Re: [R] install.packages() - old version deleted, new version did not install
On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Jon __ 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] Alternative to extended recode sintax? Bug?
All right, I get it now: lubridate's week() define weeks from Thursday till the following Wednesday. You'd probably agree with me that it's a bit strange what it is going to do over the turn of the year: y - as.POSIXct(c(2010-12-27,2010-12-28,2010-12-29,2010-12-30,2010-12-31,2011-01-01,2011-01-02,2011-01-03,2011-01-04,2011-01-05,2011-01-06,2011-01-07,2011-01-08,2011-01-09,2011-01-10,2011-01-11,2010-01-12,2010-01-13,2010-01-14)) week(y) [1] 52 52 52 53 53 1 1 1 1 1 1 2 2 2 2 2 2 2 3 Why would the first week of the year be made of 6 days and the turn from week 1 to week 2 on the night between Thursday and Friday and not Wednesday and Friday like every other week? Cheers, Luca Il giorno 19/dic/2010, alle ore 18.14, Uwe Ligges ha scritto: On 19.12.2010 13:20, David Winsemius wrote: On Dec 19, 2010, at 5:11 AM, Luca Meyer wrote: Something goes wrong with the week function of the lubridate package: x= as.POSIXct(factor(c(2010-12-15 17:28:27, + 2010-12-15 17:32:34, + 2010-12-15 18:48:39, + 2010-12-15 19:25:00, + 2010-12-16 08:00:00, + 2010-12-16 08:25:49, + 2010-12-16 09:00:00))) require(lubridate) weekdays(x) [1] Mercoledì Mercoledì Mercoledì Mercoledì Giovedì Giovedì Giovedì week(x) [1] 50 50 50 50 51 51 51 But 2010-12-15 is a Wednesday and 2010-12-16 is a Thursday. Together with the description of ?week this shows that lubridate's week() function works as documented rather than as expected by Luca Meyer. Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R is not running well
Hi, I'm working on windows 7. Recently I install the latest R and also use together with Tinn-R but it is not working well. I got the following message. I have tried saving the Rprofile as suggested by the R forum, but still it is not running. What should I do? source(.trPaths[5], echo=TRUE, max.deparse.length=150) Error in source(.trPaths[5], echo = TRUE, max.deparse.length = 150) : object '.trPaths' not found [[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] incremental learning for LOESS time series model
Hi All, I am currently working on some time series data, I know I can use LOESS/ARIMA model. The data is written to a vector whose length is 1000, which is a queue, updating every 15 minutes, Thus the old data will pop out while the new data push in the vector. I can rerun the whole model on a scheduler, e.g. retrain the model every 15 minutes, that is, Use the whole 1000 value to train The LOESS model, However it is very inefficient, as every time only one value is insert while another 999 vlaues still same as last time. So how can I achieve better performance? Many thanks Ying [[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] Alternative to extended recode sintax? Bug?
On 20.12.2010 06:54, Luca Meyer wrote: All right, I get it now: lubridate's week() define weeks from Thursday till the following Wednesday. You'd probably agree with me that it's a bit strange what it is going to do over the turn of the year: y- as.POSIXct(c(2010-12-27,2010-12-28,2010-12-29,2010-12-30,2010-12-31,2011-01-01,2011-01-02,2011-01-03,2011-01-04,2011-01-05,2011-01-06,2011-01-07,2011-01-08,2011-01-09,2011-01-10,2011-01-11,2010-01-12,2010-01-13,2010-01-14)) week(y) [1] 52 52 52 53 53 1 1 1 1 1 1 2 2 2 2 2 2 2 3 Why would the first week of the year be made of 6 days and the turn from week 1 to week 2 on the night between Thursday and Friday and not Wednesday and Friday like every other week? Well, it's the definition in that week() function from that package, if you don't like that definition, choose another one. I have not said that I like it, just that it seems to work as documented. Uwe Cheers, Luca Il giorno 19/dic/2010, alle ore 18.14, Uwe Ligges ha scritto: On 19.12.2010 13:20, David Winsemius wrote: On Dec 19, 2010, at 5:11 AM, Luca Meyer wrote: Something goes wrong with the week function of the lubridate package: x= as.POSIXct(factor(c(2010-12-15 17:28:27, + 2010-12-15 17:32:34, + 2010-12-15 18:48:39, + 2010-12-15 19:25:00, + 2010-12-16 08:00:00, + 2010-12-16 08:25:49, + 2010-12-16 09:00:00))) require(lubridate) weekdays(x) [1] Mercoledì Mercoledì Mercoledì Mercoledì Giovedì Giovedì Giovedì week(x) [1] 50 50 50 50 51 51 51 But 2010-12-15 is a Wednesday and 2010-12-16 is a Thursday. Together with the description of ?week this shows that lubridate's week() function works as documented rather than as expected by Luca Meyer. Uwe Ligges __ 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] Layout of mulitpage conditioned lattice plots
David Winsemius wrote:
Here's my latest guess at what you may want:
pdf(file=multpage.pdf)
xyplot(val~time|subj + comp, data=dt,type=l,
layout=c(3,5, 3),
skip=rep(c(rep(FALSE,13), TRUE, TRUE), 3) )
dev.off()
Not really, but skip was the right idea. I added another idea of Deepayan
from a cited thread, first to plot all, then to update indexed parts with a
computed skip.
The code has become a bit lengthy because I added a more flexible
orphan-avoiding scheme.
Dieter
library(lattice)
# Distribute panels on page, so that each panel has the same size,
# even on last page
# Use adjustCol to adjust colPerPage to avoid orphans on the last page
#
# - adjustedColPerPage
-
adjustedColPerPage = function(colPerPage, ncols){
# Allow for 20% or plus/minus 2
searchRange = max(2L,as.integer(colPerPage*0.2))
colsPerPage = (colPerPage-searchRange):(colPerPage+searchRange)
nColLast = ncols %% colsPerPage
nPages = (ncols %/% colsPerPage)+ as.integer(nColLast!=0)
# Prefer solution with equal number on a page
matchPage = which(nColLast==0)
if (length(matchPage) 0) {
colsPerPage[matchPage[which.min(abs(matchPage-searchRange))]]
} else {
colsPerPage[which.max(nColLast)] # not perfect
}
}
# - xyPaged
--
xyPaged = function(x, adjustCol = FALSE, colPerPage = 5,main=NULL) {
nrows = nlevels(x$comp) # This is not very general
ncols = nlevels(x$subj) #
if (adjustCol) # try to get an alternative layout that fits the pages
better
{
colPerPage = adjustedColPerPage(colPerPage,ncols)
main = paste(main, usedCol= ,colPerPage)
}
p = xyplot(val~time|subj+comp, data=x,type=l,
layout = c(colPerPage,nrows),main=main)
# http://r-project.markmail.org/thread/rcztoawll5kduw4x
page = 1
for (fromCol in seq(1,ncols,by=colPerPage)){
toCol = min(fromCol+colPerPage-1,ncols)
showCol = toCol %% colPerPage
skip = rep(FALSE,colPerPage)
if (showCol != 0) skip[(showCol+1):colPerPage] = TRUE
print(update(p[fromCol:toCol],skip=skip,sub=page))
page = page +1
}
}
# Test
testFrame = expand.grid(adjustCol=c(FALSE,TRUE),
nsubj=c(5,11,13),colPerPage=c(5,9,14) )
pdf(file=multpage.pdf)
for (i in 1:nrow(testFrame)) {
test = testFrame[i,]
dt = expand.grid(time=1:20,comp=LETTERS[1:3],subj=letters[1:test$nsubj])
dt$val = rnorm(nrow(dt))
with (test, xyPaged(dt,adjustCol, colPerPage,
main=paste(nsubj=,test$nsubj, requestedCol= ,colPerPage)))
}
dev.off()
--
View this message in context:
http://r.789695.n4.nabble.com/Layout-of-mulitpage-conditioned-lattice-plots-tp3094581p3095284.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] barplot: width of label
On 12/20/2010 02:13 AM, fransiepansiekevertje wrote:
Hello,
I try to make barplots with rather wide labels. A simplified example of
this:
x- c(12, 33, 56, 67, 15, 66)
names(x)- c('Richard with a long surname','Minnie with a long
name,'Albert','Helen','Joe','Kingston')
barplot(x, las = 2)
Now the label 'Richard with a long surname' is too long to fit beneath the
bars. A simple solution would be enlarge the space for the labels by
positioning the bar region higher. But I cannot find how to do this. Please
Help!
Hi Frans,
Does this do what you want?
library(plotrix)
barp(x,names.arg=names(x),staxx=TRUE)
Jim
__
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] contourplot help
Hello I'm using the following call to create a contourplot: library(lattice) m - as.matrix(read.table(data.txt)) contourplot(m[,3] ~ m[,2] * -m[,1], at = c(1e-6, 1e-5, 1e-4, 1e-3, 1e-2, 1e-1), scales = list(x = list(log = 10, labels = c(1, 10, 100), at = c(1, 10, 100)), y = list(labels = c(14, 12, 10, 8, 6, 4, 2)), at = c(-14, -12, -10, -8, -6, -4, -2)), labels = c(expression(10^-6), expression(10^-5), expression(10^-4), expression(10^-3), expression(10^-2), expression(10^-1), expression(10^0)), xlim = c(0.75, 10^2), xlab = Out-degree, ylab = In-degree) Which gives the the output in the file below http://ompldr.org/vNm4xag/contour.eps As it can be seen, the level labels are not displayed nicely because there's not enough room for them. Also, the 10^-1 label is not displayed. Is there a way for me to hardcode the position of each label? I tried setting labels = F and then calling text() for each one, but that doesn't work. If that's not possible, one option would be to color each level line differently and then add a legend. Is it possible to do that? Finally, how can I remove the tick marks from the top and right axes? Thanks, Andre __ 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] For-loop
Hi,
I have the following problem:
I have a data.frame with 36 sample sites (colums) for which I have covariates
in 3 categories: Area, Month and River. Each Area consists of 3 rivers, which
were sampled over 3 month. Now I want to fuse River 1-3 for one area in one
month. To get a data.frame with 12 colums.
I am trying to do a for loop (which may be a complicated solution, but I
don't see an easier way), which is not working, apparently because a[,ij] or
a[,c(i,j)] is not working as a definition of the matrix with a double condition
in the colums.
How can I make it work or what would be an easier solution?
Thank you for your help,
Anne
data=data.frame(matrix(1:99,nrow=5,ncol=36))
colnames(data)=c(paste(plot,1:36))
cov=data.frame(rep(1:3,12),c(rep(Jan,12),rep(Feb,12),rep(Mar,12)),rep(c(1,1,1,2,2,2,3,3,3,4,4,4),3))
dimnames(cov)=list(colnames(data),c(River,Month,Area))
###loop###
a=matrix(nrow=dim(data)[1],ncol=length(levels(factor(cov$Month)))*length(levels(factor(cov$Area
for(i in 1:length(levels(factor(cov$Month
{
for(j in 1:length(levels(factor(cov$Area
{
a[,ij]=as.numeric(rowSums(data[,factor(cov$Month)==levels(factor(cov$Month))[i]factor(cov$Area)==levels(factor(cov$Area))[j]]))
}
}
__
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] transposing panel data
I am currently trying to transpose some large panel data set ie transposing
multiple rows in into a single column. instead the transpose functionality
transposes all rows into columns. my sample data set looks like below:
ACCT_NUM
ACCOUNT_NAME
TRAN_AMT
DATE
EMPLOYER
101913
GK
7489
30-Apr-10
PENSION
101913
GK
7489
30-May-10
PENSION
101913
GK
7489
30-Jun-10
PENSION
101913
GK
7489
31-Jul-10
PENSION
101913
GK
7489
31-Jan-10
PENSION
101913
GK
7489
28-Feb-10
PENSION
101913
GK
7489
31-Mar-10
PENSION
101913
GK
7489
30-Oct-10
PENSION
101913
GK
7489
30-Aug-10
PENSION
101913
GK
7489
23-Sep-10
PENSION
101913
GK
53772
30-Sep-10
MKU
101920
CKM
22739.7
31-Jul-10
POLICE
101920
CKM
18327.6
30-Sep-10
POLICE
101920
CKM
8840.4
31-Mar-10
POLICE
101920
CKM
8986.25
24-May-10
POLICE
101920
CKM
34252.2
30-Jun-10
POLICE
101920
CKM
8913.35
30-Apr-10
 POLICE
Â
Ideally i would like a row to have one row of ACCT_NUM and TranAmount and
Date be transposed to columns.
