Re: [R] file reading problem unique to windows. Err info: Error in file(file, ifelse(append, a, w)). cannot open the connection
We don't have any of the information asked for in the posting guide, such as your version of R, reproducible example But please try R-patched, since this might be • download.file() could leave the destination file open if the URL was not able to be opened. (PR#14414) (If you had followed the posting guide you would have tried R-patched before posting ) On Wed, 24 Nov 2010, Yong Wang wrote: Dear List I asked this question before, got some tips but can't get it solved. Where? You didn't give a reference, and it would have helped the helpers. it is clear now that this problem only occurs when run on windows (I tested it on windows XP) nothing wrong at all when run on unix. unfortunately, sometimes I have to run it on windows, I appreciate any suggestion on how to circumvent this problem when run on windows. below is the problem description I submitted before. # I am running a loop downloading web pages and save the html to a temporary file (use download.file() ) then read (using readLines) it in for processing; finally write useful info from each processed page to a unique file the problem is once the loop runs up to somewhere near 5000, it will throw out an err like below and won't go further. Error in file(file, ifelse(append, a, w)) : cannot open the connection - In the meantime, a request for new connection won't be successful, for example, a request for the help page of file will trigger err below --- ?file Error in gzfile(file, rb) : cannot open the connection In addition: Warning message: In gzfile(file, rb) : cannot open compressed file 'C:/PROGRA~1/R/R-211~1.1/library/stats/help/aliases.rds', probable reason 'Too many open files' --- I am not sure if the problem is too many connections not closed. since I close the file connection after each readLines. checking with showConnections(all=T) does not show excessive connections and closeAllConnections() does not help. Can any one help me on this? Any answer highly appreciated. yong __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to find a row index in a matrix or a data frame ?
Dear all, this looks pretty much a standard problem, but I couldn't find a satisfying and understandable solution. (A) Given a data frame (or matrix), e.g. x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5)) and a row of this data frame, e.g. r - c(2, 5) I need to find one row index i (or all such indices) such that r is at the i-th row in x, that is, the expression all(x[i,]==as.list(r)) evaluates to TRUE. I can not evaluate an expression like x[x[,1]==2 x[,2]==5,] because I do not know in advance how many columns x will have. Basically, thus, I'm looking for an equivalent of vectorfind in Scilab. (B) Which would be the most appropriate data type for x, matrix or data frame or another type? (C) What will be better, searching for rows or searching for columns? Thank you for your help! Best, Mirko __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find a row index in a matrix or a data frame ?
Mirko,
Here is a solution - I am sure other R mentors would find a more efficient
one
x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5))
x - rbind(x,x)
r - c(2,5)
(rowfind=apply(x,1,FUN=function(row,add){isTRUE(all.equal(row,add,check.attributes
= FALSE))},add=r))
1 2 3 4 5 6
FALSE TRUE TRUE FALSE TRUE TRUE
(rowfindindex=(1:nrow(x))[rowfind])
[1] 2 3 5 6
Hope this help,
Eric
2010/11/25 Luedde, Mirko mirko.lue...@sap.com
Dear all,
this looks pretty much a standard problem, but I couldn't find a
satisfying and understandable solution.
(A) Given a data frame (or matrix), e.g.
x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5))
and a row of this data frame, e.g.
r - c(2, 5)
I need to find one row index i (or all such indices) such that r
is at the i-th row in x, that is, the expression
all(x[i,]==as.list(r))
evaluates to TRUE. I can not evaluate an expression like
x[x[,1]==2 x[,2]==5,]
because I do not know in advance how many columns x will have.
Basically, thus, I'm looking for an equivalent of vectorfind in
Scilab.
(B) Which would be the most appropriate data type for x, matrix or
data frame or another type?
(C) What will be better, searching for rows or searching for columns?
Thank you for your help!
Best, Mirko
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eric Lecoutre
Consultant - Business Decision
Business Intelligence Customer Intelligence
[[alternative HTML version deleted]]
__
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Re: [R] Problem with plotting diagnostics - Error in object$coefficients : $ operator is invalid for atomic vectors
this problem seems to only exist in R 2.12.0 but not in R 2.11.1. any ideas? a bug? -- dr. katharina manderscheid soziologisches seminar universität luzern kasernenplatz 3 6000 luzern 7 tel. ++41 41 228 4657 web: http://www.unilu.ch/deu/dr.-katharina-manderscheid_346380.aspx -Ursprüngliche Nachricht- Von: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] Gesendet: Mittwoch, 17. November 2010 16:33 An: Manderscheid Katharina Cc: 'r-help@r-project.org' Betreff: Re: [R] Problem with plotting diagnostics - Error in object$coefficients : $ operator is invalid for atomic vectors On 17/11/2010 10:28 AM, Manderscheid Katharina wrote: hi all, after fitting a multiple linear regression model- lm(y ~ a + b+ c+d) i wanted to plot diagnostics plot(model) but get the error message Error in object$coefficients : $ operator is invalid for atomic vectors. which does not make a lot of sense, since there is no $ - i am working with an attached dataset. can anyone help, please?? thanks a lot, kat I just tried those lines (with fake data for a,b,c,d and y) and got no error message. I was using R 2.12.0. I think you need to show us a reproducible example, and the sessionInfo() to go with it, to help with this. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find a row index in a matrix or a data frame ?
rows - which(apply(mapply(x, r, FUN===), MARGIN=1, FUN=all)); /H On Thu, Nov 25, 2010 at 1:59 AM, Luedde, Mirko mirko.lue...@sap.com wrote: Dear all, this looks pretty much a standard problem, but I couldn't find a satisfying and understandable solution. (A) Given a data frame (or matrix), e.g. x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5)) and a row of this data frame, e.g. r - c(2, 5) I need to find one row index i (or all such indices) such that r is at the i-th row in x, that is, the expression all(x[i,]==as.list(r)) evaluates to TRUE. I can not evaluate an expression like x[x[,1]==2 x[,2]==5,] because I do not know in advance how many columns x will have. Basically, thus, I'm looking for an equivalent of vectorfind in Scilab. (B) Which would be the most appropriate data type for x, matrix or data frame or another type? (C) What will be better, searching for rows or searching for columns? Thank you for your help! Best, Mirko __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running R from SAS
Ziad, Also you didn't mention which version of SAS and which modules. In case you are working with 9.2 and have at disposal IML, you can have a look at http://support.sas.com/documentation/cdl/en/imlsstat/63545/HTML/default/viewer.htm#statr_toc.htm Eric 2010/11/24 ziad.elmou...@tnsglobal.com Hello All, I am interested in running an R program with several random seeds. One approach is to launch the program from SAS. The recommended approach is to use the X command as shown below: OPTIONS XWAIT XSYNC; X r.exe --no-save --quiet c:\temp\r\program.r c:\temp\r\program.log; However, this does not seem to work for me. Does anyone know how to launch an R program from SAS? Thank you in advance. Ziad Elmously ziad.elmou...@tnsglobal.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eric Lecoutre Consultant - Business Decision Business Intelligence Customer Intelligence [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find a row index in a matrix or a data frame ?
Hi all, thank you for quick help. Any ideas on (better) efficiency (with other data types)? Best, Mirko -Ursprüngliche Nachricht- Von: henrik.bengts...@gmail.com [mailto:henrik.bengts...@gmail.com] Im Auftrag von Henrik Bengtsson Gesendet: Donnerstag, 25. November 2010 11:25 An: Luedde, Mirko Cc: r-help@r-project.org Betreff: Re: [R] how to find a row index in a matrix or a data frame ? rows - which(apply(mapply(x, r, FUN===), MARGIN=1, FUN=all)); /H On Thu, Nov 25, 2010 at 1:59 AM, Luedde, Mirko mirko.lue...@sap.com wrote: Dear all, this looks pretty much a standard problem, but I couldn't find a satisfying and understandable solution. (A) Given a data frame (or matrix), e.g. x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5)) and a row of this data frame, e.g. r - c(2, 5) I need to find one row index i (or all such indices) such that r is at the i-th row in x, that is, the expression all(x[i,]==as.list(r)) evaluates to TRUE. I can not evaluate an expression like x[x[,1]==2 x[,2]==5,] because I do not know in advance how many columns x will have. Basically, thus, I'm looking for an equivalent of vectorfind in Scilab. (B) Which would be the most appropriate data type for x, matrix or data frame or another type? (C) What will be better, searching for rows or searching for columns? Thank you for your help! Best, Mirko __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find a row index in a matrix or a data frame ?
Try this: which((apply(x, 1, toString) %in% toString(r))) On Thu, Nov 25, 2010 at 7:59 AM, Luedde, Mirko mirko.lue...@sap.com wrote: Dear all, this looks pretty much a standard problem, but I couldn't find a satisfying and understandable solution. (A) Given a data frame (or matrix), e.g. x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5)) and a row of this data frame, e.g. r - c(2, 5) I need to find one row index i (or all such indices) such that r is at the i-th row in x, that is, the expression all(x[i,]==as.list(r)) evaluates to TRUE. I can not evaluate an expression like x[x[,1]==2 x[,2]==5,] because I do not know in advance how many columns x will have. Basically, thus, I'm looking for an equivalent of vectorfind in Scilab. (B) Which would be the most appropriate data type for x, matrix or data frame or another type? (C) What will be better, searching for rows or searching for columns? Thank you for your help! Best, Mirko __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in improving the style of R plots
On 25/11/2010 5:38 AM, pilchat wrote: Dear all, I am running a MCMC on my data, and I want to plot the results of the simulation. I want to dedicate a page in the PS file for each element in my sample, and in each page I need to plot 8 quadrants. In attachment you find my first attemp (just the first page). Here are my troubles (I'm sorry if they are stupid, but I am a newby with R): -) I need to substitute the greek letters wherever you find the full spelling (I also need to substitute Surface with the greek Sigma letter) See ?plotmath. The general idea is that you would use xlab = expression(alpha) instead of xlab = alpha -)in the top plots I need to write the variable=median+-stddeviation (units) in a more readable way: some characters overlap and I want to approximate the numebrs to the second decimal digit. This is how I have been trying so far: text(min(mchain[1,]),max(ndensity$y),substitute( Log(Surface)[med] ==nmed%+-%nstd,list(nmed=nmed,nstd=nstd) ),pos=4 ) You have some strange font problems. I would guess that you produced the plots by copying from the screen or another device: don't do that, it gets the spacing wrong. Use pdf() or postscript() to open the device and plot directly to it in the desired final size. -) plot (c): ticks, annotations and labels overlap, but they shouldn't. Same problem as above. Here is the code: persp(d,col=fcol,zlim=zlim,theta=theta,phi=phi,zlab=density,xlab=xlab,ylab=ylab,main=,sub=(c),ticktype=detailed) -) plot (e) and (f): I need to cancel plot (f) and place plot (e) at the center of the bottom row The layout() function might help with this. -) when I close the R session, the content of the PS file disappears. How can I retain the graphics output after closing R? I think you are misunderstanding what's going on. R deletes temporary files when it closes, but you wouldn't normally create a plot in a temporary file. Duncan Murdoch Can you help me in getting what I wish? Any help is greatly appreciated. Thanks a lot Gaetano __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] \Sweaveopts error
I have a file 4lmetc.Rnw, intended for inclusion in a LaTeX document,
that starts:
\SweaveOpts{engine=R, keep.source=TRUE}
\SweaveOpts{eps=FALSE, prefix.string=snArt/4lmetc}
The attempt to process the file through Sweave generates the error:
Sweave(4lmetc)
Writing to file 4lmetc.tex
Processing code chunks ...
1 : keep.source term verbatim
Error in file(srcfile$filename, open = rt, encoding = encoding) :
cannot open the connection
In addition: Warning message:
In file(srcfile$filename, open = rt, encoding = encoding) :
cannot open file '4lmetc': No such file or directory
The same file processes through Stangle() without problems.
If I comment out the \Sweaveopts lines, there is no problem,
except that I do not get the options that I want.
This processed fine in R-2.11.1
sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets grid methods
[8] base
other attached packages:
[1] lattice_0.19-13 DAAG_1.02 randomForest_4.5-36
[4] rpart_3.1-46MASS_7.3-8 reshape_0.8.3
[7] plyr_1.2.1 proto_0.3-8
loaded via a namespace (and not attached):
[1] ggplot2_0.8.8 latticeExtra_0.6-14
Is there a workaround?
John Maindonald email: john.maindon...@anu.edu.au
phone : +61 2 (6125)3473fax : +61 2(6125)5549
Centre for Mathematics Its Applications, Room 1194,
John Dedman Mathematical Sciences Building (Building 27)
Australian National University, Canberra ACT 0200.
http://www.maths.anu.edu.au/~johnm
__
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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] Generalized linear models with categorical data
I got a question about using a GLZ with categorical x categorical data. Below there is a data set I want to know the influence of treatments (CONT, and LPS2H LPS24H) on the categories of pigmentation of the right testis of an amphibian. From these data, I used the function glm with binomial family (logit). But in the result (see below) is not possible to know the influence of the three treatments on the categories, only the treatments as a whole. Could someone help me? Thanks in advance td=read.table(file.choose(), h=T) td Tratamento Categoria 1CONT Cat.2 2CONT Cat.2 3CONT Cat.2 4CONT Cat.2 5CONT Cat.3 6CONT Cat.3 7CONT Cat.3 8CONT Cat.3 9CONT Cat.3 10 CONT Cat.3 11 LPS2h Cat.2 12 LPS2h Cat.2 13 LPS2h Cat.2 14 LPS2h Cat.3 15 LPS2h Cat.3 16 LPS2h Cat.3 17 LPS2h Cat.3 18 LPS2h Cat.3 19 LPS2h Cat.3 20 LPS2h Cat.3 21 LPS24h Cat.1 22 LPS24h Cat.1 23 LPS24h Cat.1 24 LPS24h Cat.1 25 LPS24h Cat.1 26 LPS24h Cat.2 27 LPS24h Cat.2 28 LPS24h Cat.2 29 LPS24h Cat.2 30 LPS24h Cat.2 mod.test=glm(Categoria~Tratamento, family=binomial(logit), data=td) mod.test Call: glm(formula = Categoria ~ Tratamento, family = binomial(logit), data = td) Coefficients: (Intercept) TratamentoLPS24h TratamentoLPS2h 2.057e+01-2.057e+01 1.558e-08 Degrees of Freedom: 29 Total (i.e. Null); 27 Residual Null Deviance:27.03 Residual Deviance: 13.86 AIC: 19.86 anova(mod.test, test=Chisq) Analysis of Deviance Table Model: binomial, link: logit Response: Categoria Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(|Chi|) NULL 29 27.034 Tratamento 2 13.17127 13.863 0.001380 ** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 -- Atenciosamente, *Diogo Borges Provete* == Biólogo Mestre em Biologia Animal (UNESP) Laboratório de Ecologia Animal Departamento de Zoologia e Botânica Instituto de Biociências, Letras e Ciências Exatas Universidade Estadual Paulista - UNESP São José do Rio Preto-SP Brazil Rua Cristóvão Colombo, 2265 Jardim Nazareth - 15054-000 *Skype*: diogoprovete *MSN*: diogop...@yahoo.com.br *Personal web page https://sites.google.com/site/diogoprovetepage/* Traduza conosco: American Journal Experts http://www.journalexperts.com/br/ D-Lang Soluções linguisticashttp://www.d-lang.com.br/site/sitept/index.htm == -- Atenciosamente, *Diogo Borges Provete* == Biólogo Mestre em Biologia Animal (UNESP) Laboratório de Ecologia Animal Departamento de Zoologia e Botânica Instituto de Biociências, Letras e Ciências Exatas Universidade Estadual Paulista - UNESP São José do Rio Preto-SP Brazil Rua Cristóvão Colombo, 2265 Jardim Nazareth - 15054-000 *Skype*: diogoprovete *MSN*: diogop...@yahoo.com.br *Personal web page https://sites.google.com/site/diogoprovetepage/* Traduza conosco: American Journal Experts http://www.journalexperts.com/br/ D-Lang Soluções linguisticashttp://www.d-lang.com.br/site/sitept/index.htm == [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] \Sweaveopts error
On 25/11/2010 6:34 AM, John Maindonald wrote:
I have a file 4lmetc.Rnw, intended for inclusion in a LaTeX document,
that starts:
I think this may have been fixed in the patched version. Could you give
it a try to confirm? If not, please send me a simplified version of the
file, and I'll see what's going wrong.
