file1,3/4/2006 22:22,email7,Japan file1,3/4/2006 22:29,email9,India file1,3/4/2006 23:06,email6,United States file1,3/4/2006 23:33,email6,United States file5,3/4/2006 23:44,email10,China file1,3/5/2006 0:13,email9,India file2,3/5/2006 0:52,email8,Croatia file2,3/5/2006 0:54,email8,Croatia file2,3/5/2006 1:10,email5,India file6,3/5/2006 2:17,email9,India file2,3/5/2006 2:24,email11,Italy file7,3/5/2006 2:36,email12,Italy file8,3/5/2006 2:52,email12,Italy file2,3/5/2006 3:09,email13,United Kingdom file2,3/5/2006 4:02,email14,India file2,3/5/2006 4:07,email14,India file2,3/5/2006 4:14,email14,India file2,3/5/2006 4:37,email5,India file2,3/5/2006 4:44,email15,Belgium file1,3/5/2006 5:02,email9,India file1,3/5/2006 5:24,email16,Taiwan file2,3/5/2006 6:06,email17,Saudi Arabia file2,3/5/2006 7:32,email17,Saudi Arabia file2,3/5/2006 8:12,email18,Brazil file2,3/5/2006 8:26,email18,Brazil file2,3/5/2006 9:49,email19,United Kingdom file1,3/5/2006 10:49,email11,Italy file1,3/5/2006 11:16,email13,United Kingdom file1,3/5/2006 11:16,email13,United Kingdom file1,3/5/2006 11:45,email13,United Kingdom file1,3/5/2006 14:34,email20,Australia file9,3/5/2006 14:56,email20,Australia file9,3/5/2006 14:56,email20,Australia file5,3/5/2006 16:43,email21,United States file1,3/5/2006 17:17,email7,Japan file2,3/5/2006 17:26,email22,Japan file2,3/5/2006 17:27,email22,Japan file2,3/5/2006 17:33,email23,China file1,3/5/2006 17:45,email22,Japan file2,3/5/2006 17:45,email22,Japan file2,3/5/2006 17:59,email23,China file1,3/5/2006 18:27,email24,Japan