Thanks
Kind regards,
Tabitha Mundia , Project Management Office, Equity Bank
Limited,P.O. Box 75104-00200
Head Office, Upper hill, NHIF BLDG, 14th Floor, Nairobi, Kenya
Direct Extension : +254732112721Â Â Mobile: +254722309538
Email: tabitha.mun...@equitybank.co.keskype ID: twamaeYahoo ID:
tabygath...@yahoo.com
An idea not coupled with action will never get any bigger than the brain cell
it occupied.
Arnold Glasgow
..
Attempt something large enough that failure is guaranteedâ¦unless God steps
in!
--- On Mon, 12/20/10, Kohleth Chia kohl...@gmail.com wrote:
From: Kohleth Chia kohl...@gmail.com
Subject: [R] package arules - 'transpose' of the transactions
To: r-help@r-project.org
Date: Monday, December 20, 2010, 8:41 AM
Suppose this is my list of
transactions:
set.seed(200)
tran=random.transactions(100,3)
inspect(tran)
 items  transactionID
1 {item80}Â Â Â Â trans1
2 {item8,
   item20}    trans2
3 {item28}Â Â Â Â trans3
I want to get the 'transpose' of the data, i.e.
 transactionID items
1 {trans2}Â Â Â Â item8
2 {trans2}Â Â Â Â item20
3 {trans3}Â Â Â Â item28
4 {trans1}Â Â Â Â item80
I tried converting tran into a matrix, then transpose it, then convert it
back to transactions. But my dataset is actually very very large, so I
wonder if there is any faster method?
Thanks
--
KC
   [[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.
[R] Modifying effect display (effects package)
Hi, I am using the effects package and plot() to produce an effect plot for a higher order term from a GLM. However, I need to modify the graph it produces by removing the tick marks and axes from the right and top of the graph, and moving the remaining tick marks inside the axes. Does anybody know how to do this? I can't find a way to modify the plot. Thanks Anna __ 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] Metafor package
I have some question about metafor package. I'm interest to perform a random effect meta-analysis of proportion (single group summary of prevalence of disease in a population as reported by different study) It ask: 1. PFT: The Freeman-Tukey double arcsine transformed proportion is reported to be equal to 1/2*(asin(sqrt(xi/(ni+1))) + asin(sqrt((xi+1)/(ni+1. Hovewer, i also found the same formula but without 1/2*. 2. how vi, the corresponding (estimated) sampling variance is calculated? (i.e. the formula used to estimate vi) 3. it is possible to return a forest plot with the backtransformed value of proportion, confidenc interval and weight and not with the transformed value after performing rma.uni? Many thanks Mario Petretta Dipartimento di Medicina Clinica Scienze Cardiovascolari e Immunologiche Facoltà di Medicina e Chirurgia Università di Napoli Federico II 081 - 7462233 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R is not running well
On 20.12.2010 10:29, Roslina Zakaria wrote: Hi, I'm working on windows 7. Recently I install the latest R and also use together with Tinn-R but it is not working well. I got the following message. I have tried saving the Rprofile as suggested by the R forum, but still it is not running. What should I do? source(.trPaths[5], echo=TRUE, max.deparse.length=150) Error in source(.trPaths[5], echo = TRUE, max.deparse.length = 150) : object '.trPaths' not found R is running well, Tinn-R is not. Please ask the Tinn-R maintainers or read the list archives where this has been discussed before. Uwe Ligges [[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] install.packages() - old version deleted, new version did not install
On 20.12.2010 09:41, Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Which suggests there may be another R session that has the package loaded (i.e. the dll locked). Just close all your R sessions and try again. Uwe Ligges Jon __ 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] install.packages() - old version deleted, new version did not install
Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Jon Did you have it open at the time? Windows won't let open files be removed, so that could have caused the problem. If it's not that, it could be a permissions problem. Have you tried running R as administrator for the install? 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] How to read japanese characters from a database/file
Hi, I am using rodbc package to connect with Sql Server 2005. I have a table which contains some japanese characters. Now I want to use these japanese characters as labels in simple x y plot? Here is the data in the table サービスデスク 35355 データネットワーク 8417 I want to read this data from database table and plot it using R and give xlabels as japanese words . How is it possible in R ?? -- View this message in context: http://r.789695.n4.nabble.com/How-to-read-japanese-characters-from-a-database-file-tp3095466p3095466.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] Alternative to extended recode sintax? Bug?
On Dec 20, 2010, at 12:54 AM, Luca Meyer wrote: All right, I get it now: lubridate's week() define weeks from Thursday till the following Wednesday. You'd probably agree with me that it's a bit strange what it is going to do over the turn of the year: y - as.POSIXct(c(2010-12-27,2010-12-28,2010-12-29,2010-12-30,2010-12-31,2011-01-01,2011-01-02,2011-01-03,2011-01-04,2011-01-05,2011-01-06,2011-01-07,2011-01-08,2011-01-09,2011-01-10,2011-01-11,2010-01-12,2010-01-13,2010-01-14)) week(y) [1] 52 52 52 53 53 1 1 1 1 1 1 2 2 2 2 2 2 2 3 Why would the first week of the year be made of 6 days and the turn from week 1 to week 2 on the night between Thursday and Friday and not Wednesday and Friday like every other week? weeks in lubridate start on whatever day of the week is the first of that year. If you want a Monday starting day (or the option to change to another starting day), then package chron has such facilities. Cheers, Luca Il giorno 19/dic/2010, alle ore 18.14, Uwe Ligges ha scritto: On 19.12.2010 13:20, David Winsemius wrote: On Dec 19, 2010, at 5:11 AM, Luca Meyer wrote: Something goes wrong with the week function of the lubridate package: x= as.POSIXct(factor(c(2010-12-15 17:28:27, + 2010-12-15 17:32:34, + 2010-12-15 18:48:39, + 2010-12-15 19:25:00, + 2010-12-16 08:00:00, + 2010-12-16 08:25:49, + 2010-12-16 09:00:00))) require(lubridate) weekdays(x) [1] Mercoledì Mercoledì Mercoledì Mercoledì Giovedì Giovedì Giovedì week(x) [1] 50 50 50 50 51 51 51 But 2010-12-15 is a Wednesday and 2010-12-16 is a Thursday. Together with the description of ?week this shows that lubridate's week() function works as documented rather than as expected by Luca Meyer. Uwe Ligges 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] Odp: For-loop
Hi
r-help-boun...@r-project.org napsal dne 20.12.2010 11:48:51:
Hi,
I have the following problem:
I have a data.frame with 36 sample sites (colums) for which I have
covariates
in 3 categories: Area, Month and River. Each Area consists of 3 rivers,
which
were sampled over 3 month. Now I want to fuse River 1-3 for one area in
one
month. To get a data.frame with 12 colums.
I am trying to do a for loop (which may be a complicated solution, but
I
don't see an easier way), which is not working, apparently because
a[,ij] or a
[,c(i,j)] is not working as a definition of the matrix with a double
condition
in the colums.
How can I make it work or what would be an easier solution?
Thank you for your help,
Anne
data=data.frame(matrix(1:99,nrow=5,ncol=36))
colnames(data)=c(paste(plot,1:36))
cov=data.frame(rep(1:3,12),c(rep(Jan,12),rep(Feb,12),rep(Mar,12)),rep(c
(1,1,1,2,2,2,3,3,3,4,4,4),3))
dimnames(cov)=list(colnames(data),c(River,Month,Area))
###loop###
a=matrix(nrow=dim(data)[1],ncol=length(levels(factor(cov$Month)))*length
(levels(factor(cov$Area
for(i in 1:length(levels(factor(cov$Month
{
for(j in 1:length(levels(factor(cov$Area
{
a[,ij]=as.numeric(rowSums(data[,factor(cov$Month)==levels(factor(cov$Month))
[i]factor(cov$Area)==levels(factor(cov$Area))[j]]))
}
}
I am not exactly sure what you want to do. What operation is fuse? If it
is sum so having you data you can do
area-rep(1:12, each=3)
data.t-t(data)
aggregate(data.t, list(area), sum)
Group.1 V1 V2 V3 V4 V5
11 18 21 24 27 30
22 63 66 69 72 75
33 108 111 114 117 120
44 153 156 159 162 165
55 198 201 204 207 210
66 243 246 249 252 255
77 189 192 195 198 102
88 36 39 42 45 48
99 81 84 87 90 93
10 10 126 129 132 135 138
11 11 171 174 177 180 183
12 12 216 219 222 225 228
t(aggregate(data.t, list(area), sum))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
Group.1123456789101112
V118 63 108 153 198 243 189 36 81 126 171 216
V221 66 111 156 201 246 192 39 84 129 174 219
V324 69 114 159 204 249 195 42 87 132 177 222
V427 72 117 162 207 252 198 45 90 135 180 225
V530 75 120 165 210 255 102 48 93 138 183 228
but then there is Month value, which is not apparent from your example.
Maybe
t(aggregate(data.t, list(area, data.t$Month), sum))
Could do the trick but you probably need to show us maybe str and/or head
of your real data.
Regards
Petr
__
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] For-loop
try this:
# create indexing vector to select 3 adjacent columns
indx - sapply(seq(1, 36, 3), seq, length = 3)
# process each row of the table
ans - t(apply(data, 1, function(.row){
+ # use indx to sum up the columns
+ apply(indx, 2, function(.indx){
+ sum(.row[.indx])
+ })
+ }))
ans
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 18 63 108 153 198 243 189 36 81 126 171 216
[2,] 21 66 111 156 201 246 192 39 84 129 174 219
[3,] 24 69 114 159 204 249 195 42 87 132 177 222
[4,] 27 72 117 162 207 252 198 45 90 135 180 225
[5,] 30 75 120 165 210 255 102 48 93 138 183 228
On Mon, Dec 20, 2010 at 5:48 AM, Anne-Christine Mupepele
anne-chr@web.de wrote:
Hi,
I have the following problem:
I have a data.frame with 36 sample sites (colums) for which I have covariates
in 3 categories: Area, Month and River. Each Area consists of 3 rivers, which
were sampled over 3 month. Now I want to fuse River 1-3 for one area in one
month. To get a data.frame with 12 colums.
I am trying to do a for loop (which may be a complicated solution, but I
don't see an easier way), which is not working, apparently because a[,ij] or
a[,c(i,j)] is not working as a definition of the matrix with a double
condition in the colums.
How can I make it work or what would be an easier solution?