Duncan Murdoch
\SweaveOpts{engine=R, keep.source=TRUE}
\SweaveOpts{eps=FALSE, prefix.string=snArt/4lmetc}
The attempt to process the file through Sweave generates the error:
Sweave(4lmetc)
Writing to file 4lmetc.tex
Processing code chunks ...
1 : keep.source term verbatim
Error in file(srcfile$filename, open = rt, encoding = encoding) :
cannot open the connection
In addition: Warning message:
In file(srcfile$filename, open = rt, encoding = encoding) :
cannot open file '4lmetc': No such file or directory
The same file processes through Stangle() without problems.
If I comment out the \Sweaveopts lines, there is no problem,
except that I do not get the options that I want.
This processed fine in R-2.11.1
sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets grid methods
[8] base
other attached packages:
[1] lattice_0.19-13 DAAG_1.02 randomForest_4.5-36
[4] rpart_3.1-46MASS_7.3-8 reshape_0.8.3
[7] plyr_1.2.1 proto_0.3-8
loaded via a namespace (and not attached):
[1] ggplot2_0.8.8 latticeExtra_0.6-14
Is there a workaround?
John Maindonald email: john.maindon...@anu.edu.au
phone : +61 2 (6125)3473fax : +61 2(6125)5549
Centre for Mathematics Its Applications, Room 1194,
John Dedman Mathematical Sciences Building (Building 27)
Australian National University, Canberra ACT 0200.
http://www.maths.anu.edu.au/~johnm
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] RODBC
Hi, I am running the RODBC examples form the help guide. I am trying to UPDATE a table in an Access data base but I am having an error. library(RODBC) library(termstrc) path = getwd() setwd(getwd()) dbName = data.mdb pathdbname = paste(path,/,dbName,sep=) accesChannel = odbcConnectAccess(pathdbname, uid = , pwd = ) sqlSave(accesChannel, USArrests, rownames = state, addPK=TRUE) sqlFetch(accesChannel , USArrests, rownames = state) # get the lot foo - cbind(state=row.names(USArrests), USArrests)[1:3, c(1,3)] foo[1:3,2] - sqlUpdate(accesChannel , foo, USArrests) The sqlSave and sqlFetch command seem to work fine. foo state Assault Alabama Alabama Alaska Alaska Arizona Arizona sqlUpdate(accesChannel , foo, USArrests) Error in sqlUpdate(accesChannel, foo, USArrests) : cannot update 'USArrests' without unique column I am using R 2.12.0(2010-10-15) Using Microsoft access 2003. Furthermore, the sqlColumns(accesChannel , USArrests) returns the following information sqlColumns(accesChannel , USArrests) TABLE_CAT TABLE_SCHEM TABLE_NAME 1 C:\\ARTIFICALDESKTOP\\CurrentDownloads\\termstrc\\dataNA USArrests 2 C:\\ARTIFICALDESKTOP\\CurrentDownloads\\termstrc\\dataNA USArrests 3 C:\\ARTIFICALDESKTOP\\CurrentDownloads\\termstrc\\dataNA USArrests 4 C:\\ARTIFICALDESKTOP\\CurrentDownloads\\termstrc\\dataNA USArrests 5 C:\\ARTIFICALDESKTOP\\CurrentDownloads\\termstrc\\dataNA USArrests COLUMN_NAME DATA_TYPE TYPE_NAME COLUMN_SIZE BUFFER_LENGTH DECIMAL_DIGITS 1 state12 VARCHAR 255 510 NA 2 Murder 8DOUBLE 53 8 NA 3 Assault 4 INTEGER 10 4 0 4UrbanPop 4 INTEGER 10 4 0 5Rape 8DOUBLE 53 8 NA NUM_PREC_RADIX NULLABLE REMARKS COLUMN_DEF SQL_DATA_TYPE SQL_DATETIME_SUB 1 NA1NA NA12 NA 2 21NA NA 8 NA 3 101NA NA 4 NA 4 101NA NA 4 NA 5 21NA NA 8 NA CHAR_OCTET_LENGTH ORDINAL_POSITION IS_NULLABLE ORDINAL 1 5101 YES 1 2NA2 YES 2 3NA3 YES 3 4NA4 YES 4 5NA5 YES 5 Any ideas as of what might I have missed? Thanks, Jorge [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an equivalent to predict(..., type=linear) of a Proportional hazard model for a Cox model instead?
I manage to achieve similar results with a Cox model as follows but I don't really understand why we have to take the inverse of the linear prediction with the Cox model and why we do not need to divide by the number of days in the year anymore? Am I getting a similar result out of pure luck? thanks in advance, Ben # MASS example with the proportional hazard model par(mfrow = c(1, 2)); (aids.ps - survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83), levels = levels(Aidsp$state)), T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age = 0:82), se = T, type = linear) plot(0:82, exp(zz$fit)/365.25, type = l, ylim = c(0, 2), xlab = age, ylab = expected lifetime (years)) lines(0:82, exp(zz$fit+1.96*zz$se.fit)/365.25, lty = 3, col = 2) lines(0:82, exp(zz$fit-1.96*zz$se.fit)/365.25, lty = 3, col = 2) rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015) # same example but with a Cox model instead (aids.pscp - coxph(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW, 83), levels = levels(Aidsp$state)), T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age = 0:82), se = T, type = lp) plot(0:82, 1/exp(zzcp$fit), type = l, ylim = c(0, 2), xlab = age, ylab = expected lifetime (years)) lines(0:82, 1/exp(zzcp$fit+1.96*zzcp$se.fit), lty = 3, col = 2) lines(0:82, 1/exp(zzcp$fit-1.96*zzcp$se.fit), lty = 3, col = 2) rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] looking for the RMySQL package for R 2.12.0 under XP
Thank you for all your answers.
I've looked at RODBC but it seems that you have to declare the database with
the administration tools before using it . With RMySQL, you can do it
directly in the R-script which is very convenient for me as I regularly
treat new databases.
For those who already used rtool :
- are the administration rights are requested (to install or/and to use it)
?
- after the installation, install.packages('RMySQL',type='source') is the
only things to do or more skills in computer are needed ?
I will have a look at the R-sig-DB list if I can't do it alone.
Again thanks for your help and advices.
Ptit Bleu
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[R] difficulty setting the random = argument to lme()
My small brain is having trouble getting to grips with lme() I wonder if anyone can help me correctly set the random = argument to lme() for this kind of setup with (I think) 9 variance/covariance components ... Study.1 Study.2 ... Study.10 Treatment.A: subject: 1 2 3 4 5 6 etc. 28 29 30 Treatment.B: subject: 31 32 3334 35 36 58 59 60 A variable is measured at 2 fixed sites (A and B) on each subject so we have fixed effects :- between-Treatments between-sites (A and B) Treatment*site interaction and we have random effects :- study effects at site A study effects at site B correlation between site A and site B study effects study*treatment interaction effects at site A study*treatment interaction effects at site B correlation between site A and B study*treatment interaction effects residual (between-subject) effects at site A residual (between-subject) effects at site B correlation between site A and B residuals (between-subject) effects My problem is formulating the random = argument to give estimates of all 9 random components ... Hope someone can help ... Robert Kinley Study: Pos tissue VC, Neg tissue VC, Pos/Neg tissue correlation Study*Group:Pos tissue VC, Neg tissue VC, Pos/Neg tissue correlation Residual (animal): Pos tissue VC, Neg tissue VC, Pos/Neg tissue correlation [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] t-stat for the coefficients of an ARIMA model
Dear all, I am fitting a time series using the following command: Ts.arima-arima(x,c(2,1,2)) where x is a time series. What the function returns is perfectly fine but I was wondering if I could access to the t-stat of the coefficients I got from the arima function. Any help would be greatly appreciated. Samuel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find a row index in a matrix or a data frame ?
On Thu, Nov 25, 2010 at 4:59 AM, Luedde, Mirko mirko.lue...@sap.com wrote: Dear all, this looks pretty much a standard problem, but I couldn't find a satisfying and understandable solution. (A) Given a data frame (or matrix), e.g. x - data.frame(A=c(1, 2, 2), B=c(4, 5, 5)) and a row of this data frame, e.g. r - c(2, 5) I need to find one row index i (or all such indices) such that r is at the i-th row in x, that is, the expression all(x[i,]==as.list(r)) evaluates to TRUE. I can not evaluate an expression like x[x[,1]==2 x[,2]==5,] because I do not know in advance how many columns x will have. Basically, thus, I'm looking for an equivalent of vectorfind in Scilab. (B) Which would be the most appropriate data type for x, matrix or data frame or another type? (C) What will be better, searching for rows or searching for columns? Try this: which(apply(t(x) == r, 2, all)) or this: which(colSums((t(x) != r) 0) == 0) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] txtProgressBar strange behavior in R 2.12.0
I've seen it, but not reliably enough to reproduce. The issue is that there isn't really support for '\r' (CR) in Rgui, and so it was sometimes getting junk characters. However, I think that this is probably fixed in R-patched = r53662, so please try an update in a day or two. On Mon, 22 Nov 2010, Viechtbauer Wolfgang (STAT) wrote: I believe nobody has responded to far, so maybe this is not a wide-spread issue. However, I have also encountered this since upgrading to R 2.12.0 (Windows 7, 64-bit). In my simulations where I use txtProgressBar(), the problem usually disappears after the bar has progressed to a certain amount, but it's quite strange nonetheless. The characters that appear are gibberish and include some Asian symbols. Here is a screenshot: http://www.wvbauer.com/screenshot.jpg sessionInfo(): R version 2.12.0 (2010-10-15) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base No idea what could be causing this. Best, -- Wolfgang Viechtbauer Department of Psychiatry and Neuropsychology School for Mental Health and Neuroscience Maastricht University, P.O. Box 616 6200 MD Maastricht, The Netherlands Tel: +31 (43) 368-5248 Fax: +31 (43) 368-8689 Web: http://www.wvbauer.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jalla Sent: Saturday, November 20, 2010 23:49 To: r-help@r-project.org Subject: [R] txtProgressBar strange behavior in R 2.12.0 Hi, I am running R 2.12.0 (windows). example(txtProgressBar) gives me some funny screen output with all kinds of special characters appearing and disappearing. It's happening on two different mashines since vs. 2.12.0. Is this a known issue? Best, Jalla -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lifting Wavelet Transform
( I'm top posting for clarity, hotmail decided not to original text and I'm not going to even try to outguess it LOL) I have a non-definitive negative answer : I found two packages but it isn't obvious either expose anything related to lifting. I haven't worked with these lately and so not sure I missed obvious synonyms etc. http://cran.r-project.org/web/packages/wavelets/index.html http://cran.r-project.org/web/packages/waveslim/index.html and normally I would have documentation available for grepping for various terms but quick scan of pdf didn't show anything related to lifting. AFAIK, the wavelet designs are limited to fixed types/lengths and there is no explicit method for deriving new ones and the WT impl is not obvious on a quick read but looks like filters just take coefs and time series and output mra. http://en.wikipedia.org/wiki/Lifting_scheme The Daubechies Sweldens paper seems to be available online etc. Date: Wed, 24 Nov 2010 20:46:30 -0800 From: assaed...@yahoo.com To: r-help@r-project.org Subject: [R] Lifting Wavelet Transform Hi R users Thanks in advance Is lifting wavelet transform implemented in R? If so, which package or codes can be used for performing that? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Filled contour plot showing labeled isolines?
jt306 wrote: Is it possible to create a contour plot with the isolines labeled. I know you can do this with Matlab. Argh! I tried creating a filled contour plot, then using par(new=T), followed by overlaying the contour plot on top. However, the placement of the filled contour plot and the contour plot do not align correctly. Any suggestions would be appreciated. Thanks, Jon Jon, I don't use filled.contour but image and then contour. Try this. image(volcano) contour(volcano,add=TRUE) Is this what you want? Regards, Fiona -- View this message in context: http://r.789695.n4.nabble.com/Filled-contour-plot-showing-labeled-isolines-tp3056437p3058974.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Filled contour plot showing labeled isolines?
If you usre filled.contour, then use contour() as a part of the function
supplied as the axis parameter, you can correctly overlay contours on
the colour contour plot.
Steve e
-Original Message-
From: Fiona Berryman f.berry...@wlv.ac.uk
To: r-help@r-project.org
Sent: 11/25/2010 13:57:06
Subject: Re: [R] Filled contour plot showing labeled isolines?
jt306 wrote:
Is it possible to create a contour plot with the isolines labeled. I
know
you can do this with Matlab. Argh!
I tried creating a filled contour plot, then using par(new=T),
followed by
overlaying the contour plot on top. However, the placement of the
filled
contour plot and the contour plot do not align correctly. Any
suggestions
would be appreciated.
Thanks,
Jon
Jon,
I don't use filled.contour but image and then contour.
Try this.
image(volcano)
contour(volcano,add=TRUE)
Is this what you want?
Regards,
Fiona
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[R] What to do if geoRglm results showed that a non-spacial model fits better?
Hi R-floks: Working in geoRglm, it shows me, according to AIC criterion, that the non-spacial model describes the process in a better way. It's the first time that I'm facing up to. These are my results: OP2003Seppos.AICnonsp-OP2003Seppos.AICsp #[1] -4 (OP2003Seppos.lf0.p-exp(OP2003Seppos.lf0$beta)/(1+exp(OP2003Seppos.lf0$beta))) #P non spatial #[1] 0.9717596 (OP2003Seppos.lf.p-exp(OP2003Seppos.lf$beta)/(1+exp(OP2003Seppos.lf$beta))) #P spatial #[1] 0.9717596 It must what have an important influence at kriging, because it shows as following: OP2003Sepposbin.krig-glsm.krige(OP2003Seppos.tune,loc=OP2003Seppospro.pred.grid,bor=OP2003Sepposbor) #glsm.krige: Prediction for a generalised linear spatial model #There are 50 or mode advices (use warnings() to see them) # warnings() #Warning messages: #1: In asympvar(kpl.result$predict, messages = FALSE) ... : # value of argument lag.max is not suffiently long #2: In asympvar(kpl.result$predict, messages = FALSE) ... : # value of argument lag.max is not suffiently long I don't understand the warnings. Help me, please. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an equivalent to predict(..., type=linear) of a Proportional hazard model for a Cox model instead?