Thank you for your help,
Anne
data=data.frame(matrix(1:99,nrow=5,ncol=36))
colnames(data)=c(paste(plot,1:36))
cov=data.frame(rep(1:3,12),c(rep(Jan,12),rep(Feb,12),rep(Mar,12)),rep(c(1,1,1,2,2,2,3,3,3,4,4,4),3))
dimnames(cov)=list(colnames(data),c(River,Month,Area))
###loop###
a=matrix(nrow=dim(data)[1],ncol=length(levels(factor(cov$Month)))*length(levels(factor(cov$Area
for(i in 1:length(levels(factor(cov$Month
{
for(j in 1:length(levels(factor(cov$Area
{
a[,ij]=as.numeric(rowSums(data[,factor(cov$Month)==levels(factor(cov$Month))[i]factor(cov$Area)==levels(factor(cov$Area))[j]]))
}
}
__
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
Data Munger Guru
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] transposing panel data
taby gathoni tabieg at yahoo.com writes: I am currently trying to transpose some large panel data set ie transposing multiple rows in into a single column. instead the transpose functionality transposes all rows into columns. my sample data set looks like below: Try the melt and cast functions in the reshape (or reshape2) packages (reshape2 may be faster). It may take a bit of work to get the syntax right (try it out on a small subset of your data). Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turning a Variable into String
Thank you very much to the both of you
Paolo
On 20 December 2010 00:35, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 19/12/2010 7:21 PM, Paolo Rossi wrote:
I would like to know how to turn a variable into a string. I have tried
as.symbol and as.name but it doesnt work for what I'd like to do
Essentially, I'd like to feed the function below with two variables. This
works fine in the bit working out number of elements in each variable.
In the print(sprintf(OK with %s and %s\n, var1, var2)) line I would
like
var1 and var2 to be magically substituted with a string containing the
name
of var1 and name of var2.
The name of var1 is var1, so I assume you mean the expression passed to
your function and bound to var1. In that case, what you want is
deparse(substitute(var1))
Watch out: if the expression is really long, that can be a vector with
more than one element. See ?deparse for ways to deal with that.
Duncan Murdoch
Thanks in advance
Paolo
haveSameLength- function(var1, var2) {
if (length(var1)==length(var2))
{
print(sprintf(OK with %s and %s\n, var1, var2))
} else {
print(Problems!!)
}
}
[[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.
[[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] install.packages() - old version deleted, new version did not install
On 12/20/2010 1:43 PM, Duncan Murdoch wrote: Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Jon Did you have it open at the time? Windows won't let open files be removed, so that could have caused the problem. If it's not that, it could be a permissions problem. Have you tried running R as administrator for the install? Yes, I had it open. In this case it was intentional to give a reproducible example in case something had changed in the new version, in other cases I have had to wait for 2 days before I could reinstall a package. It seems the .dll is the one causing the problem, so wouldnt it be possible to test if this file can be unlinked before trying to unlink the complete directory in utils:::unpackPkgZip? Then the package should be left untouched if it is in use, and not partly deleted as today. I know that it is possible to avoid this problem by not installing a package in use, but 1) it seems only to affect packages with .dll's, so some packages can be reinstalled while in use 2) you dont always know if a dependent package will download a new version of an installed package Best wishes, Jon __ 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] install.packages() - old version deleted, new version did not install
On 12/20/2010 1:30 PM, Uwe Ligges wrote: On 20.12.2010 09:41, Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Which suggests there may be another R session that has the package loaded (i.e. the dll locked). Just close all your R sessions and try again. Uwe Ligges Yes, I know. The question was whether there is another way of reinstalling the missing parts, or if there is a way of avoiding that the package gets partly deleted when install.packages is called with the package itself or a dependent package. It seems like the answer to the first is no, but a fix for the second one could save some trouble. Cheers, Jon __ 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] Defining Two-dimensional Separable Variance-Covariance Structure in nlme Package
*Hello All,* * * *I'd highly appreciate if someone can explain the way of defining Two-dimensional Separable Variance-Covariance Structure in nlme Package. Thanks* * * * * -- * Muhammad Yaseen * [[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] box-and-whisker plots based on summary not data
Matthew Vernon m.c.ver...@warwick.ac.uk writes: Is it possible to produce box-and-whisker plots given that I have the median, interquartile and 5/95th centile values, but not the data from which they come? The answer, which came from Steve Ellison (thanks!) is to note that stats is the only bit of the z argument to bxp that has to be present, so given a data file with columns year,mean,median,centiles, something like this works: # transform data so the rows are the various centiles in order # there's probably a more elegant approach! bsr - matrix(allobs[,c(4,5,3,6,7)],nrow=5,byrow=TRUE) # column 1 is the year, which is our factor bsbox - list( stats=bsr,names=allobs$V1) bxp(bsbox,xlab=Year,ylab=Number of cattle per batch) I thought I'd follow-up so the answer to my question will be in the archives. Thanks, Matthew -- Matthew Vernon, Research Fellow Ecology and Epidemiology Group, University of Warwick http://blogs.warwick.ac.uk/mcvernon __ 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] install.packages() - old version deleted, new version did not install
On 20/12/2010 9:03 AM, Jon Olav Skoien wrote: On 12/20/2010 1:43 PM, Duncan Murdoch wrote: Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Jon Did you have it open at the time? Windows won't let open files be removed, so that could have caused the problem. If it's not that, it could be a permissions problem. Have you tried running R as administrator for the install? Yes, I had it open. In this case it was intentional to give a reproducible example in case something had changed in the new version, in other cases I have had to wait for 2 days before I could reinstall a package. It seems the .dll is the one causing the problem, so wouldnt it be possible to test if this file can be unlinked before trying to unlink the complete directory in utils:::unpackPkgZip? Then the package should be left untouched if it is in use, and not partly deleted as today. I don't know. Perhaps we could try to rename the folder; if that fails, abort the whole thing. If that succeeds but something later fails, then remove all the new stuff and restore the old folder. Do you know of a better test? Duncan Murdoch I know that it is possible to avoid this problem by not installing a package in use, but 1) it seems only to affect packages with .dll's, so some packages can be reinstalled while in use 2) you dont always know if a dependent package will download a new version of an installed package Best wishes, Jon __ 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] box-and-whisker plots based on summary not data
On Dec 20, 2010, at 8:33 AM, Matthew Vernon wrote: Matthew Vernon m.c.ver...@warwick.ac.uk writes: Is it possible to produce box-and-whisker plots given that I have the median, interquartile and 5/95th centile values, but not the data from which they come? Please note that the whiskers of the default BWP are NOT at the 5th and 95th percentiles although I would have guessed that they were too before I looked a closely at the code and the help pages. They are at the most extreme data points which do not exceed 1.5 times the box values which in turn are at versions of the first and third quartiles. (So , no, you cannot get what boxplot would have given unless you have the IQR values and all of the data outside those values. You can, of course, make up your own approach and use the boxplot machinery for plotting as long as you inform your audience what you are plotting.) The answer, which came from Steve Ellison (thanks!) is to note that stats is the only bit of the z argument to bxp that has to be present, so given a data file with columns year,mean,median,centiles, something like this works: # transform data so the rows are the various centiles in order # there's probably a more elegant approach! bsr - matrix(allobs[,c(4,5,3,6,7)],nrow=5,byrow=TRUE) # column 1 is the year, which is our factor bsbox - list( stats=bsr,names=allobs$V1) bxp(bsbox,xlab=Year,ylab=Number of cattle per batch) I thought I'd follow-up so the answer to my question will be in the archives. Thanks, Matthew 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] install.packages() - old version deleted, new version did not install
On 20.12.2010 15:19, Duncan Murdoch wrote: On 20/12/2010 9:03 AM, Jon Olav Skoien wrote: On 12/20/2010 1:43 PM, Duncan Murdoch wrote: Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Jon Did you have it open at the time? Windows won't let open files be removed, so that could have caused the problem. If it's not that, it could be a permissions problem. Have you tried running R as administrator for the install? Yes, I had it open. In this case it was intentional to give a reproducible example in case something had changed in the new version, in other cases I have had to wait for 2 days before I could reinstall a package. It seems the .dll is the one causing the problem, so wouldnt it be possible to test if this file can be unlinked before trying to unlink the complete directory in utils:::unpackPkgZip? Then the package should be left untouched if it is in use, and not partly deleted as today. I don't know. Perhaps we could try to rename the folder; if that fails, abort the whole thing. If that succeeds but something later fails, then remove all the new stuff and restore the old folder. Do you know of a better test? I think we had something like that in the past which did not work properly on network shares and we had to change the way it works for that reason. Uwe Duncan Murdoch I know that it is possible to avoid this problem by not installing a package in use, but 1) it seems only to affect packages with .dll's, so some packages can be reinstalled while in use 2) you dont always know if a dependent package will download a new version of an installed package Best wishes, Jon __ 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] install.packages() - old version deleted, new version did not install
On 20/12/2010 9:26 AM, Uwe Ligges wrote: On 20.12.2010 15:19, Duncan Murdoch wrote: On 20/12/2010 9:03 AM, Jon Olav Skoien wrote: On 12/20/2010 1:43 PM, Duncan Murdoch wrote: Jon Olav Skoien wrote: On 12/17/2010 6:22 PM, Duncan Murdoch wrote: On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Could you try out the new 2.12.1 release? I recall hearing that something like this had changed, but I can't spot the NEWS item right now. Duncan Murdoch It seems it didnt change yet... I installed 2.12.1 (on a different computer, still Windows, but Vista and 32 bit), and after installing and loading sp in one session, I opened a new session and got: R version 2.12.1 (2010-12-16) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i386-pc-mingw32/i386 (32-bit) install.packages(sp) Installing package(s) into ‘C:\Users\Jon\Documents/R/win-library/2.12’ (as ‘lib’ is unspecified) --- Please select a CRAN mirror for use in this session --- provo con l'URL 'http://cran.at.r-project.org/bin/windows/contrib/2.12/sp_0.9-76.zip' Content type 'application/zip' length 997444 bytes (974 Kb) URL aperto downloaded 974 Kb package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' The downloaded packages are in C:\Users\Jon\AppData\Local\Temp\RtmpCTJeBk\downloaded_packages library(sp) Errore in library(sp) : non c'è alcun pacchetto chiamato 'sp' The error message is the same as earlier, there is no package called sp, the attempt to install it again removed the old version except for the .dll. Jon Did you have it open at the time? Windows won't let open files be removed, so that could have caused the problem. If it's not that, it could be a permissions problem. Have you tried running R as administrator for the install? Yes, I had it open. In this case it was intentional to give a reproducible example in case something had changed in the new version, in other cases I have had to wait for 2 days before I could reinstall a package. It seems the .dll is the one causing the problem, so wouldnt it be possible to test if this file can be unlinked before trying to unlink the complete directory in utils:::unpackPkgZip? Then the package should be left untouched if it is in use, and not partly deleted as today. I don't know. Perhaps we could try to rename the folder; if that fails, abort the whole thing. If that succeeds but something later fails, then remove all the new stuff and restore the old folder. Do you know of a better test? I think we had something like that in the past which did not work properly on network shares and we had to change the way it works for that reason. I just took a look at the code, and I see that something like that is still there. I haven't tried it yet, so I don't know if it is in use. Duncan Murdoch Uwe Duncan Murdoch I know that it is possible to avoid this problem by not installing a package in use, but 1) it
Re: [R] moving average with gaps in time series
Thank you for the suggestion Jannis. I ended up using the function
'running' in package gregmisc. 'rollyapply' was not able to process all
the data on my computer. Happy holidays to you and the R community
Jannis bt_jan...@yahoo.de
12/16/2010 06:56 PM
To
josef.kar...@phila.gov
cc
r-help@r-project.org
Subject
Re: [R] moving average with gaps in time series
Hi Josef,
is there any particular reason why you want to use your own function?
Have a look at the stats functions or special timeseries functions for
R. I am sure you will find something that calculates an ordinary moving
average (and a bunch of fancier stuff).
'filter' (stats) for example can be used for moving averages. You could
also use 'rollmean' from package zoo.
Not sure how to handle the missing values here, though (both ways most
probably work with NAs). You could try 'rollaply 'from package zoo with
a function that sums the valid values in the window and returns TRUE in
case all are valid and FALSE if they are not. Not sure whether this is
faster than your solution though.
Are you sure you want to compute a moving average with your missing
values specification? Or do you just want to aggregate your values to
hourly means, with those values removed where not all values are valid?
(in that case 'aggregate' would be your function)
HTH
Jannis
josef.kar...@phila.gov schrieb:
I have a time series with interval of 2.5 minutes, or 24 observations
per
hour. I am trying to find a 1 hr moving average, looking backward, so
that moving average at n = mean(n-23 : n)
The time series has about 1.5 million rows, with occasional gaps due to
poor data quality. I only want to take a 1 hour moving average for
those
periods that are complete, i.e. have 24 observations in the previous
hour.
The data is in 3 columns
Value DateTimeinterval
For example:
Value - rnorm (100, 50, 3) #my data has 1.5 million rows; using 100
here
for simple example
DateTime - seq(from = 915148800, to=915156150, by =150) #time steps
1:50 at 150 second intervals
DateTime [51] - 915156450 #skip one time step;
DateTime[52:100] - seq(from = 915156600, to =915163800, by = 150)
#resume
time steps of 150 seconds
x - cbind (Value, DateTime)
x - as.data.frame(x)
x$DateTime -as.POSIXct(x$DateTime, origin=1970-01-01, tz=GMT)
x1 - x[-c(1:23), ] #trimming x to create direct comparison of
DateTimes
in x and x1
x[,3] -difftime(x1[,2], x[,2], units=mins) #ignore warning
message
colnames(x) [3] - interval
x[24:nrow(x),3] - x[1:(nrow(x)-23),3] #set interval to be the
number
of minutes between n-23 and n.
#57.5 indicates no gaps in the previous hour up to and including
n.
#57.5 indicates a gap in the previous n-23 rows.
x[1:23,3] - 0/0 #NaN assigned to first 23 rows so as not to take
average
of first hour.
#as expected, row 51: 73 indicates a gap, i.e. interval 57.5
index - which (x [,3] == 57.5) #which rows have no gaps in previous
hour
#loop to calculate 1 hour moving average, only for periods which are
complete
for (i in 1:length(index)) {
x [index[i],4] - mean(x[index[i]-23,1] : x[index[i],1])
}
#This loop works on this simple example; but this takes VERY long time
to
run on x with 1.5 million rows. Over 1 hour running and still not
complete. I also tried increasing memory.limit to 4095, but still very
slow.
Any suggestions to make this run faster? I thought about using the lag
function but could not get it to work
[[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] lower/upper case question
toupper() and tolower()
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Erin Hodgess
Sent: Friday, December 17, 2010 4:10 PM
To: R help
Subject: [R] lower/upper case question
Dear R People:
Is there a function to convert a character string to all uppercase or
all lowercase please?
I'm sure that I've used one before but I'm drawing a blank.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences University of Houston -
Downtown
mailto: erinm.hodg...@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.
***
This message is for the named person's use only. It may\...{{dropped:20}}
__
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] dotchart for matrix data
On 18/12/2010, e-letter inp...@gmail.com wrote: On 18/12/2010, Peter Ehlers ehl...@ucalgary.ca wrote: On 2010-12-18 07:50, e-letter wrote: Ben Bolker Sat, 18 Dec 2010 07:07:24 -0800 [... snip ...] I am trying to create a chart like this (http://www.b-eye-network.com/images/content/Fig4_3.jpg); so this is not possible using R? That looks an awful lot like what lattice's dotplot would produce. So: have you tried dotplot() as Ben has suggested? If one set of value ranges for 10-20 and another set ranges form 1000-1500, how to adjust the graph such that: with categories on the ordinate (y-axis), can the bottom abscissa (x-axis bottom) be set with a scale suitable for the data set of range 1000-1500? can the top abscissa (x-top) be set with a scale 10-20? or is it better practice to change the scale of the data set 1000-1500 by two orders of magnitude? how is it possible to control the order of the category variables? For example, if the graph shows: a b c d Is it possible to change to a d b c Thanks __ 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] install.packages() - old version deleted, new version did not install
On 20/12/2010 9:29 AM, Duncan Murdoch wrote: On 20/12/2010 9:26 AM, Uwe Ligges wrote: On 20.12.2010 15:19, Duncan Murdoch wrote: On 20/12/2010 9:03 AM, Jon Olav Skoien wrote: [ lots deleted ] Yes, I had it open. In this case it was intentional to give a reproducible example in case something had changed in the new version, in other cases I have had to wait for 2 days before I could reinstall a package. It seems the .dll is the one causing the problem, so wouldnt it be possible to test if this file can be unlinked before trying to unlink the complete directory in utils:::unpackPkgZip? Then the package should be left untouched if it is in use, and not partly deleted as today. I don't know. Perhaps we could try to rename the folder; if that fails, abort the whole thing. If that succeeds but something later fails, then remove all the new stuff and restore the old folder. Do you know of a better test? I think we had something like that in the past which did not work properly on network shares and we had to change the way it works for that reason. I just took a look at the code, and I see that something like that is still there. I haven't tried it yet, so I don't know if it is in use. After a closer look I see that the code is there for source installs, but not binary installs. For binary installs the procedure is: unzip the package to a temporary directory look at the DESCRIPTION to find the package name (since you can't trust the .zip name) abort if the package is in the search list. (This is the test that could be stronger, because it doesn't detect a package in use in another instance of R.) remove the old package if it is there. (This is the step that removes everything but the old DLL when the DLL is in use.) copy the new package into place. (This fails because of the collision with the old DLL.) I'll look into strengthening the test at the 3rd step with something along the lines described above. From Uwe's description, this might still not be perfect, but it should help with local installs. 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] contourplot help
On Dec 20, 2010, at 5:36 AM, Andre Nathan wrote: Hello I'm using the following call to create a contourplot: library(lattice) m - as.matrix(read.table(data.txt)) contourplot(m[,3] ~ m[,2] * -m[,1], at = c(1e-6, 1e-5, 1e-4, 1e-3, 1e-2, 1e-1), scales = list(x = list(log = 10, labels = c(1, 10, 100), at = c(1, 10, 100)), y = list(labels = c(14, 12, 10, 8, 6, 4, 2)), at = c(-14, -12, -10, -8, -6, -4, -2)), labels = c(expression(10^-6), expression(10^-5), expression(10^-4), expression(10^-3), expression(10^-2), expression(10^-1), expression(10^0)), xlim = c(0.75, 10^2), xlab = Out-degree, ylab = In-degree) Which gives the the output in the file below http://ompldr.org/vNm4xag/contour.eps As it can be seen, the level labels are not displayed nicely because there's not enough room for them. Also, the 10^-1 label is not displayed. Is there a way for me to hardcode the position of each label? I don't see it, but that doesn't mean much unless it is echoed by Sarkar or Andrews. I tried setting labels = F and then calling text() for each one, but that doesn't work. Right, text() is a base graphics function. There are, however, lattice functions that supply the same functionality: ?ltext # = l(attice)text ltext(x, y = NULL, labels = seq_along(x), col, alpha, cex, srt = 0, lineheight, font, fontfamily, fontface, adj = c(0.5, 0.5), pos = NULL, offset = 0.5, ...) You could first see if you can better results by not automatically labeling the 10^-6 level, because the -2, -3, -5 levels look ok. That might let the 10^-4 label move to a bit less hashed up position and then you could put 10^-1 and 10^-6 where you want them If that's not possible, one option would be to color each level line differently and then add a legend. Is it possible to do that? Finally, how can I remove the tick marks from the top and right axes? ..., scales=list(tck=c(1,0)), ... actually you would add the tck list to your existing scales. Found on the ?xyplot page, but tested with a modified volcano example from : ?contourplot Thanks, Andre 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] R.matlab memory use
Hi Ben, Thanks for your reply. My data structure is about 2 x 2000 so one order of magnitude the one you tried. I have no problem saving and reading smaller data structures (even large ones, just not his large) between octave and R using octave's save -7 (which saves MATLAB v5 files) and R.matlab's readMat. And I can save in text format in octave and read in R using read.octave (from package foreign) so it's not a big deal. I was just surprised that R.matlab needed more memory than I have (I have 3GB on this machine). Thanks, Stefano On Sun, Dec 19, 2010 at 10:54 PM, Ben Bolker bbol...@gmail.com wrote: Stefano Ghirlanda dr.ghirlanda at gmail.com writes: I am trying to load into R a MATLAB format file (actually, as saved by octave). The file is about 300kB but R complains with a memory allocation error: library(Rcompression) library(R.matlab) Loading required package: R.oo Loading required package: R.methodsS3 R.methodsS3 v1.2.0 (2010-03-13) successfully loaded. See ?R.methodsS3 for help. R.oo v1.7.2 (2010-04-13) successfully loaded. See ?R.oo for help. R.matlab v1.3.1 (2010-04-20) successfully loaded. See ?R.matlab for help. f - readMat(freq.mat) Error: cannot allocate vector of size 296.5 Mb On the other hand, if I save the same data in ascii format (from octave: save -text), resulting in a 75MB file, then I can load it without problems with the read.octave() function from package foreign. Is this a known issue or am I doing something wrong? My R version is: This is not a package I'm particularly familiar with, but: what commands did you use to save the file in octave? Based on 'help save' I think that 'save' by default would get you an octave format file ... you might have to do some careful reading in ?readMat (in R) and 'help save' (in octave) to figure out the correspondence between octave/MATLAB and R/MATLAB. If possible, try saving a small file and see if it works; if you still don't know what's going on, post that file somewhere for people to try. I was able to save -6 save.mat in octave and readMat(save.mat) in R successfully, saving a vector of integers from 1 to 1 million (which took about 7.7 Mb) __ 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. -- Stefano Ghirlanda www.intercult.su.se/~stefano - drghirlanda.wordpress.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 on timeseries
Thanks Frederic. that solved the problem. Thanks again. Cameron -- View this message in context: http://r.789695.n4.nabble.com/help-on-timeseries-tp3076866p3095633.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] After heteroskedasticity correction, how can I get new confidential interval?
I just corrected std.error of my 'model'(Multi Regression). Then how can I get new t and p-values? Isn't there any R command which shows new t and p values? -- View this message in context: http://r.789695.n4.nabble.com/After-heteroskedasticity-correction-how-can-I-get-new-confidential-interval-tp3095643p3095643.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] How to optimize function parameters?
Hi,
I have a dataset and I want to fit a function to it.
The function is variogram model (http://en.wikipedia.org/wiki/Variogram)
The variogram model is defined by three parameters and I want them to be
automatically optimized for real time data.
I tried to use gafit {gafit} for this, but there are some data
configuration, where optimal results given by gafit() are negative, which is
not correct and cannot be used for further calculating. gafit() does not
enable me to set a range of possible results (at least I did not succeded in
doing so), therefore I am looking for another solution (Powell´s gradient
descent possibly?).
Can anyone give me a hint where to look?
Thanks
Zbynek
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-optimize-function-parameters-tp3095603p3095603.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] How to optimize function parameters?
Without an example, I'm not sure how much help I can be. Does your
Variogram calculation return a single value that could be minimized? Or
could it be coerced to be so?
I would suggest, if you haven't done so, a look at ?nlm and ?nlminb.
Alternatively, the CRAN package 'optimx' serves as a wrapper to many
optimization algorithms and offers methods to compare them side-by-side.