On Nov 25, 2010, at 7:27 AM, Ben Rhelp wrote: I manage to achieve similar results with a Cox model as follows but I don't really understand why we have to take the inverse of the linear prediction with the Cox model Different parameterization. You can find expanded answer(s) in the archives and in the documentation of survreg.distributions. and why we do not need to divide by the number of days in the year anymore? Here I'm guessing (since you don't offer enough evidence to confirm) that the difference is in the time scales used in your Aidsp$survtime versus some other example to which you are comparing . data(Aidsp) Warning message: In data(Aidsp) : data set 'Aidsp' not found Am I getting a similar result out of pure luck? thanks in advance, Ben # MASS example with the proportional hazard model par(mfrow = c(1, 2)); (aids.ps - survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83), levels = levels(Aidsp$state)), T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age = 0:82), se = T, type = linear) plot(0:82, exp(zz$fit)/365.25, type = l, ylim = c(0, 2), xlab = age, ylab = expected lifetime (years)) lines(0:82, exp(zz$fit+1.96*zz$se.fit)/365.25, lty = 3, col = 2) lines(0:82, exp(zz$fit-1.96*zz$se.fit)/365.25, lty = 3, col = 2) rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015) # same example but with a Cox model instead (aids.pscp - coxph(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW, 83), levels = levels(Aidsp$state)), T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age = 0:82), se = T, type = lp) plot(0:82, 1/exp(zzcp$fit), type = l, ylim = c(0, 2), xlab = age, ylab = expected lifetime (years)) lines(0:82, 1/exp(zzcp$fit+1.96*zzcp$se.fit), lty = 3, col = 2) lines(0:82, 1/exp(zzcp$fit-1.96*zzcp$se.fit), lty = 3, col = 2) rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015) David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] overlap cdf plots and add colors and etc
On 2010-11-24 22:24, Roslina Zakaria wrote: Hi Jorge, I tried but still it does not work. Thank you for your time. Jorge's code works perfectly well. If you prefer lines() over plot(, add = TRUE), then use lines(): plot( ecdf(rnorm(15, sd=3)), verticals=TRUE, col=black) lines(ecdf(rnorm(11, sd=3)), verticals=TRUE, col=red) legend(topright, legend = c(observed,fitted), col = c(black,red), pch=c(NA,NA), lty = c(1, 1), lwd=c(3,3),bty=n, pt.cex=2) (Note that topright is probably the worst choice you can make. And why set pt.cex when you don't have points??) The key thing is that you can't specify your two colours in the first plot() call. At that point, R has no idea that you want to add to the plot. (I understand that Greg Snow is working on a mind-reading package, but so far it can probably read only his mind, not yours.) Here is a (very brief) reminder of how to post: 1. Use an informative subject line. You've done that. Good! 2. *Never* use the phrase it does not work. That is meaningless. Be specific about your problem. 3. Provide *reproducible* code. We don't have your 'datobs' or 'gam_sum_gen'. 4. Try to make the code *minimal*. It's not likely that anyone cares what labels you use for your axes/title (unless that's the problem). And nobody wants to see reams of data; use rnorm(), etc, or built-in data if possible. 5. Figure out how to set your mail program to send plain text. I know the above (and more) is in the posting guide, but it seems that nobody wants to read that quite brief document. Peter Ehlers From: Jorge Ivan Velezjorgeivanve...@gmail.com Cc: r-help@r-project.org Sent: Thu, November 25, 2010 4:46:37 PM Subject: Re: [R] overlap cdf plots and add colors and etc Hi Roslina, Try par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(rnorm(100))) plot(ecdf(rnorm(100)), add = TRUE, col = 2) HTH, Jorge On Thu, Nov 25, 2010 at 12:18 AM, Roslina Zakaria wrote: Hi r-users, I would like to overlap 2 ecdf plots. I tried this below and it gives me two plots of ecdf but just both just in black. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(datobs)) lines(ecdf(gam_sum_gen)) Then I try to add colors etc and also the legend but fail. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(datobs),main =CDF of the sum for winter season-Hume,cex.axis=1.2,xlab=Rainfall (mm), pch = pch,verticals=TRUE,col=c(black,red), lty=c(1,1),ylab=Cumulative probability, xlim=c(0,800),lwd=1) lines(ecdf(gam_sum_gen)) legend(topright, legend = c(observed,fitted), col = c(black,red), pch=c(NA,NA), lty = c(1, 1), lwd=c(3,3),bty=n, pt.cex=2) box() I'm not sure why it doesn't show up at all. Thank you for any help given. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to change value of y axis from log relative Hazard to relative Hazard
On Nov 25, 2010, at 2:00 AM, jsntxt wrote: http://r.789695.n4.nabble.com/file/n3058505/file.csv file.csv Hi, Rusers I have a problem in making a rcspline.plot with a Hmisc package. My data is in the upload attachment. My programme as follows: library(Hmisc) A-read.csv(file.csv,header=TRUE) attach(A) rcspline .plot (factor ,Time ,model = cox ,xrange = c (0,3 ),ylim = c (-1,2 ),event = event ,nk = 4 ,knots = c (0.8,1.0,1.5,2.0 ),showknots=TRUE,plotcl=FALSE,statloc=none,subset=SEX==2,lty=2) The plot could be made but its y axis represents the value of log relative Hazard, is there any method to change value of y axis from log relative Hazard to relative Hazard? I have never used that function. For the task you are facing, Harrell provides tools in the form of Predict with the fun argument = exp and the usual ploting functions for rms -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Applying function to elements of matrices in a list
Hello R-help, Please cc me on all responses, as I only receive summary emails from this list. I'm wondering if anybody has any tips on how to accomplish this efficiently. I have a list of matrices, and I'm trying to get the mean of the [i,j]'th element of each matrix in a list. So if I have a list of matrices, say x - list(a=matrix(rnorm(4),nrow=2),b=matrix(rnorm(4),nrow=2)) How would I get a 2x2 matrix, where the i,j'th element would be the mean across the the list of each of the i,j'th elements in the list? That is, where the [1,2] element would be the average of a[1,2] and b[1,2]. Of course my list and matrices are much larger, and I was hoping there would be some trick with lapply that I may be missing here. Thanks, Brian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Applying function to elements of matrices in a list
try this: Reduce(+, x) / length(x) Best, Dimitris On 11/25/2010 3:42 PM, statmobile wrote: Hello R-help, Please cc me on all responses, as I only receive summary emails from this list. I'm wondering if anybody has any tips on how to accomplish this efficiently. I have a list of matrices, and I'm trying to get the mean of the [i,j]'th element of each matrix in a list. So if I have a list of matrices, say x - list(a=matrix(rnorm(4),nrow=2),b=matrix(rnorm(4),nrow=2)) How would I get a 2x2 matrix, where the i,j'th element would be the mean across the the list of each of the i,j'th elements in the list? That is, where the [1,2] element would be the average of a[1,2] and b[1,2]. Of course my list and matrices are much larger, and I was hoping there would be some trick with lapply that I may be missing here. Thanks, Brian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/biostatistiek/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Applying function to elements of matrices in a list
On 11/25/2010 09:44 AM, Dimitris Rizopoulos wrote: try this: Reduce(+, x) / length(x) Thanks Dimitris, that's very slick, I was unaware of this Reduce function. The issue, is that I actually wanted to do a trimmed mean, and if things prove possible even the median. Is there a way to apply a generic function in the manner I described? Thanks, Brian Best, Dimitris On 11/25/2010 3:42 PM, statmobile wrote: Hello R-help, Please cc me on all responses, as I only receive summary emails from this list. I'm wondering if anybody has any tips on how to accomplish this efficiently. I have a list of matrices, and I'm trying to get the mean of the [i,j]'th element of each matrix in a list. So if I have a list of matrices, say x - list(a=matrix(rnorm(4),nrow=2),b=matrix(rnorm(4),nrow=2)) How would I get a 2x2 matrix, where the i,j'th element would be the mean across the the list of each of the i,j'th elements in the list? That is, where the [1,2] element would be the average of a[1,2] and b[1,2]. Of course my list and matrices are much larger, and I was hoping there would be some trick with lapply that I may be missing here. Thanks, Brian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an equivalent to predict(..., type=linear) of a Proportional hazard model for a Cox model instead?
Hi David,
Thank you for your reply. See below for more information.
From: David Winsemius
On Nov 25, 2010, at 7:27 AM, Ben Rhelp wrote:
I manage to achieve similar results with a Cox model as follows but I don't
really understand why we have to take the inverse of the linear prediction
with
the Cox model
Different parameterization. You can find expanded answer(s) in the archives
and in the documentation of survreg.distributions.
I understand (i think) the difference in model structures between a Cox (coxph)
and Proportional hazard model (survreg).
and why we do not need to divide by the number of days in the year
anymore?
Here I'm guessing (since you don't offer enough evidence to confirm) that
the
difference is in the time scales used in your Aidsp$survtime versus some
other
example to which you are comparing .
Both models are run from the same data, so I am not expecting any differences
in
time scales.
To get similar results, I need to actually run the following equations:
expected_lifetime_in_years = exp(fit)/365.25--- Linear
prediction of the Proportional hazard model
expected_lifetime_in_years = 1/exp(fit)--- Linear
prediction of the Cox model
where fit come from the linear prediction of each models, respectively.
Actually, in the code below, I re-run the models and predictions based on a
yearly sampling time (instead of daily).
Again, to get similar results, I now need to actually run the following
equations:
expected_lifetime_in_years = exp(fit) --- Linear
prediction of the Proportional hazard model
expected_lifetime_in_years = 1/exp(fit)--- Linear
prediction of the Cox model
I think I understand the logic behind the results of the proportional hazard
model, but not for the prediction of the Cox model.
Thank you for your help. I hope this is not a too stupid hole in my logic.
Here is the self contained R code to produce the charts:
library(MASS);
library(survival);
#Same data but parametric fit
make.aidsp - function(){
cutoff - 10043 # July 1987 in julian days
btime - pmin(cutoff, Aids2$death) - pmin(cutoff, Aids2$diag)
atime - pmax(cutoff, Aids2$death) - pmax(cutoff, Aids2$diag)
survtime - btime + 0.5*atime
status - as.numeric(Aids2$status)
data.frame(survtime, status = status - 1, state = Aids2$state,
T.categ = Aids2$T.categ, age = Aids2$age, sex = Aids2$sex)
}
Aidsp - make.aidsp()
# MASS example with the proportional hazard model
par(mfrow = c(1, 2));
(aids.ps - survreg(Surv(survtime + 0.9, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83), levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age =
0:82), se = T, type = linear)
plot(0:82, exp(zz$fit)/365.25, type = l, ylim = c(0, 2), xlab = age, ylab =
expected lifetime (years))
lines(0:82, exp(zz$fit+1.96*zz$se.fit)/365.25, lty = 3, col = 2)
lines(0:82, exp(zz$fit-1.96*zz$se.fit)/365.25, lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015)
# Same example but with a Cox model instead
(aids.pscp - coxph(Surv(survtime + 0.9, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW, 83), levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age =
0:82), se = T, type = lp)
plot(0:82, 1/exp(zzcp$fit), type = l, ylim = c(0, 2), xlab = age, ylab =
expected lifetime (years))
lines(0:82, 1/exp(zzcp$fit+1.96*zzcp$se.fit), lty = 3, col = 2)
lines(0:82, 1/exp(zzcp$fit-1.96*zzcp$se.fit), lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015)
# Change the sampling time from daily to yearly
par(mfrow = c(1, 1));
(aids.ps - survreg(Surv((survtime + 0.9)/365.25, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83), levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age =
0:82), se = T, type = linear)
plot(0:82, exp(zz$fit), type = l, ylim = c(0, 2), xlab = age, ylab =
expected lifetime (years))
(aids.pscp - coxph(Surv((survtime + 0.9)/365.25, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW, 83), levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)), age =
0:82), se = T, type = lp)
lines(0:82, 1/exp(zzcp$fit),col=4)
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Re: [R] Help on running regression by grouping firms
Hi: You could try something like this: For illustration, I'll use a data frame that was presented in a recent post to the ggplot2 group. The poster wanted regressions by individual, but you can add more than one grouping variable to the code I show below. It uses the plyr package. library(plyr) ds_test - structure(list(individual = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L), .Label = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30), class = factor), time = c(0L, 1671L, 1896L, 0L, 105L, 196L, 384L, 582L, 797L, 998L, 1419L, 0L, 290L, 451L, 752L, 0L, 487L, 619L, 820L, 0L, 384L, 463L, 832L, 932L, 1322L, 1688L, 0L, 101L, 390L, 0L, 746L, 761L, 899L, 1118L, 1236L, 1375L, 0L, 544L, 870L, 927L, 1117L, 1870L, 0L, 326L, 383L, 573L, 1326L, 0L, 1572L, 1592L), size = c(2, 2.6, 2.6, 1.2, 1.4, 1.5, 1.6, 1.7, 1.8, 2, 2.2, 1.3, 1.6, 1.5, 1.5, 2.8, 2.8, 2.4, 2.9, 2.1, 2.4, 2.4, 2.4, 2.3, 2.5, 2.4, 6, 5.8, 5.4, 1.1, 1.6, 1.5, 1.5, 1.5, 2.3, 2.3, 3.2, 4.1, 4, 3.9, 4.1, 4.3, 1.2, 2.1, 2.2, 2.2, 3, 2.2, 3, 3.9)), .Names = c(individual, time, size), row.names = c(NA, 50L), class = data.frame) # Run models by individual and put the results into a list. The advantage # is that one can extract multiple pieces from each component of the list, # if so desired, by writing simple extraction functions using plyr. dlply() is an # apply-like function: the first letter indicates that the input object (first # argument) is a data frame and that the output object after executing # the function is a list (in this case, a list of lists). The anonymous function # in the call performs the desired operation on each generic data subset x. mods - dlply(ds_test, .(individual), function(x) lm(size ~ time, data = x)) # This function does the actual work within subgroup; since the number # of residuals will vary from group to group, the output of the calling # function has to be a list object of residuals, one component per individual. # The outer function do.call() is intended to collapse the list object into a # vector, and the resulting vector can be attached to the original data frame # with $: res - function(x) resid(x) ds_test$u - do.call(c, llply(mods, res)) In your case, where you have multiple grouping factors, you may have to be a little more careful, but the strategy is the same. You could possibly reduce it to a one-liner (untested): ds_test$u - do.call(c, dlply(ds_test, .(individual), function(x) resid(lm(size ~ time, data = x HTH, Dennis On Wed, Nov 24, 2010 at 4:56 PM, Ray Zhang lz...@sfu.ca wrote: Hi there, I have a huge data set with multiple firms years and other firm characteristics. I want to run a regression on the dependent variable and other explanatory variables and calculate the residual terms by grouping the firms in same year and same industry. What I want to do is to divide my obseravtion into sub sample that contains the observation with same fiscal year(FYEAR=1990) and same firm characteristic (Industry =1) and run the regression and put the residual back to the observation by creating a new column. I want to do that for multiple years and multiple firms. I wonder is that any easy command with out creating multiple loops? Ray [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Applying function to elements of matrices in a list
On 2010-11-25 07:06, statmobile wrote: On 11/25/2010 09:44 AM, Dimitris Rizopoulos wrote: try this: Reduce(+, x) / length(x) Thanks Dimitris, that's very slick, I was unaware of this Reduce function. The issue, is that I actually wanted to do a trimmed mean, and if things prove possible even the median. Is there a way to apply a generic function in the manner I described? You could use the abind function in the abind package to convert your list to a 3d array and then use apply on that: require(abind) xa - abind(x, along=3) apply(xa, 1:2, mean, trim=0.3) Peter Ehlers Thanks, Brian Best, Dimitris On 11/25/2010 3:42 PM, statmobile wrote: Hello R-help, Please cc me on all responses, as I only receive summary emails from this list. I'm wondering if anybody has any tips on how to accomplish this efficiently. I have a list of matrices, and I'm trying to get the mean of the [i,j]'th element of each matrix in a list. So if I have a list of matrices, say x- list(a=matrix(rnorm(4),nrow=2),b=matrix(rnorm(4),nrow=2)) How would I get a 2x2 matrix, where the i,j'th element would be the mean across the the list of each of the i,j'th elements in the list? That is, where the [1,2] element would be the average of a[1,2] and b[1,2]. Of course my list and matrices are much larger, and I was hoping there would be some trick with lapply that I may be missing here. Thanks, Brian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help: program efficiency
hey guys,
I am working on a function to make a duplicated value unique. For example,
the original vector would be like : a = c(2,1,1,3,3,3,4)
I'll like to transform it into:
a.nodup = 2, 1.01, 1.02, 3.01, 3.02, 3.03, 4
basically, find the duplicates and assign a unique value by adding a small
amount and keep it in order.
I come up with the following codes, but it runs slow if t is large. Is there
a better way to do it?
nodup = function(t)
{
t.index=0
t.dup=duplicated(t)
for (i in 2:length(t))
{
if (t.dup[i]==T)
t.index=t.index+0.01
else t.index=0
t[i]=t[i]+t.index
}
return(t)
}
--
View this message in context:
http://r.789695.n4.nabble.com/help-program-efficiency-tp3059079p3059079.html
Sent from the R help mailing list archive at Nabble.com.