Good luck,
Jon
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
Is the room still a room when its empty? Does the room,
the thing itself have purpose? Or do we, what's the word... imbue it.
- Jubal Early, Firefly
r-help-boun...@r-project.org wrote on 12/20/2010 09:39:49 AM:
[image removed]
[R] How to optimize function parameters?
zbynek.jano...@gmail.com
to:
r-help
12/20/2010 10:43 AM
Sent by:
r-help-boun...@r-project.org
Hi,
I have a dataset and I want to fit a function to it.
The function is variogram model (http://en.wikipedia.org/wiki/Variogram)
The variogram model is defined by three parameters and I want them to be
automatically optimized for real time data.
I tried to use gafit {gafit} for this, but there are some data
configuration, where optimal results given by gafit() are negative,
which is
not correct and cannot be used for further calculating. gafit() does not
enable me to set a range of possible results (at least I did not
succeded in
doing so), therefore I am looking for another solution (Powell´s
gradient
descent possibly?).
Can anyone give me a hint where to look?
Thanks
Zbynek
--
View this message in context: http://r.789695.n4.nabble.com/How-to-
optimize-function-parameters-tp3095603p3095603.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] Alternative to extended recode sintax? Bug?
Right, I appreciate the first day of the year start date. I am just wondering why then the cut off day is not the same for the rest of the year...but it's all right to use other packages. Thanks, Luca Il giorno 20/dic/2010, alle ore 14.16, David Winsemius ha scritto: On Dec 20, 2010, at 12:54 AM, Luca Meyer wrote: All right, I get it now: lubridate's week() define weeks from Thursday till the following Wednesday. You'd probably agree with me that it's a bit strange what it is going to do over the turn of the year: y - as.POSIXct(c(2010-12-27,2010-12-28,2010-12-29,2010-12-30,2010-12-31,2011-01-01,2011-01-02,2011-01-03,2011-01-04,2011-01-05,2011-01-06,2011-01-07,2011-01-08,2011-01-09,2011-01-10,2011-01-11,2010-01-12,2010-01-13,2010-01-14)) week(y) [1] 52 52 52 53 53 1 1 1 1 1 1 2 2 2 2 2 2 2 3 Why would the first week of the year be made of 6 days and the turn from week 1 to week 2 on the night between Thursday and Friday and not Wednesday and Friday like every other week? weeks in lubridate start on whatever day of the week is the first of that year. If you want a Monday starting day (or the option to change to another starting day), then package chron has such facilities. Cheers, Luca Il giorno 19/dic/2010, alle ore 18.14, Uwe Ligges ha scritto: On 19.12.2010 13:20, David Winsemius wrote: On Dec 19, 2010, at 5:11 AM, Luca Meyer wrote: Something goes wrong with the week function of the lubridate package: x= as.POSIXct(factor(c(2010-12-15 17:28:27, + 2010-12-15 17:32:34, + 2010-12-15 18:48:39, + 2010-12-15 19:25:00, + 2010-12-16 08:00:00, + 2010-12-16 08:25:49, + 2010-12-16 09:00:00))) require(lubridate) weekdays(x) [1] Mercoledì Mercoledì Mercoledì Mercoledì Giovedì Giovedì Giovedì week(x) [1] 50 50 50 50 51 51 51 But 2010-12-15 is a Wednesday and 2010-12-16 is a Thursday. Together with the description of ?week this shows that lubridate's week() function works as documented rather than as expected by Luca Meyer. Uwe Ligges 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] install.packages() - old version deleted, new version did not install
This sounds like a good solution for the case I described in my first email. Thanks a lot! Jon On 12/20/2010 4:05 PM, Duncan Murdoch wrote: On 20/12/2010 9:29 AM, Duncan Murdoch wrote: On 20/12/2010 9:26 AM, Uwe Ligges wrote: On 20.12.2010 15:19, Duncan Murdoch wrote: On 20/12/2010 9:03 AM, Jon Olav Skoien wrote: [ lots deleted ] Yes, I had it open. In this case it was intentional to give a reproducible example in case something had changed in the new version, in other cases I have had to wait for 2 days before I could reinstall a package. It seems the .dll is the one causing the problem, so wouldnt it be possible to test if this file can be unlinked before trying to unlink the complete directory in utils:::unpackPkgZip? Then the package should be left untouched if it is in use, and not partly deleted as today. I don't know. Perhaps we could try to rename the folder; if that fails, abort the whole thing. If that succeeds but something later fails, then remove all the new stuff and restore the old folder. Do you know of a better test? I think we had something like that in the past which did not work properly on network shares and we had to change the way it works for that reason. I just took a look at the code, and I see that something like that is still there. I haven't tried it yet, so I don't know if it is in use. After a closer look I see that the code is there for source installs, but not binary installs. For binary installs the procedure is: unzip the package to a temporary directory look at the DESCRIPTION to find the package name (since you can't trust the .zip name) abort if the package is in the search list. (This is the test that could be stronger, because it doesn't detect a package in use in another instance of R.) remove the old package if it is there. (This is the step that removes everything but the old DLL when the DLL is in use.) copy the new package into place. (This fails because of the collision with the old DLL.) I'll look into strengthening the test at the 3rd step with something along the lines described above. From Uwe's description, this might still not be perfect, but it should help with local installs. 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] Alternative to extended recode sintax? Bug?
On Dec 20, 2010, at 10:58 AM, Luca Meyer wrote: Right, I appreciate the first day of the year start date. I am just wondering why then the cut off day is not the same for the rest of the year...but it's all right to use other packages. Are you saying it shifts within the year? I am not seeing that: require(lubridate) weekdays(as.POSIXct(2010-01-01)+(0:8)*24*60*60) [1] FridaySaturday SundayMondayTuesday Wednesday [7] Thursday FridaySaturday week(as.POSIXct(2010-01-01)+(0:8)*24*60*60) [1] 1 1 1 1 1 1 2 2 2 Looks to be incrementing weeks between Wed and Thurs at the beginning of the year just as it did in your example. I admit that I thought that it should be shifting at the Thursday - Friday divide, but setting a zero point can be ambiguous. I thought if it were Midnight Thursday-Friday that all of Thurdays would be in week 1. But at least it appears consistent. Thanks, Luca Il giorno 20/dic/2010, alle ore 14.16, David Winsemius ha scritto: On Dec 20, 2010, at 12:54 AM, Luca Meyer wrote: All right, I get it now: lubridate's week() define weeks from Thursday till the following Wednesday. You'd probably agree with me that it's a bit strange what it is going to do over the turn of the year: y - as.POSIXct(c(2010-12-27,2010-12-28,2010-12-29,2010-12-30,2010-12-31,2011-01-01,2011-01-02,2011-01-03,2011-01-04,2011-01-05,2011-01-06,2011-01-07,2011-01-08,2011-01-09,2011-01-10,2011-01-11,2010-01-12,2010-01-13,2010-01-14)) week(y) [1] 52 52 52 53 53 1 1 1 1 1 1 2 2 2 2 2 2 2 3 Why would the first week of the year be made of 6 days and the turn from week 1 to week 2 on the night between Thursday and Friday and not Wednesday and Friday like every other week? weeks in lubridate start on whatever day of the week is the first of that year. If you want a Monday starting day (or the option to change to another starting day), then package chron has such facilities. Cheers, Luca Il giorno 19/dic/2010, alle ore 18.14, Uwe Ligges ha scritto: On 19.12.2010 13:20, David Winsemius wrote: On Dec 19, 2010, at 5:11 AM, Luca Meyer wrote: Something goes wrong with the week function of the lubridate package: x= as.POSIXct(factor(c(2010-12-15 17:28:27, + 2010-12-15 17:32:34, + 2010-12-15 18:48:39, + 2010-12-15 19:25:00, + 2010-12-16 08:00:00, + 2010-12-16 08:25:49, + 2010-12-16 09:00:00))) require(lubridate) weekdays(x) [1] Mercoledì Mercoledì Mercoledì Mercoledì Giovedì Giovedì Giovedì week(x) [1] 50 50 50 50 51 51 51 But 2010-12-15 is a Wednesday and 2010-12-16 is a Thursday. Together with the description of ?week this shows that lubridate's week() function works as documented rather than as expected by Luca Meyer. Uwe Ligges David Winsemius, MD West Hartford, CT 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] After heteroskedasticity correction, how can I get new confidential interval?
It is hard to say without knowing more about the type of model you are running. A good place to look would be at the coeftest() function in the package lmtest. Andrew Miles On Dec 20, 2010, at 10:04 AM, JoonGi wrote: I just corrected std.error of my 'model'(Multi Regression). Then how can I get new t and p-values? Isn't there any R command which shows new t and p values? -- View this message in context: http://r.789695.n4.nabble.com/After-heteroskedasticity-correction-how-can-I-get-new-confidential-interval-tp3095643p3095643.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.
[R] ideas, modeling highly discrete time-series data
Hello all, First of all, thanks so those of you who helped me a week or so ago managing a time series with varying gaps between the data series in 'R'. (My final preferred solution was to use its function then forecast(Arima( ) ). ) My next question is a general statistical question where I'd like some advice, for those willing / able to proffer any wisdom: - I need to predict using this same time series, where the *data* are highly discrete. E.g., I will have values like 1e5, 2.2e5, and 3.6e5, but I will never have 1.3e5 or 1.8e5, etc. - I could simply leave these values as discrete, similar to a binomial distribution, but then I am not sure how to use time series tricks like arima above. For time-series analyses that I know of, an assumption of an approximately normal distribution is expected. No simple normalization (e.g., log(values) ) works, since the non-normality arises from the highly discrete distribution more than any drastic asymmetry in the population spread. - I could leave the values as they are an work with a model where the assumption is violated... I am not sure how sensitive a model such as arima is on the population distribution - Or I could... (here's where I am hoping for some collective genius). Thanks in advance for any help! If this isn't the best forum, since I know this is not specifically an 'R' question, please let me know of a better forum to post such a question. Thanks! Mike Telescopes and bathyscaphes and sonar probes of Scottish lakes, Tacoma Narrows bridge collapse explained with abstract phase-space maps, Some x-ray slides, a music score, Minard's Napoleanic war: The most exciting frontier is charting what's already here. -- xkcd -- Help protect Wikipedia. Donate now: http://wikimediafoundation.org/wiki/Support_Wikipedia/en [[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] Passing parameter to a function
I am trying to pass a couple of variable names to a xtabs formula:
tab - function(x,y){
xtabs(time~x+y, data=D)
}
But when I run:
tab(A,B)
I get:
Error in eval(expr, envir, enclos) : object A not found
I am quite sure that there is some easy way out, but I have tried with
different combinations of deparse(), substitute(), eval(), etc without success,
can someone help?
Thanks,
Luca
Luca Meyer
www.lucameyer.com
IBM SPSS Statistics release 19.0.0
R version 2.12.1 (2010-12-16)
Mac OS X 10.6.5 (10H574) - kernel Darwin 10.5.0
__
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] Alternative to extended recode sintax? Bug?
Yes, I am seeing that at the end of 2010-beginning 2011. Try: weekdays(as.POSIXct(2010-12-25)+(0:20)*24*60*60) week(as.POSIXct(2010-12-25)+(0:20)*24*60*60) Week 1 (2011) is made up of 6 days Luca Il giorno 20/dic/2010, alle ore 17.54, David Winsemius ha scritto: On Dec 20, 2010, at 10:58 AM, Luca Meyer wrote: Right, I appreciate the first day of the year start date. I am just wondering why then the cut off day is not the same for the rest of the year...but it's all right to use other packages. Are you saying it shifts within the year? I am not seeing that: require(lubridate) weekdays(as.POSIXct(2010-01-01)+(0:8)*24*60*60) [1] FridaySaturday SundayMondayTuesday Wednesday [7] Thursday FridaySaturday week(as.POSIXct(2010-01-01)+(0:8)*24*60*60) [1] 1 1 1 1 1 1 2 2 2 Looks to be incrementing weeks between Wed and Thurs at the beginning of the year just as it did in your example. I admit that I thought that it should be shifting at the Thursday - Friday divide, but setting a zero point can be ambiguous. I thought if it were Midnight Thursday-Friday that all of Thurdays would be in week 1. But at least it appears consistent. Thanks, Luca Il giorno 20/dic/2010, alle ore 14.16, David Winsemius ha scritto: On Dec 20, 2010, at 12:54 AM, Luca Meyer wrote: All right, I get it now: lubridate's week() define weeks from Thursday till the following Wednesday. You'd probably agree with me that it's a bit strange what it is going to do over the turn of the year: y - as.POSIXct(c(2010-12-27,2010-12-28,2010-12-29,2010-12-30,2010-12-31,2011-01-01,2011-01-02,2011-01-03,2011-01-04,2011-01-05,2011-01-06,2011-01-07,2011-01-08,2011-01-09,2011-01-10,2011-01-11,2010-01-12,2010-01-13,2010-01-14)) week(y) [1] 52 52 52 53 53 1 1 1 1 1 1 2 2 2 2 2 2 2 3 Why would the first week of the year be made of 6 days and the turn from week 1 to week 2 on the night between Thursday and Friday and not Wednesday and Friday like every other week? weeks in lubridate start on whatever day of the week is the first of that year. If you want a Monday starting day (or the option to change to another starting day), then package chron has such facilities. Cheers, Luca Il giorno 19/dic/2010, alle ore 18.14, Uwe Ligges ha scritto: On 19.12.2010 13:20, David Winsemius wrote: On Dec 19, 2010, at 5:11 AM, Luca Meyer wrote: Something goes wrong with the week function of the lubridate package: x= as.POSIXct(factor(c(2010-12-15 17:28:27, + 2010-12-15 17:32:34, + 2010-12-15 18:48:39, + 2010-12-15 19:25:00, + 2010-12-16 08:00:00, + 2010-12-16 08:25:49, + 2010-12-16 09:00:00))) require(lubridate) weekdays(x) [1] Mercoledì Mercoledì Mercoledì Mercoledì Giovedì Giovedì Giovedì week(x) [1] 50 50 50 50 51 51 51 But 2010-12-15 is a Wednesday and 2010-12-16 is a Thursday. Together with the description of ?week this shows that lubridate's week() function works as documented rather than as expected by Luca Meyer. Uwe Ligges David Winsemius, MD West Hartford, CT 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] Complicated nls formula giving singular gradient message
Though my topic is slightly old already, I feel that it is necessary to post an update on my situation. I ended being able to estimate the parameters for this problem without having to worry as much about initial parameter estimates using AD Model Builder. It calculates the exact gradient using automatic differentiation so it's able to avoid the singular gradient problem nls can give. I also used the R package PBSadmb, which allowed me to run AD Model Builder and retrieve the results from within R. Then I could do what I liked with the results: generate graphs, more analysis, etc. Thanks to everyone who helped, Jared On Wed, Dec 15, 2010 at 8:47 AM, dave fournier da...@otter-rsch.com wrote: Jared Blashka wrote: Hi, Can you write a little note to the R list saying something like Re: SOLVED[R] Complicated nls formula giving singular gradient message I was able to estimate the parameters for this problem using AD Mode Builder which calculates the exact gradient for you using automatic differentiation and is thus able to avoid the singular gradient problem I encountered in nls. That way other R users who might be able to take advantage of the software will hear about it. Cheers, Dave Dave, That's exactly what I was looking for! Thanks for all your help! Jared On Tue, Dec 14, 2010 at 7:13 AM, dave fournier da...@otter-rsch.commailto: da...@otter-rsch.com wrote: Jared Blashka wrote: The source code for that is in jared.tpl I changed from least squares to a concentrated likelihood so that you could get estimated std devs via the delta method. they are in jared.std I rescaled the parameters so that the condition number of the Hessian is close to 1. You can see the eigenvalues of the Hessian in jared.eva. Your data are in jared.dat and the initial parameter values are in jared.pin. The parameter estimates with their estiamted std devs are: index name value std dev 1 NS 1.1254e-02 7.1128e-03 2 LogKi -8.8933e+00 8.2411e-02 3 LogKi -5.2005e+00 9.2179e-02 4 LogKi -7.2677e+00 7.7047e-02 5 BMax 2.1226e+05 5.1699e+03 ~How does it look? Cheers, Dave Dave - AD Model Builder looks like a great tool that I can use, but I'm curious if it can also perform global parameter estimations across multiple data sets. In regards to the example I have provided, I have two similar data sets that also need to be analyzed, but the values for NS and BMax between the three data sets should be the same. Each data set has a unique LogKi value however. In R, I accomplished this by merging the three data sets and adding an additional field for each data point that identified which set it was originally from. Then in the regression formula I specified the LogKi term as a vector: LogKi[dset]. The results of the regression gave me one value each for NS and BMax, but three LogKi values. I haven't had much time to look through the AD Model Builder documentation yet, but are you aware if such an analysis method is possible? Here's one such example of a data set all -structure(list(X = c(-13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5), Y = c(3146L, 3321L, 2773L, 2415L, 2183L, 1091L, 514L, 191L, 109L, 65L, 54L, 50L, 3288L, 3243L, 2826L, 2532L, 2060L, 896L, 517L, 275L, 164L, 106L, 202L, 53L, 3146L, 3502L, 2658L, 3038L, 3351L, 3238L, 2935L, 3212L, 3004L, 3088L, 2809L, 1535L, 3288L, 2914L, 2875L, 2489L, 3104L, 2771L, 2861L, 3309L, 2997L, 2361L, 2687L, 1215L, 3224L, 3131L, 3126L, 2894L, 2495L, 2935L, 2516L, 2994L, 3074L, 3008L, 2780L, 1454L, 3146L, 2612L, 2852L, 2774L, 2663L, 3097L, 2591L, 2295L, 1271L, 1142L, 646L, 68L, 3288L, 2606L, 2838L, 1320L, 2890L, 2583L, 2251L, 2155L, 1164L, 695L, 394L, 71L, 3224L, 2896L, 2660L, 2804L, 2762L, 2525L, 2615L, 1904L, 1364L, 682L, 334L, 64L), dset = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)), .Names = c(X,
Re: [R] Passing parameter to a function
On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:
tab- function(x,y){
xtabs(time~x+y, data=D)
}
But when I run:
tab(A,B)
I get:
Error in eval(expr, envir, enclos) : object A not found
I am quite sure that there is some easy way out, but I have tried with
different combinations of deparse(), substitute(), eval(), etc without success,
can someone help?
I assume that A and B are columns in D? If so, you could use
tab(D$A, D$B)
to get what you want. If you really want tab(A,B) to work, you'll need
to do messy work with substitute, e.g. in the tab function, something like
fla - substitute(time ~ x + y, list(x = substitute(x), y = substitute(y))
xtabs(fla, data=D)
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] package survey
JS == Joel Schwartz j...@joelschwartz.com on Sat, 18 Dec 2010 19:09:24 -0800 writes: -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Saturday, December 18, 2010 5:54 PM To: Joel Schwartz Cc: r-help@r-project.org Subject: Re: [R] package survey On Dec 18, 2010, at 8:11 PM, Joel Schwartz wrote: and does anyone know if it is possible to find the codes for functions in survey package? Yes, you can find the code by doing the following: 1) Go to the CRAN R package list (http://cran.r-project.org/web/packages/ ), scroll down to the survey package link and click on it. 2) Scroll down to the Downloads section and download the package source file. The R folder in this file contains the code for the functions in the package. You can of course follow an analogous procedure to get the code for other packages. There might be an easier or quicker way to do it from within R but ,if there is, I haven't learned it yet. (I suspect Joel knows this.) If the package is loaded, you can just type the name of the function at the console. svyhist # produces about a half-page of code. JS Yes, I should have suggested that option as well. It's JS probably the quickest way if you just want the code for JS one or a few functions. But if you want the code for JS most or all functions in a package (including ones for JS which you might not know the name off the top of your JS head) is there some way of sending the code for all JS functions in particular package to a .r file from the JS command line with one or two lines of code? Note that only the source package contains the real source code. What you get when you print the object is the result of parsing the original code and print()ing it. In almost all cases, this will have lost all comments, and will not format the same way the original authors have formatted the original source ... which for good R programmers is typically better / easier to read and understand than the parsed+deparsed version. Martin Maechler, ETH Zurich (and R Core Team) __ 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] sum with times
RRJ == Ronaldo Reis Junior chrys...@gmail.com on Sun, 19 Dec 2010 12:10:54 -0200 writes: RRJ Hi, Forget, the chron package work with this No, don't forget, rather solve the problem: Date / dateTime classes are native R entities; extension package 'chron' objects are not. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R.matlab memory use
Stefano Ghirlanda dr.ghirlanda at gmail.com writes: Hi Ben, Thanks for your reply. My data structure is about 2 x 2000 so one order of magnitude the one you tried. I have no problem saving and reading smaller data structures (even large ones, just not his large) between octave and R using octave's save -7 (which saves MATLAB v5 files) and R.matlab's readMat. And I can save in text format in octave and read in R using read.octave (from package foreign) so it's not a big deal. I was just surprised that R.matlab needed more memory than I have (I have 3GB on this machine). Thanks, Stefano On Sun, Dec 19, 2010 at 10:54 PM, Ben Bolker bbolker at gmail.com wrote: Stefano Ghirlanda dr.ghirlanda at gmail.com writes: I am trying to load into R a MATLAB format file (actually, as saved by octave). The file is about 300kB but R complains with a memory allocation error: [snip] A few more questions about what's going on: it makes sense that your ASCII format file is about 75 M (i.e., 10x larger than mine); presumably your data set is highly redundant so that it can be compressed to 300Kb? Any chance you can post your data somewhere, or figure out a way to generate a sample data set that is similar (i.e. starts at about 75 M and compresses to 300Kb when saved appropriately in octave)? Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing parameter to a function
Hi Duncan,
Yes, A and B are columns in D. Having said that I and trying to avoid
tab(D$A,D$B)
and I would prefer:
tab(A,B)
Unfortunately the syntax you suggest is giving me the same error:
Error in eval(expr, envir, enclos) : object A not found
I have tried to add some deparse() but I have got the error over again. The
last version I have tried:
function(x,y){
z - substitute(time ~ x + y, list(x = deparse(substitute(x)), y =
deparse(substitute(y
xtabs(z, data=D)
gives me another error:
Error in terms.formula(formula, data = data) :
formula models not valid in ExtractVars
Any idea on how I should modify the function to make it work?
Thanks,
Luca
Il giorno 20/dic/2010, alle ore 19.28, Duncan Murdoch ha scritto:
On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:
tab- function(x,y){
xtabs(time~x+y, data=D)
}
But when I run:
tab(A,B)
I get:
Error in eval(expr, envir, enclos) : object A not found
I am quite sure that there is some easy way out, but I have tried with
different combinations of deparse(), substitute(), eval(), etc without
success, can someone help?