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[R] moving xlabels in lattice
Dear R users, I am trying to move the xlab string on my xyplot, without success, I would like it to shifted down, would one of you know a way ? Thanks for reading Colin -- View this message in context: http://r.789695.n4.nabble.com/moving-xlabels-in-lattice-tp3059092p3059092.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Spearman's Power
I am trying to get some idea of the power of the Spearman's Rank correlation I have carried out. Is there a way I can calculate power post-hoc? I know that confidence intervals are preferred by many people, but how do I generate those for a Spearman's? Thanks Lewis Dean [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] About searching criteria
Hi folks, I need to search the dataset on data with name on heading; Run conc density I look at ??help.search and could not resolve; help.search(pattern, fields = c(alias, concept, title) What shall I replace pattern? I suppose replacing alias, concept, title with Run, conc, density ? Please help. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] moving xlabels in lattice
On Nov 25, 2010, at 10:00 AM, statquant2 wrote: Dear R users, I am trying to move the xlab string on my xyplot, without success, I would like it to shifted down, would one of you know a way ? A quick and dirty way would be to stick a \n in front of your x-label. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] t-stat for the coefficients of an ARIMA model
Dear all, I am fitting a time series using the following command: Ts.arima-arima(x,c(2,1,2)) where x is a time series. What the function returns is perfectly fine but I was wondering if I could access to the t-stat of the coefficients I got from the arima function. Any help would be greatly appreciated. Samuel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] moving xlabels in lattice
Hi Colin, I wish I could do the same with a persp plot, since the labels overlap the ticks and the annotations. Unfortunately, I don't know the answer to your question, but I'll stay tuned on this thread. Good luck Gaetano On 11/25/2010 04:00 PM, statquant2 wrote: Dear R users, I am trying to move the xlab string on my xyplot, without success, I would like it to shifted down, would one of you know a way ? Thanks for reading Colin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help: program efficiency
one way is the following:
a - c(2,1,1,3,3,3,4)
d - unlist(sapply(rle(a)$length, function (x)
if (x 1) seq(0.01, by = 0.01, len = x) else 0))
a + d
I hope it helps.
Best,
Dimitris
On 11/25/2010 3:49 PM, randomcz wrote:
hey guys,
I am working on a function to make a duplicated value unique. For example,
the original vector would be like : a = c(2,1,1,3,3,3,4)
I'll like to transform it into:
a.nodup = 2, 1.01, 1.02, 3.01, 3.02, 3.03, 4
basically, find the duplicates and assign a unique value by adding a small
amount and keep it in order.
I come up with the following codes, but it runs slow if t is large. Is there
a better way to do it?
nodup = function(t)
{
t.index=0
t.dup=duplicated(t)
for (i in 2:length(t))
{
if (t.dup[i]==T)
t.index=t.index+0.01
else t.index=0
t[i]=t[i]+t.index
}
return(t)
}
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/
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[R] Is there an implementation for URL Encoding (/format) in R?
Hello all, I would like some R function that can translate a string to a URL encoding (see here: http://www.w3schools.com/tags/ref_urlencode.asp) Is it implemented? (I wasn't able to find any reference to it) Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Generalized linear models with categorical data
Hi: On Thu, Nov 25, 2010 at 3:53 AM, Diogo B. Provete dbprov...@gmail.comwrote: I got a question about using a GLZ with categorical x categorical data. Below there is a data set I want to know the influence of treatments (CONT, and LPS2H LPS24H) on the categories of pigmentation of the right testis of an amphibian. From these data, I used the function glm with binomial family (logit). But in the result (see below) is not possible to know the influence of the three treatments on the categories, only the treatments as a whole. Could someone help me? You have three response categories, which makes the response multinomial, not binomial. with(td, table(Tratamento, Categoria)) Categoria Tratamento Cat.1 Cat.2 Cat.3 CONT 0 4 6 LPS24h 5 5 0 LPS2h 0 3 7 Why is it that for each treatment, the responses only fall into two of the three categories? Is this by happenstance or by design? Methinks some information is lacking at the moment...but I'm pretty sure the binomial model is not correct in its current manifestation. Dennis Thanks in advance td=read.table(file.choose(), h=T) td Tratamento Categoria 1CONT Cat.2 2CONT Cat.2 3CONT Cat.2 4CONT Cat.2 5CONT Cat.3 6CONT Cat.3 7CONT Cat.3 8CONT Cat.3 9CONT Cat.3 10 CONT Cat.3 11 LPS2h Cat.2 12 LPS2h Cat.2 13 LPS2h Cat.2 14 LPS2h Cat.3 15 LPS2h Cat.3 16 LPS2h Cat.3 17 LPS2h Cat.3 18 LPS2h Cat.3 19 LPS2h Cat.3 20 LPS2h Cat.3 21 LPS24h Cat.1 22 LPS24h Cat.1 23 LPS24h Cat.1 24 LPS24h Cat.1 25 LPS24h Cat.1 26 LPS24h Cat.2 27 LPS24h Cat.2 28 LPS24h Cat.2 29 LPS24h Cat.2 30 LPS24h Cat.2 mod.test=glm(Categoria~Tratamento, family=binomial(logit), data=td) mod.test Call: glm(formula = Categoria ~ Tratamento, family = binomial(logit), data = td) Coefficients: (Intercept) TratamentoLPS24h TratamentoLPS2h 2.057e+01-2.057e+01 1.558e-08 Degrees of Freedom: 29 Total (i.e. Null); 27 Residual Null Deviance:27.03 Residual Deviance: 13.86 AIC: 19.86 anova(mod.test, test=Chisq) Analysis of Deviance Table Model: binomial, link: logit Response: Categoria Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(|Chi|) NULL 29 27.034 Tratamento 2 13.17127 13.863 0.001380 ** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 -- Atenciosamente, *Diogo Borges Provete* == Biólogo Mestre em Biologia Animal (UNESP) Laboratório de Ecologia Animal Departamento de Zoologia e Botânica Instituto de Biociências, Letras e Ciências Exatas Universidade Estadual Paulista - UNESP São José do Rio Preto-SP Brazil Rua Cristóvão Colombo, 2265 Jardim Nazareth - 15054-000 *Skype*: diogoprovete *MSN*: diogop...@yahoo.com.br *Personal web page https://sites.google.com/site/diogoprovetepage/* Traduza conosco: American Journal Experts http://www.journalexperts.com/br/ D-Lang Soluções linguisticas http://www.d-lang.com.br/site/sitept/index.htm == -- Atenciosamente, *Diogo Borges Provete* == Biólogo Mestre em Biologia Animal (UNESP) Laboratório de Ecologia Animal Departamento de Zoologia e Botânica Instituto de Biociências, Letras e Ciências Exatas Universidade Estadual Paulista - UNESP São José do Rio Preto-SP Brazil Rua Cristóvão Colombo, 2265 Jardim Nazareth - 15054-000 *Skype*: diogoprovete *MSN*: diogop...@yahoo.com.br *Personal web page https://sites.google.com/site/diogoprovetepage/* Traduza conosco: American Journal Experts http://www.journalexperts.com/br/ D-Lang Soluções linguisticas http://www.d-lang.com.br/site/sitept/index.htm == [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an implementation for URL Encoding (/format) in R?
?URLencode On Thu, Nov 25, 2010 at 3:53 PM, Tal Galili tal.gal...@gmail.com wrote: Hello all, I would like some R function that can translate a string to a URL encoding (see here: http://www.w3schools.com/tags/ref_urlencode.asp) Is it implemented? (I wasn't able to find any reference to it) Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an implementation for URL Encoding (/format) in R?
Le 25/11/10 16:53, Tal Galili a écrit : Hello all, I would like some R function that can translate a string to a URL encoding (see here: http://www.w3schools.com/tags/ref_urlencode.asp) Is it implemented? (I wasn't able to find any reference to it) Thanks, Tal Perhaps ?URLencode -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/9VOd3l : ZAT! 2010 |- http://bit.ly/c6DzuX : Impressionnism with R `- http://bit.ly/czHPM7 : Rcpp Google tech talk on youtube __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] t-stat for the coefficients of an ARIMA model
On Nov 25, 2010, at 10:50 AM, Samuel Le wrote: Dear all, I am fitting a time series using the following command: Ts.arima-arima(x,c(2,1,2)) where x is a time series. What the function returns is perfectly fine but I was wondering if I could access to the t-stat of the coefficients I got from the arima function. The typical approach is to see if there is a coef function and a vcov function for your fit and to see if this gives sensible results: coef(fit)/sqrt(vcov(diag(fit))) (I looked at the docs and those functions are available for arima objects. However, it's still your responsibility to properly interpret such output. I have no substantial experience with time series analysis and as a general rule worry when the package authors choose not to provide a particular statistic. ) David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fast Two-Dimensional Optimization
Thank you for your kind comments. I tried different algorithms such as BFGS, CG, and so on. However, the choice of algorithms was not effective on reducing computation time. Instead, your second suggestion of coding the gradient of minimization function was a little bit successful to reduce computation time. It was twice faster than before. I appreciate you for your help. 2010/11/22 Rubén Roa r...@azti.es Did you try different algorithms? optim has Nelder-Mead (default), BFGS, simulated annealing, CG, etc. Depending on your problem, they can be much faster than the default. See ?optim and check the info for 'method' Check also optimx, a new wrapper for optim that can try out all methods. Another approach that may speed up calculations is to code for the gradients of the function that you're minimizing. Check argument 'gr' of optimx. HTH Dr. Rubén Roa-Ureta AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g 48395 Sukarrieta (Bizkaia) SPAIN -Mensaje original- De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] En nombre de Wonsang You Enviado el: lunes, 22 de noviembre de 2010 16:17 Para: r-help@r-project.org Asunto: [R] Fast Two-Dimensional Optimization Dear R Helpers, I have attempted optim function to solve a two-dimensional optimization problem. It took around 25 second to complete the procedure. However, I want to reduce the computation time: less than 7 second. Is there any optimization function in R which is very rapid? Best Regards, Wonsang - Wonsang You Leibniz Institute for Neurobiology -- View this message in context: http://r.789695.n4.nabble.com/R-Fast-Two-Dimensional-Optimizat ion-tp3053782p3053782.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] About searching criteria
Try this: d - data() ne - new.env() data(list = grep(\\(, d$results[,'Item'], value = TRUE, invert = TRUE), envir = ne) out - eapply(ne, names) names(which(lapply(lapply(out, '%in%', c(Run, conc, density)), sum) == 3)) On Thu, Nov 25, 2010 at 1:38 PM, Stephen Liu sati...@yahoo.com wrote: Hi folks, I need to search the dataset on data with name on heading; Run conc density I look at ??help.search and could not resolve; help.search(pattern, fields = c(alias, concept, title) What shall I replace pattern? I suppose replacing alias, concept, title with Run, conc, density ? Please help. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] moving xlabels in lattice
On 2010-11-25 07:00, statquant2 wrote: Dear R users, I am trying to move the xlab string on my xyplot, without success, I would like it to shifted down, would one of you know a way ? Thanks for reading Colin Have a look at the possible height adjustments with trellis.par.get()$layout.heights Likely candidates for adjustment are: xlab (space for the label) axis.bottom (space between axis and bottom) axis.xlab.bottom (space between axis, label, and bottom) Then try xyplot(rnorm(10) ~ 1:10, xlab = the x-label, par.settings = list( layout.heights = list( xlab = 3 # axis.xlab.padding = 3 # axis.bottom = 3 ))) Comment/uncomment to suit. Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] aftreg vs survreg loglogistic aft model (different intercept term)
Hi, I'm estimating a loglogistic aft (accelerated failure time) model, just a simple plain vanilla one (without time dependent covariates), I'm comparing the results that I obtain between aftreg (eha package) and survreg(surv package). If I don't use any covariate the results are identical , if I add covariates all the coefficients are the same until a precision of 10^4 or 10^-5 except for the intercept term(in survreg)=log(scale) (in aftreg) that are similar but with a worse precision of 10^-1 or 10^-2. Any idea of why this happens? Probably the different covariates coefficients have a so big influence on the intercept term? Thank you Immanuel -- View this message in context: http://r.789695.n4.nabble.com/aftreg-vs-survreg-loglogistic-aft-model-different-intercept-term-tp3059250p3059250.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help: program efficiency
If the input vector t is known to be ordered
(or if you only care about runs of duplicated
values, not all duplicated values) the following
is pretty quick
nodup3 - function (t) {
t + (sequence(rle(t)$lengths) - 1)/100
}
If you don't know if the the input will be ordered
then ave() will do it a bit faster than your
code
nodup2 - function (t) {
ave(t, t, FUN = function(x) x + (seq_along(x) - 1)/100)
}
E.g., for a sorted sequence of 300,000 numbers drawn with
replacement from 1:100,000 I get:
a2 - sort(sample(1:1e5, size=3e5, replace=TRUE))
system.time(v - nodup(a2))
user system elapsed
2.780.053.97
system.time(v2 - nodup2(a2))
user system elapsed
1.830.022.66
system.time(v3 - nodup3(a2))
user system elapsed
0.180.000.14
identical(v,v2) identical(v,v3)
[1] TRUE
If speed is truly an issue, the built-in sequence may
be replaced by a faster one that does the same thing:
nodup3a - function (t) {
faster.sequence - function(nvec) {
seq_len(sum(nvec)) - rep(cumsum(c(0L, nvec[-length(nvec)])),
nvec)
}
t + (faster.sequence(rle(t)$lengths) - 1)/100
}
That took 0.05 seconds on the a2 dataset and produced
identical results.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of randomcz
Sent: Thursday, November 25, 2010 6:49 AM
To: r-help@r-project.org
Subject: [R] help: program efficiency
hey guys,
I am working on a function to make a duplicated value unique.
For example,
the original vector would be like : a = c(2,1,1,3,3,3,4)
I'll like to transform it into:
a.nodup = 2, 1.01, 1.02, 3.01, 3.02, 3.03, 4
basically, find the duplicates and assign a unique value by
adding a small
amount and keep it in order.
I come up with the following codes, but it runs slow if t is
large. Is there
a better way to do it?
nodup = function(t)
{
t.index=0
t.dup=duplicated(t)
for (i in 2:length(t))
{
if (t.dup[i]==T)
t.index=t.index+0.01
else t.index=0
t[i]=t[i]+t.index
}
return(t)
}
--
View this message in context:
http://r.789695.n4.nabble.com/help-program-efficiency-tp305907
9p3059079.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an equivalent to predict(..., type=linear) of a Proportional hazard model for a Cox model instead?
On Nov 25, 2010, at 10:08 AM, Ben Rhelp wrote:
Hi David,
Thank you for your reply. See below for more information.
From: David Winsemius
On Nov 25, 2010, at 7:27 AM, Ben Rhelp wrote:
I manage to achieve similar results with a Cox model as follows
but I don't
really understand why we have to take the inverse of the linear
prediction
with
the Cox model
Different parameterization. You can find expanded answer(s) in the
archives
and in the documentation of survreg.distributions.
I understand (i think) the difference in model structures between a
Cox (coxph)
and Proportional hazard model (survreg).
I couldn't tell whether this means you decided that those citations
answered your question. If not, then refer to Therneau's or Lumley's
replies in rhelp to a similar question earlier this month.:
https://stat.ethz.ch/pipermail/r-help/2010-November/259796.html
https://stat.ethz.ch/pipermail/r-help/2010-November/259747.html
and why we do not need to divide by the number of days in the year
anymore?
Here I'm guessing (since you don't offer enough evidence to
confirm) that the
difference is in the time scales used in your Aidsp$survtime
versus some other
example to which you are comparing .
Both models are run from the same data, so I am not expecting any
differences in
time scales.