I assume that A and B are columns in D? If so, you could use
tab(D$A, D$B)
to get what you want. If you really want tab(A,B) to work, you'll need to do
messy work with substitute, e.g. in the tab function, something like
fla - substitute(time ~ x + y, list(x = substitute(x), y = substitute(y))
xtabs(fla, data=D)
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] survexp - unable to reproduce example
Dear All,
when I try to reproduce an example of survexp, taken from the help
page of survdiff, I receive the error message
Error in floor(temp) : Non-numeric argument to mathematical function
.
It seems to come from match.ratetable. I think, it has to do with
character variables in a ratetable.
I would be interested to know, if it works for others. With an older
version of survival, it worked well.
best regards,
Heinz
library(survival)
Loading required package: splines
## Example from help page of survdiff
## Expected survival for heart transplant patients based on
## US mortality tables
expect - survexp(futime ~ ratetable(age=(accept.dt - birth.dt),
+ sex=1,year=accept.dt,race=white), jasa, cohort=FALSE,
+ ratetable=survexp.usr)
Error in floor(temp) : Non-numeric argument to mathematical function
sessionInfo('survival')
R version 2.12.1 Patched (2010-12-18 r53869)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=German_Switzerland.1252 LC_CTYPE=German_Switzerland.1252
[3] LC_MONETARY=German_Switzerland.1252 LC_NUMERIC=C
[5] LC_TIME=German_Switzerland.1252
attached base packages:
character(0)
other attached packages:
[1] survival_2.36-2
loaded via a namespace (and not attached):
[1] base_2.12.1 graphics_2.12.1 grDevices_2.12.1 methods_2.12.1
[5] splines_2.12.1 stats_2.12.1 tools_2.12.1 utils_2.12.1
traceback()
2: match.ratetable(rdata, ratetable)
1: survexp(futime ~ ratetable(age = (accept.dt - birth.dt), sex = 1,
year = accept.dt, race = white), jasa, cohort = FALSE,
ratetable = survexp.usr)
__
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] sample() issue
length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Thanks Cory [[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] After heteroskedasticity correction, how can I get new confidential interval?
First of all, thanks for your guide! Let me be more specific, here. Using data(Housing) from Ecdat library I ran a regression raw.model-lprice~llot+lbed+lbath+lsto+factor(driveway)+factor(recroom)+factor(fullbase)+factor(gashw)+factor(airco)+factor(prefarea)+factor(garagepl) coeftest(lm(raw.model)) and I got heteroskedasticity in significant level 5% such as below. bptest(lm(raw.model)) So, I corrected like this. sqrt(diag(hccm(lm(raw.model),type=hc1))) This gave me the corrected std.errors. From this moment, I want to test whether my parameters are individually and jointly significant since the new std.error will change coeftest(lm(raw.model))'s t, p and F value. Then, How can I get these new t, p, F? Is there any R command for this? -- View this message in context: http://r.789695.n4.nabble.com/After-heteroskedasticity-correction-how-can-I-get-new-confidential-interval-tp3095643p3095852.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] How to optimize function parameters?
Thanks for your suggestions, I will certainly look at that To answer your question... I can calculate root square error between empirical data a those predicted by model (I used this to optimize the parameters using gafit()). -- View this message in context: http://r.789695.n4.nabble.com/How-to-optimize-function-parameters-tp3095603p3095938.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] performance issue with merge
Hi,
I'm trying to merge a couple of matrices together. The matrices are created
from a couple of input files that have very similar (but not in all cases)
values in first and second column. 1st and 2nd column are used to generate
rownames, the actually interesting value is in column 3. I'd like to merge
the different files by the rownames generated from 1st and 2nd column. Files
look like this:
ID1 100 3.14
ID1 200 2.71
ID1 300 0.92
...
ID2 100 2.45
.
etc. (sorry, I do not know how to create such files/ matrices with R
commands).
Some files do not have the full range of values in the second column, but I
don't want to loose any values during the merge. As far as I understood it I
have to merge using the all.x and all.y directives which appears to be
related to the outer join of relational dbs. This would give me something
like:
ID1_100 3.14 1.56 3.45
ID1_200 2.71 NA 1.34
ID1_300 0.92 1.22 NA
...
Merging works as long I do not set the all.x and all.y directive in the
merge command:
DF - data.frame(NULL)
count=0
filenames - list.files()
filenames
for(i in filenames) {
count-count+1
tmp - read.delim(i, header=FALSE)
rwnames - paste(tmp[,1], tmp[,2], sep=_)
tmp-tmp[,3]
tmp-as.matrix(tmp)
rownames(tmp)-rwnames
if (count == 1) {
DF - tmp
} else {
DF - merge(DF, tmp, by=row.names,all.x = TRUE, all.y = TRUE)
rownames(DF)-DF$Row.names
DF-DF[,2:ncol(DF)]
}
}
As soon as I set the all directives the script runs forever without any
effect (files sizes: a couple of million lines per file). Is it expected,
that this type of merge takes so much longer (I think it never finishes!)?
Or do I have a conceptual problem with how merge works?
Maxim
P.S.: I'd be really happy in case I receive comments how to make my easy
parsing problem a bit more straight forward in terms of how the code looks
like!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sample() issue
On Dec 20, 2010, at 2:04 PM, cory n wrote: length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Read the FAQ: 25000*(1-.55)-11250 [1] -1.818989e-12 25000*(1-.55) 11250 [1] TRUE ?round ?all.equal ?zapsmall -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing parameter to a function
Duncan's solution works for me. I strongly suspect that you have not
read the Help file for xtabs carefully enough. Does the time column
in D consist of a vector of counts?
A - factor(c(A,A,B,B))
B - factor(c(1,2,1,2))
D - data.frame(A,B)
D$time - c(10,12,14,16)
tab(A,B)
B
A1 2
A 10 12
B 14 16
-- Bert
On Mon, Dec 20, 2010 at 11:02 AM, Luca Meyer lucam1...@gmail.com wrote:
Hi Duncan,
Yes, A and B are columns in D. Having said that I and trying to avoid
tab(D$A,D$B)
and I would prefer:
tab(A,B)
Unfortunately the syntax you suggest is giving me the same error:
Error in eval(expr, envir, enclos) : object A not found
I have tried to add some deparse() but I have got the error over again. The
last version I have tried:
function(x,y){
z - substitute(time ~ x + y, list(x = deparse(substitute(x)), y =
deparse(substitute(y
xtabs(z, data=D)
gives me another error:
Error in terms.formula(formula, data = data) :
formula models not valid in ExtractVars
Any idea on how I should modify the function to make it work?
Thanks,
Luca
Il giorno 20/dic/2010, alle ore 19.28, Duncan Murdoch ha scritto:
On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:
tab- function(x,y){
xtabs(time~x+y, data=D)
}
But when I run:
tab(A,B)
I get:
Error in eval(expr, envir, enclos) : object A not found
I am quite sure that there is some easy way out, but I have tried with
different combinations of deparse(), substitute(), eval(), etc without
success, can someone help?
I assume that A and B are columns in D? If so, you could use
tab(D$A, D$B)
to get what you want. If you really want tab(A,B) to work, you'll need to
do messy work with substitute, e.g. in the tab function, something like
fla - substitute(time ~ x + y, list(x = substitute(x), y = substitute(y))
xtabs(fla, data=D)
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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
__
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] After heteroskedasticity correction, how can I get new confidential interval?
On Mon, 20 Dec 2010, JoonGi wrote: First of all, thanks for your guide! Let me be more specific, here. Using data(Housing) from Ecdat library I ran a regression raw.model-lprice~llot+lbed+lbath+lsto+factor(driveway)+factor(recroom)+factor(fullbase)+factor(gashw)+factor(airco)+factor(prefarea)+factor(garagepl) coeftest(lm(raw.model)) and I got heteroskedasticity in significant level 5% such as below. bptest(lm(raw.model)) So, I corrected like this. sqrt(diag(hccm(lm(raw.model),type=hc1))) This gave me the corrected std.errors. From this moment, I want to test whether my parameters are individually and jointly significant since the new std.error will change coeftest(lm(raw.model))'s t, p and F value. Then, How can I get these new t, p, F? Is there any R command for this? You can team up coeftest() with other vcov functions, such as hccm() from car or vcovHC() from the sandwich package. Actually, there is an example on ?coeftest for the latter. Also, there are various worked examples in vignette(sandwich, package = sandwich) Best, Z -- View this message in context: http://r.789695.n4.nabble.com/After-heteroskedasticity-correction-how-can-I-get-new-confidential-interval-tp3095643p3095852.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] sample() issue
See FAQ 7.31: http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f And try this: 25000*(1-.55) - 11250 25000*.45 - 11250 Notice a difference? 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 cory n Sent: Monday, December 20, 2010 20:04 To: r-help@r-project.org Subject: [R] sample() issue length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Thanks Cory [[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] sample() issue
On Mon, Dec 20, 2010 at 11:04 AM, cory n corynis...@gmail.com wrote: length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Let me make a wild guess: floor(25000*(1-.55)) [1] 11249 25000*(1-.55) [1] 11250 25000*(1-.55) - 11250 [1] -1.818989e-12 25000*0.45 - 11250 [1] 0 I'm guessing that the machine representation of 0.55 is off by something like 1e-16, which gets multiplied by a lot (25000) and this is enough for the floor (or whatever rounding the internal code uses) to make it 11249. The morale of the story is do not multiply non-exact numbers by huge constants or you may get small inaccuracies. Peter __ 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] sample() issue
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of cory n Sent: Monday, December 20, 2010 11:04 AM To: r-help@r-project.org Subject: [R] sample() issue length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Thanks Cory See FAQ 7.31 Then try .45 == (1-.55) Hope this is helpful, Dan Daniel J. Nordlund Washington State Department of Social and Health Services Planning, Performance, and Accountability Research and Data Analysis Division Olympia, WA 98504-5204 __ 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] sample() issue
Hi Cory, Check out http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f HTH, Jorge On Mon, Dec 20, 2010 at 2:04 PM, cory n wrote: length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Thanks Cory [[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] How to optimize function parameters?