To get similar results, I need to actually run the following
equations:
expected_lifetime_in_years = exp(fit)/365.25---
Linear
prediction of the Proportional hazard model
expected_lifetime_in_years = 1/exp(fit)
--- Linear
prediction of the Cox model
where fit come from the linear prediction of each models,
respectively.
Actually, in the code below, I re-run the models and predictions
based on a
yearly sampling time (instead of daily).
Again, to get similar results, I now need to actually run the
following
equations:
expected_lifetime_in_years = exp(fit)
--- Linear
prediction of the Proportional hazard model
expected_lifetime_in_years = 1/exp(fit)
--- Linear
prediction of the Cox model
I think I understand the logic behind the results of the
proportional hazard
model, but not for the prediction of the Cox model.
Cox models are PH models.
Thank you for your help. I hope this is not a too stupid hole in my
logic.
Here is the self contained R code to produce the charts:
library(MASS);
library(survival);
#Same data but parametric fit
make.aidsp - function(){
cutoff - 10043 # July 1987 in julian days
btime - pmin(cutoff, Aids2$death) - pmin(cutoff, Aids2$diag)
atime - pmax(cutoff, Aids2$death) - pmax(cutoff, Aids2$diag)
survtime - btime + 0.5*atime
status - as.numeric(Aids2$status)
data.frame(survtime, status = status - 1, state = Aids2$state,
T.categ = Aids2$T.categ, age = Aids2$age, sex = Aids2$sex)
}
Aidsp - make.aidsp()
# MASS example with the proportional hazard model
par(mfrow = c(1, 2));
(aids.ps - survreg(Surv(survtime + 0.9, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83),
levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)),
age =
0:82), se = T, type = linear)
plot(0:82, exp(zz$fit)/365.25, type = l, ylim = c(0, 2), xlab =
age, ylab =
expected lifetime (years))
lines(0:82, exp(zz$fit+1.96*zz$se.fit)/365.25, lty = 3, col = 2)
lines(0:82, exp(zz$fit-1.96*zz$se.fit)/365.25, lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015)
# Same example but with a Cox model instead
(aids.pscp - coxph(Surv(survtime + 0.9, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW, 83),
levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)),
age =
0:82), se = T, type = lp)
plot(0:82, 1/exp(zzcp$fit), type = l, ylim = c(0, 2), xlab =
age, ylab =
expected lifetime (years))
lines(0:82, 1/exp(zzcp$fit+1.96*zzcp$se.fit), lty = 3, col = 2)
lines(0:82, 1/exp(zzcp$fit-1.96*zzcp$se.fit), lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize = 0.015)
# Change the sampling time from daily to yearly
par(mfrow = c(1, 1));
(aids.ps - survreg(Surv((survtime + 0.9)/365.25, status) ~ state +
T.categ +
pspline(age, df=6), data = Aidsp))
zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83),
levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)),
age =
0:82), se = T, type = linear)
plot(0:82, exp(zz$fit), type = l, ylim = c(0, 2), xlab = age,
ylab =
expected lifetime (years))
(aids.pscp - coxph(Surv((survtime + 0.9)/365.25, status) ~ state +
T.categ +
pspline(age, df=6), data = Aidsp))
zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW, 83),
levels =
levels(Aidsp$state)),
Re: [R] Is there an implementation for URL Encoding (/format) in R?
On 11/25/10 7:53 AM, Tal Galili wrote: Hello all, I would like some R function that can translate a string to a URL encoding (see here: http://www.w3schools.com/tags/ref_urlencode.asp) Is it implemented? (I wasn't able to find any reference to it) I expect there are several implementations, spread across different packages. The function curlEscape() in RCurl is one. D. Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] difficulty setting the random = argument to lme()
Hi Robert: It appears to me that you have a split-plot structure, so let me see if I have it right. The 'whole-plot' experiment looks like a replicated randomized block design - the studies are the blocks, the treatments A and B are the whole-plot treatments, each of which is assigned randomly to three subjects...basically the 'between-subjects' part of the design. The within-subject treatment factor is site (let's call them 1 and 2 to avoid confusion with the treatment labels), and it makes sense to me that the two measurements are correlated within animal, but I don't quite see why it makes sense that they should be correlated between animals not treated alike in the same study. I'm not in pharma, so I might learn something by asking. I tend to look at the means model part of a design with ANOVA just because I'm old: Study 9 Treatment 1 Study * treatment9 Subjects40 (whole-plot error) Site 1 Site * Study9 Site * Treatment 1 Site * Study * Trt 9 Residual40 (split plot error) Study and subjects can reasonably be thought of as random effects. If the same sites are chosen for each subject, they would have to be fixed. The interactions with study and subject are random. I agree with the fixed effects specification so far, but some of the random interactions aren't obvious to me, although I can understand how they arise from the structure of the data. A couple of questions: (1) Are the studies expected to be correlated, and if so, was that the motivation for the replicate subjects per treatment level? I'm rather accustomed to blocks representing independent replications of the experiment, which I would have expected by the use of different subjects in each study. Or is all of this a necessary precaution in the clinical trial? (2) Since sites were the same for each subject, I can see the within-subject correlation and between-subject correlation for subjects in the same study with the same treatment, but I'm curious as to why the Site * Study interaction is relevant. The whole-plot part of this is pretty easy, but I think I (and perhaps others) would need some data to play with to work on getting the random effects and correlation structure specified properly. It also appears you will need to use lme4 rather than nlme for this problem due to the crossed random effects, which lme() can't handle. That may be the source of your trouble :) This is certainly an interesting problem; thanks for sharing it. I might also suggest that this be taken to the r-sig-mixed-models list, where you are likely to have more people who are interested in this type of problem. Cheers, Dennis On Thu, Nov 25, 2010 at 5:00 AM, Robert Kinley kinley_rob...@lilly.comwrote: My small brain is having trouble getting to grips with lme() I wonder if anyone can help me correctly set the random = argument to lme() for this kind of setup with (I think) 9 variance/covariance components ... Study.1 Study.2 ... Study.10 Treatment.A: subject: 1 2 3 4 5 6 etc. 28 29 30 Treatment.B: subject: 31 32 3334 35 36 58 59 60 A variable is measured at 2 fixed sites (A and B) on each subject so we have fixed effects :- between-Treatments between-sites (A and B) Treatment*site interaction and we have random effects :- study effects at site A study effects at site B correlation between site A and site B study effects study*treatment interaction effects at site A study*treatment interaction effects at site B correlation between site A and B study*treatment interaction effects residual (between-subject) effects at site A residual (between-subject) effects at site B correlation between site A and B residuals (between-subject) effects My problem is formulating the random = argument to give estimates of all 9 random components ... Hope someone can help ... Robert Kinley Study: Pos tissue VC, Neg tissue VC, Pos/Neg tissue correlation Study*Group:Pos tissue VC, Neg tissue VC, Pos/Neg tissue correlation Residual (animal): Pos tissue VC, Neg tissue VC, Pos/Neg tissue correlation [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lifting Wavelet Transform
Hi:
Package sos is a wonderful tool for this sort of question...
library(sos)
findFn('lifting wavelet transform')
It returned nine possible matches in two packages, adlift and nlt.
HTH,
Dennis
On Wed, Nov 24, 2010 at 8:46 PM, assaedi76 assaedi76 assaed...@yahoo.comwrote:
Hi R users
Thanks in advance
Is lifting wavelet transform implemented in R? If so, which package or
codes can be used for performing that?
assaed...@yahoo.com
Thanks
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Re: [R] delete-d jackknife
Thank you so much :) -- View this message in context: http://r.789695.n4.nabble.com/delete-d-jackknife-tp3058335p3059364.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an equivalent to predict(..., type=linear) of a Proportional hazard model for a Cox model instead?
I hit the send button on my second reply before I intended to. Since
then I have noticed that the question I thought you were asking is not
at all a good match to the Subject line of your message. There is a
type =lp in predict.coxph and that is the linear predictor, although
it is not a value that is on any transformation of time as would be
the linear predictor of an accelerated failure time estimate. If you
are looking for another method (in addition to predict.coxph) to
produce expected survival from a coxph fit, you could also look at
survexp. The ratetable argument accepts a fitted Cox model.
I have still not found a good answer to the question that I thought
the body of your second posting was posing: namely why days and years
are not handled the same in predict.survreg and predict.coxph.
--
David.
On Nov 25, 2010, at 12:35 PM, David Winsemius wrote:
On Nov 25, 2010, at 10:08 AM, Ben Rhelp wrote:
Hi David,
Thank you for your reply. See below for more information.
From: David Winsemius
On Nov 25, 2010, at 7:27 AM, Ben Rhelp wrote:
I manage to achieve similar results with a Cox model as follows
but I don't
really understand why we have to take the inverse of the linear
prediction
with
the Cox model
Different parameterization. You can find expanded answer(s) in
the archives
and in the documentation of survreg.distributions.
I understand (i think) the difference in model structures between a
Cox (coxph)
and Proportional hazard model (survreg).
I couldn't tell whether this means you decided that those citations
answered your question. If not, then refer to Therneau's or Lumley's
replies in rhelp to a similar question earlier this month.:
https://stat.ethz.ch/pipermail/r-help/2010-November/259796.html
https://stat.ethz.ch/pipermail/r-help/2010-November/259747.html
and why we do not need to divide by the number of days in the year
anymore?
Here I'm guessing (since you don't offer enough evidence to
confirm) that the
difference is in the time scales used in your Aidsp$survtime
versus some other
example to which you are comparing .
Both models are run from the same data, so I am not expecting any
differences in
time scales.
To get similar results, I need to actually run the following
equations:
expected_lifetime_in_years = exp(fit)/365.25---
Linear
prediction of the Proportional hazard model
expected_lifetime_in_years = 1/exp(fit)
--- Linear
prediction of the Cox model
where fit come from the linear prediction of each models,
respectively.
Actually, in the code below, I re-run the models and predictions
based on a
yearly sampling time (instead of daily).
Again, to get similar results, I now need to actually run the
following
equations:
expected_lifetime_in_years = exp(fit)
--- Linear
prediction of the Proportional hazard model
expected_lifetime_in_years = 1/exp(fit)
--- Linear
prediction of the Cox model
I think I understand the logic behind the results of the
proportional hazard
model, but not for the prediction of the Cox model.
Cox models are PH models.
Thank you for your help. I hope this is not a too stupid hole in my
logic.
Here is the self contained R code to produce the charts:
library(MASS);
library(survival);
#Same data but parametric fit
make.aidsp - function(){
cutoff - 10043 # July 1987 in julian days
btime - pmin(cutoff, Aids2$death) - pmin(cutoff, Aids2$diag)
atime - pmax(cutoff, Aids2$death) - pmax(cutoff, Aids2$diag)
survtime - btime + 0.5*atime
status - as.numeric(Aids2$status)
data.frame(survtime, status = status - 1, state = Aids2$state,
T.categ = Aids2$T.categ, age = Aids2$age, sex = Aids2$sex)
}
Aidsp - make.aidsp()
# MASS example with the proportional hazard model
par(mfrow = c(1, 2));
(aids.ps - survreg(Surv(survtime + 0.9, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zz - predict(aids.ps, data.frame(state = factor(rep(NSW, 83),
levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)),
age =
0:82), se = T, type = linear)
plot(0:82, exp(zz$fit)/365.25, type = l, ylim = c(0, 2), xlab =
age, ylab =
expected lifetime (years))
lines(0:82, exp(zz$fit+1.96*zz$se.fit)/365.25, lty = 3, col = 2)
lines(0:82, exp(zz$fit-1.96*zz$se.fit)/365.25, lty = 3, col = 2)
rug(Aidsp$age + runif(length(Aidsp$age), -0.5, 0.5), ticksize =
0.015)
# Same example but with a Cox model instead
(aids.pscp - coxph(Surv(survtime + 0.9, status) ~ state + T.categ +
pspline(age, df=6), data = Aidsp))
zzcp - predict(aids.pscp, data.frame(state = factor(rep(NSW,
83), levels =
levels(Aidsp$state)),
T.categ = factor(rep(hs, 83), levels = levels(Aidsp$T.categ)),
age =
0:82), se = T, type = lp)
plot(0:82, 1/exp(zzcp$fit), type = l, ylim = c(0, 2), xlab =
age, ylab =
expected lifetime (years))
lines(0:82,
[R] [libsvm] predict function error
Dear R users,
There is a error message when I run the following code. It is used to load
microarray data and use the top 1000 genes for training svm to classify test
set .
library(e1071)
Loading required package: class
f=read.table(F:\\lab\\
microarray analysis\\VEH LPS\\exprs.txt,
sep=\t,header=FALSE,col.names=c(Gene,VEH,VEH,VEH,VEH,VEH,LPS,LPS,LPS,LPS,LPS))
tf=t(f)
colnames(tf)=tf[1,]
tf=tf[-1,]
tf=as.data.frame(tf)
array - apply(tf[,2:24928],c(1,2),as.numeric)
label - as.factor(tf$ENTREZGENE)
n - nrow(array) #get sample number
permutation - sample(1:n)
array.perm - array[permutation,] # random permutation of samples
label.perm - label[permutation] # same permutation of labels
k - 5 #set cross validation steps
for (i in 1:k){
+ win - round(n/k) # size of window
+ cv - ((i-1)*win+1):min(n,i*win) # move window
+
+ CV.label.test - label.perm[cv]
+ CV.label.train - label.perm[-cv]
+
+ CV.train.neg.num -
nrow(as.data.frame(CV.label.train[CV.label.train==VEH]))
+ CV.train.pos.num -
nrow(as.data.frame(CV.label.train[CV.label.train==LPS]))
+ CV.train.num - CV.train.pos.num+CV.train.neg.num
+
+ CV.test - array.perm[cv,] # samples within window
+ CV.train - array.perm[-cv,] # samples outside window
+
+ CV.train.pos -
as.data.frame(CV.train)[as.data.frame(CV.label.train)==LPS,] # matrix of
positive samples
+ CV.train.pos.mean - apply(CV.train.pos,2,mean) # compute mean of each
column (gene)
+ CV.train.pos.var - apply(CV.train.pos,2,var) # compute variance of each
column (gene)
+
+ CV.train.neg -
as.data.frame(CV.train)[as.data.frame(CV.label.train)==VEH,] # matrix of
positive samples
+ CV.train.neg.mean - apply(CV.train.neg,2,mean) # compute mean of each
column (gene)
+ CV.train.neg.var - apply(CV.train.neg,2,var)
+
+ tscore
-(abs(CV.train.pos.mean-CV.train.neg.mean)/sqrt((CV.train.pos.num-1)*CV.train.pos.var+(CV.train.neg.num-1)*CV.train.neg.var))*sqrt(CV.train.pos.num*CV.train.neg.num*(CV.train.num-2)/CV.train.num)
+ index - order(tscore, decreasing=TRUE)
+ CV.train=CV.train[,index[1:100]]
+
+ svm.train-svm(CV.train, CV.label.train, type=C-classification,
kernel=linear, cross=5, scale=TRUE)
+ }
pred-predict(svm.train,CV.test)
Error in scale.default(newdata[, object$scaled, drop = FALSE], center =
object$x.scale$scaled:center, :
length of 'center' must equal the number of columns of 'x'
I wonder what is the reason for such kind of error.
sessionInfo()
R version 2.11.1 (2010-05-31)
i386-pc-mingw32
locale:
[1] LC_COLLATE=Chinese_People's Republic of China.936
[2] LC_CTYPE=Chinese_People's Republic of China.936
[3] LC_MONETARY=Chinese_People's Republic of China.936
[4] LC_NUMERIC=C
[5] LC_TIME=Chinese_People's Republic of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] e1071_1.5-24 class_7.3-2
Thanks a lot!
--
Best Regards,
Zhe Liu
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Re: [R] help: program efficiency
Date: Thu, 25 Nov 2010 06:49:19 -0800
From: rando...@gmail.com
To: r-help@r-project.org
Subject: [R] help: program efficiency
hey guys,
I am working on a function to make a duplicated value unique. For example,
the original vector would be like : a = c(2,1,1,3,3,3,4)
I'll like to transform it into:
a.nodup = 2, 1.01, 1.02, 3.01, 3.02, 3.03, 4
basically, find the duplicates and assign a unique value by adding a small
amount and keep it in order.
I come up with the following codes, but it runs slow if t is large. Is there
a better way to do it?
I guess I'd just make a vector of uniform or even normal random numbers
and add to your input vector. This of course is not guaranteed and adds
to uniques but you can test and repeat and it is probably closer
to what you want but I am only speculating on your objectives.
nodup = function(t)
{
t.index=0
t.dup=duplicated(t)
for (i in 2:length(t))
{
if (t.dup[i]==T)
t.index=t.index+0.01
else t.index=0
t[i]=t[i]+t.index
}
return(t)
}
--
View this message in context:
http://r.789695.n4.nabble.com/help-program-efficiency-tp3059079p3059079.html
Sent from the R help mailing list archive at Nabble.com.
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[R] R and R-related meetups at UCLA
UCLA Statistics is hosting three meetup groups that are either straight R (361 members) or R-related (GRASS, which interacts directly with R, and GPGPU, which will look at interfacing R with OpenMP, GCD, OpenCL, CUDA, ...) http://www.meetup.com/LAarea-R-usergroup http://www.meetup.com/Los-Angeles-Area-GRASS-Users-Group/ http://www.meetup.com/SoCal-GPGPU-and-Commodity-Parallel-Programming-Group/ If you are local SoCal, join, drop in. === Jan de Leeuw; Distinguished Professor and Chair, UCLA Department of Statistics; Director: UCLA Center for Environmental Statistics (CES); Editor: Journal of Multivariate Analysis, Journal of Statistical Software; US mail: 8125 Math Sciences Bldg, Box 951554, Los Angeles, CA 90095-1554 phone (310)-825-9550; fax (310)-206-5658; email: dele...@stat.ucla.edu .mac: jdeleeuw ++ aim: deleeuwjan ++ skype: j_deleeuw homepages: http://gifi.stat.ucla.edu ++ http://www.cuddyvalley.org - No matter where you go, there you are. --- Buckaroo Banzai http://gifi.stat.ucla.edu/sounds/nomatter.au - __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Cumsum with a max and min value
I have a vector of values -1, 0, 1, say a - c(0, 1, 0, 1, 1, -1, -1, -1, 0, -1, -1) I want to create a vector of the cumulative sum of this, but I need to set a maximum and minimum value for the cumsum, for example: max_value - 2 min_value - -2 the expected result would be (0, 1, 1, 2, 2, 1, 0, -1, -1, -2, -2) The only way I managed to do It, was res - vector(length=length(a)) res[1] - a[1] for ( i in 2:length(a)) res[i] - res[i-1] + a[i] * (( res[i-1] max_value a[i] 0 ) | ( res[i-1] min_value a[i] 0 )) This is certainly not the best way to do it, so any suggestions? Henrique [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Go (back) from Rd to roxygen
Hi all,
Since roxygen is a great help to document R packages, I am wondering
if there exists an approach to go back from the raw Rd files to
roxygen-documentation? E.g. turn \author{Somebody} into @author
Somebody. This sounds ridiculous, but I believe it helps in the long
term for me to maintain R packages.
Thanks!
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
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[R] How to bin/average time points?
Dear all, I am pretty new to R only having an introduction course, so please bare with me. I am doing my PhD at The Max Planck Institute of Immunobiology where I am analyzing some calorimetry data from some mice. I have a spreadsheet consisting of measurements of the respiratory exchange rate at different time points measured every 9 minutes over some days. My goal is bin/average the time points of each hour to only one measurements. E.g. [Time] - [Measurement] 12.09 - 0.730 12.18 - 0.732 12.27 - 0.743 12.36 - 0.757 12.45 - 0.781 12.54 - 0.731 -- should be averaged to fx one time point and one value, fx: 12.30 - [average of the six measurements] I know how to average the measurements in a whole column but how to average every six measurements automatically and also how to average every six time points and make a new sheet consisting of these data? I hope you guys are able to help, since we are really stuck here. I can of course do it manually but with 8000 measurements it will take lots of time. Thank you very much. Best regards, Kevin Dalgaard -- View this message in context: http://r.789695.n4.nabble.com/How-to-bin-average-time-points-tp3059509p3059509.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cumsum with a max and min value
Try this: ac - cumsum(a) ifelse(ac 2, max_value, ifelse(ac -2, min_value, ac)) On Thu, Nov 25, 2010 at 6:44 PM, henrique henri...@allianceasset.com.brwrote: I have a vector of values -1, 0, 1, say a - c(0, 1, 0, 1, 1, -1, -1, -1, 0, -1, -1) I want to create a vector of the cumulative sum of this, but I need to set a maximum and minimum value for the cumsum, for example: max_value - 2 min_value - -2 the expected result would be (0, 1, 1, 2, 2, 1, 0, -1, -1, -2, -2) The only way I managed to do It, was res - vector(length=length(a)) res[1] - a[1] for ( i in 2:length(a)) res[i] - res[i-1] + a[i] * (( res[i-1] max_value a[i] 0 ) | ( res[i-1] min_value a[i] 0 )) This is certainly not the best way to do it, so any suggestions? Henrique [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with plotting diagnostics - Error in object$coefficients : $ operator is invalid for atomic vectors
On 2010-11-25 02:25, Manderscheid Katharina wrote: this problem seems to only exist in R 2.12.0 but not in R 2.11.1. any ideas? a bug? Duncan *did* say that he was using R 2.12.0. So that's not likely to be the problem. Most of the time, when users claim that a problem exists in a new version that did not exist in an older version, it's due to a change in the user's setup or to not updating packages or to not checking the NEWS file. Since you still have not provided a *reproducible* example, it's not likely that anyone can help. Can't you make up a small example that shows exactly how you are using lm() and that will generate the error? Peter Ehlers -- dr. katharina manderscheid soziologisches seminar universität luzern kasernenplatz 3 6000 luzern 7 tel. ++41 41 228 4657 web: http://www.unilu.ch/deu/dr.-katharina-manderscheid_346380.aspx -Ursprüngliche Nachricht- Von: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] Gesendet: Mittwoch, 17. November 2010 16:33 An: Manderscheid Katharina Cc: 'r-help@r-project.org' Betreff: Re: [R] Problem with plotting diagnostics - Error in object$coefficients : $ operator is invalid for atomic vectors On 17/11/2010 10:28 AM, Manderscheid Katharina wrote: hi all, after fitting a multiple linear regression model- lm(y ~ a + b+ c+d) i wanted to plot diagnostics plot(model) but get the error message Error in object$coefficients : $ operator is invalid for atomic vectors. which does not make a lot of sense, since there is no $ - i am working with an attached dataset. can anyone help, please?? thanks a lot, kat I just tried those lines (with fake data for a,b,c,d and y) and got no error message. I was using R 2.12.0. I think you need to show us a reproducible example, and the sessionInfo() to go with it, to help with this. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Apropos the day...
Since today is American Thanksgiving, I want to thank: (a) R-core for all of their efforts to produce what is, IMHO, the best statistical software around, not simply for the convenience of doing more, better, quicker, but also because it changes the landscape in the way one thinks about data analysis. It's a fantastic playground for a statistician and I enjoy using it immensely. (b) All the folks on R-help who have asked interesting, thought-provoking questions and those who have provided even more thought-provoking and insightful answers. I have learned so much from this forum in the year or so since I rejoined it and I want to express my appreciation for that to all of you. (c) All the package developers who expand the capabilities of the software and/or simplify the workflow - you do a great service to the community with your efforts. (d) All of you who have written books or tutorials about R and those who host R-related web sites for your efforts to transfer your cumulative knowledge to the community. (e) All those kindly people who do yeoman work in the background to make this list and project run more smoothly and productively. (f) Ross Ihaka and Robert Gentleman, without whom this community would not exist. (g) Anyone else I forgot who deserves some props for their work in making this one of the top open-source success stories around. To the Americans on this list, Happy Thanksgiving, and to the rest of you, thank you for making this such a vibrant professional community. Best regards to all, Dennis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there an implementation for URL Encoding (/format) in R?
Thank you Duncan, Romain and Gustavo for referring me to both: URLencode and curlEscape I see that both functions work great for English, but fail to provide with the proper translation for Hebrew characters. For example, the word ש××× (Peace, in Hebrew) Should be this: %D7%A9%D7%9C%D7%95%D7%9D But instead, both commands translate it to: URLencode(ש×××) %f9%ec%e5%ed What do you suggest? (write it myself, or is there something pre-made) Thanks again for the help, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Thu, Nov 25, 2010 at 5:59 PM, Gustavo Carvalho gustavo.bi...@gmail.comgustavo.bio%...@gmail.com wrote: URLencode [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Apropos the day...
ditto. Spencer Graves On 11/25/2010 1:20 PM, Dennis Murphy wrote: Since today is American Thanksgiving, I want to thank: (a) R-core for all of their efforts to produce what is, IMHO, the best statistical software around, not simply for the convenience of doing more, better, quicker, but also because it changes the landscape in the way one thinks about data analysis. It's a fantastic playground for a statistician and I enjoy using it immensely. (b) All the folks on R-help who have asked interesting, thought-provoking questions and those who have provided even more thought-provoking and insightful answers. I have learned so much from this forum in the year or so since I rejoined it and I want to express my appreciation for that to all of you. (c) All the package developers who expand the capabilities of the software and/or simplify the workflow - you do a great service to the community with your efforts. (d) All of you who have written books or tutorials about R and those who host R-related web sites for your efforts to transfer your cumulative knowledge to the community. (e) All those kindly people who do yeoman work in the background to make this list and project run more smoothly and productively. (f) Ross Ihaka and Robert Gentleman, without whom this community would not exist. (g) Anyone else I forgot who deserves some props for their work in making this one of the top open-source success stories around. To the Americans on this list, Happy Thanksgiving, and to the rest of you, thank you for making this such a vibrant professional community. Best regards to all, Dennis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Spencer Graves, PE, PhD President and Chief Operating Officer Structure Inspection and Monitoring, Inc. 751 Emerson Ct. San José, CA 95126 ph: 408-655-4567 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Apropos the day...
I'll second that. Paul Hurley. On 25/11/10 21:25, Spencer Graves wrote: ditto. Spencer Graves On 11/25/2010 1:20 PM, Dennis Murphy wrote: Since today is American Thanksgiving, I want to thank: (a) R-core for all of their efforts to produce what is, IMHO, the best statistical software around, not simply for the convenience of doing more, better, quicker, but also because it changes the landscape in the way one thinks about data analysis. It's a fantastic playground for a statistician and I enjoy using it immensely. (b) All the folks on R-help who have asked interesting, thought-provoking questions and those who have provided even more thought-provoking and insightful answers. I have learned so much from this forum in the year or so since I rejoined it and I want to express my appreciation for that to all of you. (c) All the package developers who expand the capabilities of the software and/or simplify the workflow - you do a great service to the community with your efforts. (d) All of you who have written books or tutorials about R and those who host R-related web sites for your efforts to transfer your cumulative knowledge to the community. (e) All those kindly people who do yeoman work in the background to make this list and project run more smoothly and productively. (f) Ross Ihaka and Robert Gentleman, without whom this community would not exist. (g) Anyone else I forgot who deserves some props for their work in making this one of the top open-source success stories around. To the Americans on this list, Happy Thanksgiving, and to the rest of you, thank you for making this such a vibrant professional community. Best regards to all, Dennis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about importing phylogenetic tree
Hello, I am trying to import a phylogenetic tree from Mesquite into R. When I use the read.nexus command I get the following message: Warning message: In matrix(x, ncol = 2, byrow = TRUE) : data length [589] is not a sub-multiple or multiple of the number of rows [295] A phylo object is created but I am unable to plot it (when I try R freezes) and I can tell by looking at the tip labels that R is not interpreting the tree properly (in addition to my taxa names I get a bunch of numbers as tip labels). I also tried downloaded RMesquite, however it failed and I got the following message: Warning: unable to access index for repository http://R-Forge.R-project.org/bin/windows/contrib/2.11 Warning message: In getDependencies(pkgs, dependencies, available, lib) : package RMesquite is not available Any suggestions would be greatly appreciated. Thank you. Nora [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] coxph strange result
The following fit does not make sense to me, please, correct me if I have a logical error. moddowsn Call: coxph(formula = Surv(start, stop, resp) ~ sn + matfac2, data = coxsn1, method = efron) coef exp(coef) se(coef) z p sn2 0.0497 1.051 0.02030 2.450 1.4e-02 sn3 -0.0532 0.948 0.02038 -2.610 9.0e-03 sn4 -0.0410 0.960 0.01979 -2.073 3.8e-02 sn5 -0.0776 0.925 0.01954 -3.973 7.1e-05 sn6 -0.1133 0.893 0.01839 -6.161 7.2e-10 sn7 -0.1252 0.882 0.01846 -6.781 1.2e-11 sn8 -0.1222 0.885 0.01994 -6.130 8.8e-10 sn9 -0.0507 0.951 0.02047 -2.478 1.3e-02 sn10 -0.0444 0.957 0.02056 -2.159 3.1e-02 sn11 -0.0433 0.958 0.02157 -2.008 4.5e-02 sn12 -0.0114 0.989 0.02037 -0.557 5.8e-01 matfac22 -0.2599 0.771 0.01727 -15.048 0.0e+00 matfac25 -0.1804 0.835 0.00924 -19.512 0.0e+00 Likelihood ratio test=651 on 13 df, p=0 n= 253802 This would indicate that in sn6 to sn8 there is less of a chance of an event. ?? do the relative frequencies implied by the following table make any sense?? table(coxsn1$matfac2,coxsn1$sn,coxsn1$resp) , , = 0 1 2 3 4 5 6 7 8 9101112 1 3407 3177 3425 3348 3975 3564 3181 3077 2894 2610 3441 3443 2 920 1005 1142 1327 1645 1530 1330 1184 964 864 888 860 5 9036 9507 10258 11888 16826 15575 13394 12346 9938 9001 8970 8599 , , = 1 1 2 3 4 5 6 7 8 9101112 1 1453 1459 1186 1496 1295 1754 1429 1153 1106 1234 965 1532 2 312 290 330 390 454 539 479 367 295 276 256 267 5 2994 3207 3371 3629 4095 5581 5837 3844 3400 3199 2705 3084 Apparently the frequency of an event is higher in the summer months. I apologize for not being able to disclose the dataset, but think that the table provides enough to address the question. Thanks everybody. Stephen B [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on running regression by grouping firms
res - function(x) resid(x) ds_test$u - do.call(c, llply(mods, res)) I'd be a little careful with this, because there's no guarantee the results will by ordered in the same way as the input (and I'd also prefer ds_test$u - unlist(llply(mods, res)) or ds_test$u - laply(mods, res)) In your case, where you have multiple grouping factors, you may have to be a little more careful, but the strategy is the same. You could possibly reduce it to a one-liner (untested): ds_test$u - do.call(c, dlply(ds_test, .(individual), function(x) resid(lm(size ~ time, data = x Or: ds_test - ddply(ds_test, .(individual), transform, u = resid(lm(size ~ time))) which will guarantee the correct ordering. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] discriminant function analysis
Hi, I am not sure if it is more robust than a discriminant function but it is certainly capable if differentiating between species based on morphology. I used 12 measurements in my fish. What did your PCA results show? Unfortunately I haven't got round to publishing my data yet but I can send you a paper by my supervisor who did a similar analysis looking at morphological differentiation in cichlids in Africa. Chris Sent from my iPhone On 25 Nov 2010, at 20:30, Mike Gibson megalop...@hotmail.com wrote: Sorry for the delay in getting back to you. I did use a PCA. I wanted to run a discriminant function as a comparison. Is PCA more robust at detecting differences based on the morphometrics? Can you give me the details of your masters (title, school, ect) so I can reference it. Mike CC: r-help@r-project.org From: chrismco...@me.com Subject: Re: [R] discriminant function analysis Date: Tue, 16 Nov 2010 22:12:32 + To: megalop...@hotmail.com Hi, I did this exact thing for my masters, with intertidal fish, I just used a PCA? have you tried that? Sent from my iPhone On 16 Nov 2010, at 17:01, Mike Gibson megalop...@hotmail.com wrote: My objective is to look at differences in two species of fish from morphometric measurements. My morphometric measurements are head length, eye diameter, snout length, and measurements from tail to each fin. I want to use discrimanant function analyis to determine if there are differences between the two species. I am familiar with R but new to discrimannt function analysis. I want to learn how to run this analysis in R. My internet search has not been very successful. I did see refrence of the linear discriminant function (lda) in R but I am not sure if this is the correct function for me. Any advice and especially a reference to any examples would be greatly appreciated. Thanks. Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] overlap cdf plots and add colors and etc
Hi Peter and Jorge, Finally, it works. Thank you for your guidance. ## CDF plots par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(dt1),verticals=TRUE, horizontals=TRUE,pch=46,do.p=FALSE, col.hor=1, col.vert=1,lwd=2,) lines(ecdf(dt2),verticals=TRUE, horizontals=TRUE,pch=20,do.p=FALSE, col.hor=2, col.vert=2,lwd=2) legend(40, 0.8, legend = c(observed,fitted), col = c(1,2), pch=c(NA,NA), lty = c(1, 1), lwd=c(3,3),bty=n) What is plain text? Thank you. From: Peter Ehlers ehl...@ucalgary.ca Cc: Jorge Ivan Velez jorgeivanve...@gmail.com; r-help@r-project.org r-help@r-project.org Sent: Fri, November 26, 2010 12:52:56 AM Subject: Re: [R] overlap cdf plots and add colors and etc On 2010-11-24 22:24, Roslina Zakaria wrote: Hi Jorge, I tried but still it does not work. Thank you for your time. Jorge's code works perfectly well. If you prefer lines() over plot(, add = TRUE), then use lines(): plot( ecdf(rnorm(15, sd=3)), verticals=TRUE, col=black) lines(ecdf(rnorm(11, sd=3)), verticals=TRUE, col=red) legend(topright, legend = c(observed,fitted), col = c(black,red), pch=c(NA,NA), lty = c(1, 1), lwd=c(3,3),bty=n, pt.cex=2) (Note that topright is probably the worst choice you can make. And why set pt.cex when you don't have points??) The key thing is that you can't specify your two colours in the first plot() call. At that point, R has no idea that you want to add to the plot. (I understand that Greg Snow is working on a mind-reading package, but so far it can probably read only his mind, not yours.) Here is a (very brief) reminder of how to post: 1. Use an informative subject line. [[elided Yahoo spam]] 2. *Never* use the phrase it does not work. That is meaningless. Be specific about your problem. 3. Provide *reproducible* code. We don't have your 'datobs' or 'gam_sum_gen'. 4. Try to make the code *minimal*. It's not likely that anyone cares what labels you use for your axes/title (unless that's the problem). And nobody wants to see reams of data; use rnorm(), etc, or built-in data if possible. 5. Figure out how to set your mail program to send plain text. I know the above (and more) is in the posting guide, but it seems that nobody wants to read that quite brief document. Peter Ehlers From: Jorge Ivan Velezjorgeivanve...@gmail.com Cc: r-help@r-project.org Sent: Thu, November 25, 2010 4:46:37 PM Subject: Re: [R] overlap cdf plots and add colors and etc Hi Roslina, Try par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(rnorm(100))) plot(ecdf(rnorm(100)), add = TRUE, col = 2) HTH, Jorge On Thu, Nov 25, 2010 at 12:18 AM, Roslina Zakaria wrote: Hi r-users, I would like to overlap 2 ecdf plots. I tried this below and it gives me two plots of ecdf but just both just in black. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(datobs)) lines(ecdf(gam_sum_gen)) Then I try to add colors etc and also the legend but fail. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0),xaxs=i, yaxs=i) plot(ecdf(datobs),main =CDF of the sum for winter season-Hume,cex.axis=1.2,xlab=Rainfall (mm), pch = pch,verticals=TRUE,col=c(black,red), lty=c(1,1),ylab=Cumulative probability, xlim=c(0,800),lwd=1) lines(ecdf(gam_sum_gen)) legend(topright, legend = c(observed,fitted), col = c(black,red), pch=c(NA,NA), lty = c(1, 1), lwd=c(3,3),bty=n, pt.cex=2) box() I'm not sure why it doesn't show up at all. Thank you for any help given. [[alternative HTML version deleted]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coxph strange result
On Nov 25, 2010, at 5:16 PM, Bond, Stephen wrote: The following fit does not make sense to me, please, correct me if I have a logical error. moddowsn Call: coxph(formula = Surv(start, stop, resp) ~ sn + matfac2, data = coxsn1, method = efron) coef exp(coef) se(coef) z p sn2 0.0497 1.051 0.02030 2.450 1.4e-02 sn3 -0.0532 0.948 0.02038 -2.610 9.0e-03 sn4 -0.0410 0.960 0.01979 -2.073 3.8e-02 sn5 -0.0776 0.925 0.01954 -3.973 7.1e-05 sn6 -0.1133 0.893 0.01839 -6.161 7.2e-10 sn7 -0.1252 0.882 0.01846 -6.781 1.2e-11 sn8 -0.1222 0.885 0.01994 -6.130 8.8e-10 sn9 -0.0507 0.951 0.02047 -2.478 1.3e-02 sn10 -0.0444 0.957 0.02056 -2.159 3.1e-02 sn11 -0.0433 0.958 0.02157 -2.008 4.5e-02 sn12 -0.0114 0.989 0.02037 -0.557 5.8e-01 matfac22 -0.2599 0.771 0.01727 -15.048 0.0e+00 matfac25 -0.1804 0.835 0.00924 -19.512 0.0e+00 Likelihood ratio test=651 on 13 df, p=0 n= 253802 This would indicate that in sn6 to sn8 there is less of a chance of an event. ?? do the relative frequencies implied by the following table make any sense?? table(coxsn1$matfac2,coxsn1$sn,coxsn1$resp) , , = 0 1 2 3 4 5 6 7 8 910 1112 1 3407 3177 3425 3348 3975 3564 3181 3077 2894 2610 3441 3443 2 920 1005 1142 1327 1645 1530 1330 1184 964 864 888 860 5 9036 9507 10258 11888 16826 15575 13394 12346 9938 9001 8970 8599 , , = 1 1 2 3 4 5 6 7 8 910 1112 1 1453 1459 1186 1496 1295 1754 1429 1153 1106 1234 965 1532 2 312 290 330 390 454 539 479 367 295 276 256 267 5 2994 3207 3371 3629 4095 5581 5837 3844 3400 3199 2705 3084 Apparently the frequency of an event is higher in the summer months. Without knowing how the data was prepared, agreement would seem highly speculative on our parts. We haven't been told very much about this dataset apart from the cryptic variable names apparently referring to calendar months regarding some aspect of the cases. I apologize for not being able to disclose the dataset, but think that the table provides enough to address the question. I disagee. Suppose sn2 cases began observation in February and then had events in July or September. Or even supposing that observations are confined within a single month. If on average, the durations are longer within those months than in others, then you could get lower rates without having lower even frequencies. Thanks everybody. Stephen B David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Go (back) from Rd to roxygen
On 25/11/2010 3:45 PM, Yihui Xie wrote:
Hi all,
Since roxygen is a great help to document R packages, I am wondering
if there exists an approach to go back from the raw Rd files to
roxygen-documentation? E.g. turn \author{Somebody} into @author
Somebody. This sounds ridiculous, but I believe it helps in the long
term for me to maintain R packages.
I have no idea, but it should be reasonably straightforward to write
such a thing, since there's an Rd parser in the tools package. Take a
look at Rd2txt or one of the other converters that works on parsed Rd
code, and write Rd2roxygen.
Duncan Murdoch
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Re: [R] How to bin/average time points?
It would be something like this (might have to change the syntax a bit)
bin_ave=0;
while (i lenth(time)){
bin_ave[k]=mean(time(i:i+6));
k=k+1;
i=i+6;
}
if your data is in a table format replace time with mytable$time.
hope this helps,
Sachin
p.s. sorry about corporate notice
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cumsum with a max and min value
On Thu, Nov 25, 2010 at 3:44 PM, henrique henri...@allianceasset.com.br wrote: I have a vector of values -1, 0, 1, say a - c(0, 1, 0, 1, 1, -1, -1, -1, 0, -1, -1) I want to create a vector of the cumulative sum of this, but I need to set a maximum and minimum value for the cumsum, for example: max_value - 2 min_value - -2 the expected result would be (0, 1, 1, 2, 2, 1, 0, -1, -1, -2, -2) Try this: f - function(x, y) max(min(x + y, max_value), min_value) Reduce(f, a, 0, accumulate = TRUE)[-1] -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Installing RQuantLib on Win 7 64 Bit
Hello Group, I am trying out RQuantLib on a 64bit Win 7 machine. But running into installation errors install.packages(RQuantLib) Warning in install.packages(RQuantLib) : argument 'lib' is missing: using 'C:\Users\Tester\Documents/R/win64-library/2.11' Warning: unable to access index for repository http://www.stats.ox.ac.uk/pub/RWin/bin/windows64/contrib/2.11 Warning message: In getDependencies(pkgs, dependencies, available, lib) : package ‘RQuantLib’ is not available Any help with this installation? Thank you. S __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cumsum with a max and min value
Does this do it; pmin(2, pmax(-2, cumsum(a))) [1] 0 1 1 2 2 2 1 0 0 -1 -2 On Thu, Nov 25, 2010 at 3:44 PM, henrique henri...@allianceasset.com.br wrote: I have a vector of values -1, 0, 1, say a - c(0, 1, 0, 1, 1, -1, -1, -1, 0, -1, -1) I want to create a vector of the cumulative sum of this, but I need to set a maximum and minimum value for the cumsum, for example: max_value - 2 min_value - -2 the expected result would be (0, 1, 1, 2, 2, 1, 0, -1, -1, -2, -2) The only way I managed to do It, was res - vector(length=length(a)) res[1] - a[1] for ( i in 2:length(a)) res[i] - res[i-1] + a[i] * (( res[i-1] max_value a[i] 0 ) | ( res[i-1] min_value a[i] 0 )) This is certainly not the best way to do it, so any suggestions? Henrique [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Go (back) from Rd to roxygen
Since roxygen is a great help to document R packages, I am wondering
if there exists an approach to go back from the raw Rd files to
roxygen-documentation? E.g. turn \author{Somebody} into @author
Somebody. This sounds ridiculous, but I believe it helps in the long
term for me to maintain R packages.
Have a look at https://gist.github.com/d1bbd44894a99a2e1d1b for a start.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to bin/average time points?
try this:
# create times 9 minutes apart
time - seq(as.POSIXct('2010-11-25 00:00'), by = '9 min', length = 480)
mySamp - data.frame(time = time, value = sample(1:100, length(time), TRUE))
# add column to split by hour
mySamp$hour - format(mySamp$time, '%Y-%m-%d %H:30')
# compute the mean for each hour
tapply(mySamp$value, mySamp$hour, mean)
2010-11-25 00:30 2010-11-25 01:30 2010-11-25 02:30 2010-11-25 03:30
2010-11-25 04:30 2010-11-25 05:30
54.42857 59.85714 47.5 45.71429
40.28571 56.5
2010-11-25 06:30 2010-11-25 07:30 2010-11-25 08:30 2010-11-25 09:30
2010-11-25 10:30 2010-11-25 11:30
46.57143 47.14286 34.0 53.85714
50.28571 36.0
2010-11-25 12:30 2010-11-25 13:30 2010-11-25 14:30 2010-11-25 15:30
2010-11-25 16:30 2010-11-25 17:30
31.57143 44.57143 42.5 52.42857
54.14286 44.7
2010-11-25 18:30 2010-11-25 19:30 2010-11-25 20:30 2010-11-25 21:30
2010-11-25 22:30 2010-11-25 23:30
50.28571 60.57143 36.0 42.14286
65.14286 37.5
2010-11-26 00:30 2010-11-26 01:30 2010-11-26 02:30 2010-11-26 03:30
2010-11-26 04:30 2010-11-26 05:30
51.71429 58.85714 48.5 45.0
44.0 38.0
2010-11-26 06:30 2010-11-26 07:30 2010-11-26 08:30 2010-11-26 09:30
2010-11-26 10:30 2010-11-26 11:30
56.0 34.14286 64.7 51.42857
57.57143 44.5
2010-11-26 12:30 2010-11-26 13:30 2010-11-26 14:30 2010-11-26 15:30
2010-11-26 16:30 2010-11-26 17:30
65.0 59.57143 63.5 52.57143
36.85714 63.3
2010-11-26 18:30 2010-11-26 19:30 2010-11-26 20:30 2010-11-26 21:30
2010-11-26 22:30 2010-11-26 23:30
44.85714 64.85714 63.0 62.57143
62.0 57.0
2010-11-27 00:30 2010-11-27 01:30 2010-11-27 02:30 2010-11-27 03:30
2010-11-27 04:30 2010-11-27 05:30
26.71429 33.57143 37.5 67.0
47.85714 63.0
2010-11-27 06:30 2010-11-27 07:30 2010-11-27 08:30 2010-11-27 09:30
2010-11-27 10:30 2010-11-27 11:30
40.28571 46.42857 54.5 41.0
51.0 58.3
2010-11-27 12:30 2010-11-27 13:30 2010-11-27 14:30 2010-11-27 15:30
2010-11-27 16:30 2010-11-27 17:30
62.14286 52.28571 75.3 43.71429
53.14286 27.5
2010-11-27 18:30 2010-11-27 19:30 2010-11-27 20:30 2010-11-27 21:30
2010-11-27 22:30 2010-11-27 23:30
33.42857 56.85714 57.8 51.0
57.71429 38.7
On Thu, Nov 25, 2010 at 3:49 PM, DonDolowy kevin.dalga...@gmail.com wrote:
Dear all,
I am pretty new to R only having an introduction course, so please bare with
me. I am doing my PhD at The Max Planck Institute of Immunobiology where I
am analyzing some calorimetry data from some mice.
I have a spreadsheet consisting of measurements of the respiratory exchange
rate at different time points measured every 9 minutes over some days.
My goal is bin/average the time points of each hour to only one
measurements.
E.g.
[Time] - [Measurement]
12.09 - 0.730
12.18 - 0.732
12.27 - 0.743
12.36 - 0.757
12.45 - 0.781
12.54 - 0.731
-- should be averaged to fx one time point and one value, fx:
12.30 - [average of the six measurements]
I know how to average the measurements in a whole column but how to average
every six measurements automatically and also how to average every six time
points and make a new sheet consisting of these data?
I hope you guys are able to help, since we are really stuck here. I can of
course do it manually but with 8000 measurements it will take lots of time.
Thank you very much.
Best regards,
Kevin Dalgaard
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What is the problem that you are trying to solve?
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Re: [R] How to bin/average time points?
Hi Kevin Here is one way: yourData - c(0.730, 0.732, 0.743, 0.757,0.781, 0.731, 0.830, 0.832, 0.843, 0.857, 0.881, 0.831) nrGroups - 2 lengthGroups - 6 tapply(yourData, factor(rep(c(1,nrGroups), each = lengthGroups)), mean) and you will have to adjust the number of groups and if necessary the length of the groups to your needs. I am assuming that all groups are of the same length. Do you really want to average the time points? In that case you might want to consider the data structures for representing dates and times, see e.g. ?POSIXct or converting the times to minutes, say. - Niels On 25/11/10 12.49, DonDolowy wrote: Dear all, I am pretty new to R only having an introduction course, so please bare with me. I am doing my PhD at The Max Planck Institute of Immunobiology where I am analyzing some calorimetry data from some mice. I have a spreadsheet consisting of measurements of the respiratory exchange rate at different time points measured every 9 minutes over some days. My goal is bin/average the time points of each hour to only one measurements. E.g. [Time] - [Measurement] 12.09 - 0.730 12.18 - 0.732 12.27 - 0.743 12.36 - 0.757 12.45 - 0.781 12.54 - 0.731 -- should be averaged to fx one time point and one value, fx: 12.30 - [average of the six measurements] I know how to average the measurements in a whole column but how to average every six measurements automatically and also how to average every six time points and make a new sheet consisting of these data? I hope you guys are able to help, since we are really stuck here. I can of course do it manually but with8000 measurements it will take lots of time. Thank you very much. Best regards, Kevin Dalgaard __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] About searching criteria
Hi Henrique, Lot of thanks for your advice which is a little complicate to me. It involves multiple R commands: d - data() ne - new.env() data(list = grep(\\(, d$results[,'Item'], value = TRUE, invert = TRUE), envir = ne) out - eapply(ne, names) names(which(lapply(lapply(out, '%in%', c(Run, conc, density)), sum) == 3)) [1] DNase DNase Runconc density 1 1 0.04882812 0.017 2 1 0.04882812 0.018 3 1 0.19531250 0.121 4 1 0.19531250 0.124 ... It works for me. grep - is simliar to sh command grep, egrep, fgrep, rgrep - print lines matching a pattern which - differs from sh command which - locate a command Wonderful !!! B.R. Stephen L From: Henrique Dallazuanna www...@gmail.com Cc: r-help@r-project.org Sent: Fri, November 26, 2010 12:35:23 AM Subject: Re: [R] About searching criteria Try this: d - data() ne - new.env() data(list = grep(\\(, d$results[,'Item'], value = TRUE, invert = TRUE), envir = ne) out - eapply(ne, names) names(which(lapply(lapply(out, '%in%', c(Run, conc, density)), sum) == 3)) Hi folks, I need to search the dataset on data with name on heading; Run conc density I look at ??help.search and could not resolve; help.search(pattern, fields = c(alias, concept, title) What shall I replace pattern? I suppose replacing alias, concept, title with Run, conc, density ? Please help. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to unsubscribe to mailing list
How to unsubscribe to mailing list -- View this message in context: http://r.789695.n4.nabble.com/How-to-unsubscribe-to-mailing-list-tp3059708p3059708.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] get list index
Hi R-users, I have a list mylist - list(c(0.79, 0.92, 0.91, 0.86, 0.96, 0.96, 0.95, 0.94, 0.99), c(0.28, 0.45, 0.59, 0.69, 0.80, 0.87, 0.95, 0.94, 0.98), c(0.29, 0.39, 0.59, 0.69, 0.68, 0.80, 0.93, 0.95, 0.98)) Is there a way to find the index of the list element that contains the lowest value among all the other elements? As the lowest value in each element is the first, the question is actually how to find the lowest 'first' values among the list elements, and then get the index of that element. In my example the list element would be (because the value is 0.28): [[2]] [1] 0.28 0.45 0.59 0.69 0.80 0.87 0.95 0.94 0.98 and the position of course 2. I am looking for the index because I would like to subset the list afterwards (e.g. mylist[[2]]) and extract that element (i.e. the whole vector). Thanks for your help Lorenzo Lorenzo Cattarino PhD Candidate (Confirmed) Landscape Ecology and Conservation Group Centre for Spatial Environmental Research School of Geography, Planning and Environmental Management The University of Queensland Brisbane, Queensland, 4072, Australia Telephone 61-7-3365 4370, Mobile 0410884610 Email l.cattar...@uq.edu.au Internet http://www.gpem.uq.edu.au/cser http://www.gpem.uq.edu.au/cser [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] get list index
Hello Lorezo, Try this... order(sapply(mylist, min))[1] Michael On 26 November 2010 11:23, Lorenzo Cattarino l.cattar...@uq.edu.au wrote: Hi R-users, I have a list mylist - list(c(0.79, 0.92, 0.91, 0.86, 0.96, 0.96, 0.95, 0.94, 0.99), c(0.28, 0.45, 0.59, 0.69, 0.80, 0.87, 0.95, 0.94, 0.98), c(0.29, 0.39, 0.59, 0.69, 0.68, 0.80, 0.93, 0.95, 0.98)) Is there a way to find the index of the list element that contains the lowest value among all the other elements? As the lowest value in each element is the first, the question is actually how to find the lowest 'first' values among the list elements, and then get the index of that element. In my example the list element would be (because the value is 0.28): [[2]] [1] 0.28 0.45 0.59 0.69 0.80 0.87 0.95 0.94 0.98 and the position of course 2. I am looking for the index because I would like to subset the list afterwards (e.g. mylist[[2]]) and extract that element (i.e. the whole vector). Thanks for your help Lorenzo Lorenzo Cattarino PhD Candidate (Confirmed) Landscape Ecology and Conservation Group Centre for Spatial Environmental Research School of Geography, Planning and Environmental Management The University of Queensland Brisbane, Queensland, 4072, Australia Telephone 61-7-3365 4370, Mobile 0410884610 Email l.cattar...@uq.edu.au Internet http://www.gpem.uq.edu.au/cser http://www.gpem.uq.edu.au/cser [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing RQuantLib on Win 7 64 Bit
On 26 November 2010 at 07:05, Santosh Srinivas wrote: | Hello Group, | | I am trying out RQuantLib on a 64bit Win 7 machine. But running into | installation errors The error message is about as clear as it can get: | install.packages(RQuantLib) | | Warning in install.packages(RQuantLib) : | argument 'lib' is missing: using | 'C:\Users\Tester\Documents/R/win64-library/2.11' | Warning: unable to access index for repository | http://www.stats.ox.ac.uk/pub/RWin/bin/windows64/contrib/2.11 | Warning message: | In getDependencies(pkgs, dependencies, available, lib) : | package ‘RQuantLib’ is not available There is your answer: there simply is no binary package. I need a win64 development box (which I currently do not have) to build QuantLib as a win64 library so that CRAN and R-Forge can turn the RQuantLib source into a binary for you. Dirk -- Dirk Eddelbuettel | e...@debian.org | http://dirk.eddelbuettel.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] overlap histogram and density
Hi Ted, Regarding your examples, is it possible to get a smooth line for the density which overlap with the histogram? Regards, Roslina From: ted.hard...@wlandres.net ted.hard...@wlandres.net To: r-help@r-project.org Sent: Fri, November 12, 2010 6:42:31 AM Subject: Re: [R] overlap histogram and density [OOPS!!I accidentally reproduced my second example below as my third example. Now corrected. See below.] On 11-Nov-10 20:02:29, Ted Harding wrote: On 11-Nov-10 18:39:34, Roslina Zakaria wrote: Hi, Does anybody encounter the same problem when we overlap histogram and density that the density line seem to shift to the right a little bit? If you do have the same problem, what should we do to correct that? Thank you. par(mar=c(4,4,2,1.2),oma=c(0,0,0,0)) hist(datobs,prob=TRUE, main =Volume of a catchment from four stations, col=yellowgreen, cex.axis=1, xlab=rainfall, ylab=Relative frequency, ylim= c(0,.003), xlim=c(0,1200)) lines(density(dd), lwd=3,col=red) #legend(topright,c(observed,generated), # lty=c(0,1),fill=c(blue,),bty=n) legend(topright, legend = c(observed,generated), col = c(yellowgreen, red), pch=c(15,NA), lty = c(0, 1), lwd=c(0,3),bty=n, pt.cex=2) box() Thank you. In theory that is not a problem. The density() function will estimate a density whose integral over each of the intervals in the histogram is equal to the probability of that interval, and the proportion of the data expected in that interval will also be its probability. In practice, the estent to which you observe what you describe (or a displacement to the left) will depend on how your data are distributed within the intervals, and on the precision with which density() happens to estimate the true density. The following 3 cases of the same data sampled from a log-Normal distribution, illustrate different impressions of the kind that one might get, depending on the details of the histogram. Note that there is no overall effect of displacement to the right in any histogram, while the extent to which one observes it varies according to the histogram. Without knowledge of your data it is not possible to comment further on the extent to [[elided Yahoo spam]] set.seed(54321) N - 1000 X - exp(rnorm(N,sd=0.4)) dd - density(X) # A coarse histogram H - hist(X,prob=TRUE, xlim=c(-0.5,4),ylim=c(0,max(dd$y)),breaks=0.5*(0:8)) dx - unique(diff(H$breaks)) lines(dd$x,dd$y) ## A finer histogram H - hist(X,prob=TRUE, xlim=c(-0.5,4),ylim=c(0,max(dd$y)),breaks=0.25*(0:16)) dx - unique(diff(H$breaks)) lines(dd$x,dd$y) ## A still finer histogram H - hist(X,prob=TRUE, ## OOPS!! xlim=c(-0.5,4),ylim=c(0,max(dd$y)),breaks=0.25*(0:16)) xlim=c(-0.5,4),ylim=c(0,max(dd$y)),breaks=0.20*(0:20)) dx - unique(diff(H$breaks)) lines(dd$x,dd$y) Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 11-Nov-10 Time: 20:12:27 -- XFMail -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Go (back) from Rd to roxygen
Awesome! Thanks a lot!
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Thu, Nov 25, 2010 at 8:09 PM, Hadley Wickham had...@rice.edu wrote:
Since roxygen is a great help to document R packages, I am wondering
if there exists an approach to go back from the raw Rd files to
roxygen-documentation? E.g. turn \author{Somebody} into @author
Somebody. This sounds ridiculous, but I believe it helps in the long
term for me to maintain R packages.
Have a look at https://gist.github.com/d1bbd44894a99a2e1d1b for a start.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
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[R] How to save a data set as .txt on fly?
Hi folks, Win7 64bit I tried to save DNase, a data set on database, as .txt file for future use with load. I can't do it on fly; save(DNase, file=C:/Users/satimis/Documents/aaa.txt) load(file=C:/Users/satimis/Documents/aaa.txt) aaa Error: object 'aaa' not found aaa.txt Error: object 'aaa.txt' not found I must perform following steps; aaa-DNase save(aaa, file=C:/Users/satimis/Documents/aaa.txt) load(file=C:/Users/satimis/Documents/aaa.txt) aaa Runconc density 1 1 0.04882812 0.017 2 1 0.04882812 0.018 3 1 0.19531250 0.121 Is there any way doing it on fly? Besides aaa.txt can't be read direct with Notepad nor WordPad TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to save a data set as .txt on fly?
On Nov 25, 2010, at 10:45 PM, Stephen Liu wrote: Hi folks, Win7 64bit I tried to save DNase, a data set on database, as .txt file for future use with load. I can't do it on fly; save(DNase, file=C:/Users/satimis/Documents/aaa.txt) load(file=C:/Users/satimis/Documents/aaa.txt) aaa Error: object 'aaa' not found aaa.txt Error: object 'aaa.txt' not found But you didn't try: DNase# which was after all the name of the object you saved. I must perform following steps; aaa-DNase save(aaa, file=C:/Users/satimis/Documents/aaa.txt) load(file=C:/Users/satimis/Documents/aaa.txt) aaa Runconc density 1 1 0.04882812 0.017 2 1 0.04882812 0.018 3 1 0.19531250 0.121 Is there any way doing it on fly? Does the above method (typing the name of the save object rather than the file name) meet your definition of on the fly? Besides aaa.txt can't be read direct with Notepad nor WordPad Right. Try reading the Posting Guide which among many other useful things has a method for saving objects in text format. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to save a data set as .txt on fly?
Hi David, But you didn't try: DNase# which was after all the name of the object you saved. Sorry I don't follow. I can't do it with following steps: DNase save(DNase, file=C:/Users/satimis/Documents/dnase.txt) load(file=C:/Users/satimis/Documents/dnase.txt) dnase Error: object 'dnase' not found dnase.txt Error: object 'dnase.txt' not found sink command can make it sink(dnase.txt, append=TRUE, split=TRUE) DNase sink() to load the file read.table(file=C:/Users/satimis/Documents/dnase.txt) Besides the file created can be read with Notpad and WordPad Does the above method (typing the name of the save object rather than the file name) meet your definition of on the fly? Could you pls explain in more detail? Thanks Right. Try reading the Posting Guide which among many other useful things has a method for saving objects in text format. Yes. I'm curious to know why the .txt file created in this way can't be read with Notpad and WordPad? B.R. Stephen L - Original Message From: David Winsemius dwinsem...@comcast.net To: Stephen Liu sati...@yahoo.com Cc: r-help@r-project.org Sent: Fri, November 26, 2010 11:56:23 AM Subject: Re: [R] How to save a data set as .txt on fly? On Nov 25, 2010, at 10:45 PM, Stephen Liu wrote: Hi folks, Win7 64bit I tried to save DNase, a data set on database, as .txt file for future use with load. I can't do it on fly; save(DNase, file=C:/Users/satimis/Documents/aaa.txt) load(file=C:/Users/satimis/Documents/aaa.txt) aaa Error: object 'aaa' not found aaa.txt Error: object 'aaa.txt' not found But you didn't try: DNase# which was after all the name of the object you saved. I must perform following steps; aaa-DNase save(aaa, file=C:/Users/satimis/Documents/aaa.txt) load(file=C:/Users/satimis/Documents/aaa.txt) aaa Runconc density 1 1 0.04882812 0.017 2 1 0.04882812 0.018 3 1 0.19531250 0.121 Is there any way doing it on fly? Does the above method (typing the name of the save object rather than the file name) meet your definition of on the fly? Besides aaa.txt can't be read direct with Notepad nor WordPad Right. Try reading the Posting Guide which among many other useful things has a method for saving objects in text format. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] \Sweaveopts error
Yes, that has fixed the problem. (2010-11-24 r53659)
Thanks.
John Maindonald email: john.maindon...@anu.edu.au
phone : +61 2 (6125)3473fax : +61 2(6125)5549
Centre for Mathematics Its Applications, Room 1194,
John Dedman Mathematical Sciences Building (Building 27)
Australian National University, Canberra ACT 0200.
http://www.maths.anu.edu.au/~johnm
On 25/11/2010, at 10:56 PM, Duncan Murdoch wrote:
On 25/11/2010 6:34 AM, John Maindonald wrote:
I have a file 4lmetc.Rnw, intended for inclusion in a LaTeX document,
that starts:
I think this may have been fixed in the patched version. Could you give it a
try to confirm? If not, please send me a simplified version of the file, and
I'll see what's going wrong.
Duncan Murdoch
\SweaveOpts{engine=R, keep.source=TRUE}
\SweaveOpts{eps=FALSE, prefix.string=snArt/4lmetc}
The attempt to process the file through Sweave generates the error:
Sweave(4lmetc)
Writing to file 4lmetc.tex
Processing code chunks ...
1 : keep.source term verbatim
Error in file(srcfile$filename, open = rt, encoding = encoding) :
cannot open the connection
In addition: Warning message:
In file(srcfile$filename, open = rt, encoding = encoding) :
cannot open file '4lmetc': No such file or directory
The same file processes through Stangle() without problems.
If I comment out the \Sweaveopts lines, there is no problem,
except that I do not get the options that I want.
This processed fine in R-2.11.1
sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets grid methods
[8] base
other attached packages:
[1] lattice_0.19-13 DAAG_1.02 randomForest_4.5-36
[4] rpart_3.1-46MASS_7.3-8 reshape_0.8.3
[7] plyr_1.2.1 proto_0.3-8
loaded via a namespace (and not attached):
[1] ggplot2_0.8.8 latticeExtra_0.6-14
Is there a workaround?
John Maindonald email: john.maindon...@anu.edu.au
phone : +61 2 (6125)3473fax : +61 2(6125)5549
Centre for Mathematics Its Applications, Room 1194,
John Dedman Mathematical Sciences Building (Building 27)
Australian National University, Canberra ACT 0200.
http://www.maths.anu.edu.au/~johnm
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