If you can somehow add the sign of the output to the root square error, nlm, nlminb, and many options within optimx accept bounds. -- Jonathan P. Daily Technician - USGS Leetown Science Center 11649 Leetown Road Kearneysville WV, 25430 (304) 724-4480 Is the room still a room when its empty? Does the room, the thing itself have purpose? Or do we, what's the word... imbue it. - Jubal Early, Firefly r-help-boun...@r-project.org wrote on 12/20/2010 01:13:53 PM: [image removed] Re: [R] How to optimize function parameters? zbynek.jano...@gmail.com to: r-help 12/20/2010 02:35 PM Sent by: r-help-boun...@r-project.org Thanks for your suggestions, I will certainly look at that To answer your question... I can calculate root square error between empirical data a those predicted by model (I used this to optimize the parameters using gafit()). -- View this message in context: http://r.789695.n4.nabble.com/How-to- optimize-function-parameters-tp3095603p3095938.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] Complicated nls formula giving singular gradient message
Jared: You realize, of course, that just because you get estimates of the parameters from the software is no guarantee that the estimates mean anything? Nor does it mean that they mean nothing, I hasten to add. If, as one might suspect, the model is overparameterized, the estimates may be so imprecise that they are effectively useless -- but the fitted values may nevertheless (__Especially__ if overarameterized) fit your data very well. The model just won't fit future data. In other words, you may have a well-fitting, scientifically meaningless model. Cheers, Bert On Mon, Dec 20, 2010 at 10:26 AM, Jared Blashka evilamaran...@gmail.com wrote: Though my topic is slightly old already, I feel that it is necessary to post an update on my situation. I ended being able to estimate the parameters for this problem without having to worry as much about initial parameter estimates using AD Model Builder. It calculates the exact gradient using automatic differentiation so it's able to avoid the singular gradient problem nls can give. I also used the R package PBSadmb, which allowed me to run AD Model Builder and retrieve the results from within R. Then I could do what I liked with the results: generate graphs, more analysis, etc. Thanks to everyone who helped, Jared On Wed, Dec 15, 2010 at 8:47 AM, dave fournier da...@otter-rsch.com wrote: Jared Blashka wrote: Hi, Can you write a little note to the R list saying something like Re: SOLVED [R] Complicated nls formula giving singular gradient message I was able to estimate the parameters for this problem using AD Mode Builder which calculates the exact gradient for you using automatic differentiation and is thus able to avoid the singular gradient problem I encountered in nls. That way other R users who might be able to take advantage of the software will hear about it. Cheers, Dave Dave, That's exactly what I was looking for! Thanks for all your help! Jared On Tue, Dec 14, 2010 at 7:13 AM, dave fournier da...@otter-rsch.commailto: da...@otter-rsch.com wrote: Jared Blashka wrote: The source code for that is in jared.tpl I changed from least squares to a concentrated likelihood so that you could get estimated std devs via the delta method. they are in jared.std I rescaled the parameters so that the condition number of the Hessian is close to 1. You can see the eigenvalues of the Hessian in jared.eva. Your data are in jared.dat and the initial parameter values are in jared.pin. The parameter estimates with their estiamted std devs are: index name value std dev 1 NS 1.1254e-02 7.1128e-03 2 LogKi -8.8933e+00 8.2411e-02 3 LogKi -5.2005e+00 9.2179e-02 4 LogKi -7.2677e+00 7.7047e-02 5 BMax 2.1226e+05 5.1699e+03 ~ How does it look? Cheers, Dave Dave - AD Model Builder looks like a great tool that I can use, but I'm curious if it can also perform global parameter estimations across multiple data sets. In regards to the example I have provided, I have two similar data sets that also need to be analyzed, but the values for NS and BMax between the three data sets should be the same. Each data set has a unique LogKi value however. In R, I accomplished this by merging the three data sets and adding an additional field for each data point that identified which set it was originally from. Then in the regression formula I specified the LogKi term as a vector: LogKi[dset]. The results of the regression gave me one value each for NS and BMax, but three LogKi values. I haven't had much time to look through the AD Model Builder documentation yet, but are you aware if such an analysis method is possible? Here's one such example of a data set all -structure(list(X = c(-13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5, -13, -11, -10, -9.5, -9, -8.5, -8, -7.5, -7, -6.5, -6, -5), Y = c(3146L, 3321L, 2773L, 2415L, 2183L, 1091L, 514L, 191L, 109L, 65L, 54L, 50L, 3288L, 3243L, 2826L, 2532L, 2060L, 896L, 517L, 275L, 164L, 106L, 202L, 53L, 3146L, 3502L, 2658L, 3038L, 3351L, 3238L, 2935L, 3212L, 3004L, 3088L, 2809L, 1535L, 3288L, 2914L, 2875L, 2489L, 3104L, 2771L, 2861L, 3309L, 2997L, 2361L, 2687L, 1215L, 3224L, 3131L, 3126L, 2894L, 2495L, 2935L, 2516L,
Re: [R] Passing parameter to a function
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Luca Meyer
Sent: Monday, December 20, 2010 11:03 AM
To: Duncan Murdoch
Cc: R-help@r-project.org
Subject: Re: [R] Passing parameter to a function
Hi Duncan,
Yes, A and B are columns in D. Having said that I and trying to avoid
tab(D$A,D$B)
and I would prefer:
tab(A,B)
Unfortunately the syntax you suggest is giving me the same error:
Error in eval(expr, envir, enclos) : object A not found
You didn't show what you did, but Duncan's suggestion
works for me when it is in a function:
f0 - function (a, b) {
aExpr - substitute(a)
bExpr - substitute(b)
formula - substitute(time ~ e1 + e2, list(e1 = aExpr, e2 = bExpr))
xtabs(formula, data = D)
}
D - data.frame(time=2^(0:9),A=rep(1:2,each=5),B=rep(11:13,c(3,3,4)))
z - f0(A, B)
print(z)
B
A11 12 13
1 7 24 0
2 0 32 960
xtabs(time~A+B, data=D)
B
A11 12 13
1 7 24 0
2 0 32 960
The main problem with f0() is that the call attribute of
xtabs' output is always xtabs(formula=formula, data=D).
If you want the formula expanded to its value you have to
do more tricks. E.g., use call() and eval() to pass
some arguments by value and some by name:
f1 - function (a, b) {
aExpr - substitute(a)
bExpr - substitute(b)
formula - substitute(time ~ e1 + e2, list(e1 = aExpr, e2 = bExpr))
# pass formula by value and D by name
theCall - call(xtabs, formula, data = quote(D))
# may have to use enclos= argument to get D from right place
eval(theCall)
}
or use substitute a fourth time to patch up the call
attribute:
f2 - function (a, b) {
aExpr - substitute(a)
bExpr - substitute(b)
formula - substitute(time ~ e1 + e2, list(e1 = aExpr, e2 = bExpr))
retval - xtabs(formula, data = D)
attr(retval, call) - do.call(substitute, list(attr(retval,
call), list(formula = formula)))
retval
}
z - f2(A,B) # f0 does the same
attr(z, call)
xtabs(formula = time ~ A + B, data = D)
z
B
A11 12 13
1 7 24 0
2 0 32 960
I have tried to add some deparse() but I have got the error
over again. The last version I have tried:
function(x,y){
z - substitute(time ~ x + y, list(x =
deparse(substitute(x)), y = deparse(substitute(y
xtabs(z, data=D)
gives me another error:
Error in terms.formula(formula, data = data) :
formula models not valid in ExtractVars
Look at what 'z' is in the above:
g -
+ function(x,y){
+ z - substitute(time ~ x + y, list(x =
+ deparse(substitute(x)), y = deparse(substitute(y
+ z
+ }
print(g(A,B))
time ~ A + B
The strings A and B don't make sense there. They need
to be names (a.k.a. symbols). Remove the calls to deparse()
and it will work.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Any idea on how I should modify the function to make it work?
Thanks,
Luca
Il giorno 20/dic/2010, alle ore 19.28, Duncan Murdoch ha scritto:
On 20/12/2010 1:13 PM, Luca Meyer wrote:
I am trying to pass a couple of variable names to a xtabs formula:
tab- function(x,y){
xtabs(time~x+y, data=D)
}
But when I run:
tab(A,B)
I get:
Error in eval(expr, envir, enclos) : object A not found
I am quite sure that there is some easy way out, but I
have tried with different combinations of deparse(),
substitute(), eval(), etc without success, can someone help?
I assume that A and B are columns in D? If so, you could use
tab(D$A, D$B)
to get what you want. If you really want tab(A,B) to work,
you'll need to do messy work with substitute, e.g. in the tab
function, something like
fla - substitute(time ~ x + y, list(x = substitute(x), y =
substitute(y))
xtabs(fla, data=D)
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.
Re: [R] sample() issue
Trying to make the fastest referal to FAQ 7.31 is becoming quite a sport around here. David beat us all to it on this round. Chances are there will be many more rounds to come. Best, Wolfgang Original Message From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jorge Ivan Velez Sent: Monday, December 20, 2010 20:42 To: cory n Cc: R mailing list Subject: Re: [R] sample() issue Hi Cory, Check out http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f HTH, Jorge On Mon, Dec 20, 2010 at 2:04 PM, cory n wrote: length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 length(sample(25000, 11250)) [1] 11250 length(sample(25000, 25000*.45)) [1] 11250 So the question is, why do I get 11249 out of the first command and not 11250? I can't figure this one out. Thanks Cory __ 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] Sine Regression in R
Hi everyone, I am trying to fit a sine function on one year of wind data. I have two questions below. Looking around on the net I managed to get the following: Sine Equation: y = a + b * sin( c + d*x ) b is the amplitude, c is the phase shift, d is something deal with periodicty of data*.* This can be linearised by sin( c+dx ) = cos(c) * sin(dx) + sin(c) * cos(dx). If one calls dx = x1 y = a + b * sin( c + d*x ) y = a + b * [cos(c) * sin(x1) + sin(c) * cos(x1)] y = a + b * cos(c) * sin(x1) + b * sin(c) * cos(x1) y = a + b1 * sin(x1) + b2 * cos(x1) where b1 = b * cos(c) and b2 = b * sin(c) This works fine for me as I am not interested in the value for b and c. By trial and error I also detemined that the sensibel value for d is 1/58. I have 366 days of data and want to fit a single sine onto it. *First question*: 1/ 58 = (2 pi) / 366. I guess 366 is the period of my data, I have 366 days. so *d = (2 pi) / Period*. Is this correct. *Second question*: I reckon that all of this looks very DIY. Is there a better way to do this in R? Maybe through a package I dont know? Thanks in advance Paolo [[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] Sine Regression in R
Hi:
Try using package sos to address your question re which packages perform
sine/trig regression:
library(sos)
findFn('sine regression')# 12 matches on my system
findFn('trigonometric regression') # 9 matches
The results point largely to the mFilter, oce, circStats and circular
packages. It's a place to start...
HTH,
Dennis
On Mon, Dec 20, 2010 at 12:01 PM, Paolo Rossi
statmailingli...@googlemail.com wrote:
Hi everyone,
I am trying to fit a sine function on one year of wind data. I have two
questions below.
Looking around on the net I managed to get the following:
Sine Equation: y = a + b * sin( c + d*x )
b is the amplitude, c is the phase shift, d is something deal with
periodicty of data*.*
This can be linearised by sin( c+dx ) = cos(c) * sin(dx) + sin(c) *
cos(dx).
If one calls dx = x1
y = a + b * sin( c + d*x )
y = a + b * [cos(c) * sin(x1) + sin(c) * cos(x1)]
y = a + b * cos(c) * sin(x1) + b * sin(c) * cos(x1)
y = a + b1 * sin(x1) + b2 * cos(x1)
where b1 = b * cos(c) and b2 = b * sin(c)
This works fine for me as I am not interested in the value for b and c. By
trial and error I also detemined that the sensibel value for d is 1/58. I
have 366 days of data and want to fit a single sine onto it.
*First question*: 1/ 58 = (2 pi) / 366. I guess 366 is the period of my
data, I have 366 days. so *d = (2 pi) / Period*. Is this correct.
*Second question*: I reckon that all of this looks very DIY. Is there a
better way to do this in R? Maybe through a package I dont know?
Thanks in advance
Paolo
[[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] breaking an inner loop in R
kayj wrote:
Hi All,
is there a way to break an inner loop in R? for example
for ( i in 1: 100){
if ( condition){
break
}
}
I have search and I have seen exampels where they add a semicolon after
break, do I need to do that? Also when I type ?break in R in returns
nothing on this command?
thanks for your help
?break
Also see section 3.2.2 of the R language manual (R-lang.pdf, version 2.12.0)
Berend
--
View this message in context:
http://r.789695.n4.nabble.com/breaking-an-inner-loop-in-R-tp3095607p3096109.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] Metafor package
Dear Mario, 1) Whether to use 1/2* or not is arbitary. It's just a multiplicative constant. 2) vi - 1/(4*ni + 2) If one leave off the 1/2* part, then it would be vi - 1/(ni + 1/2). 3) Yes, but it takes a bit more work. Fortunately, the same issue was raised not very long ago, so the following post should essentially answer your question: https://stat.ethz.ch/pipermail/r-help/2010-December/263286.html 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 petre...@unina.it Sent: Monday, December 20, 2010 13:26 To: r-help@r-project.org Subject: [R] Metafor package I have some question about metafor package. I'm interest to perform a random effect meta-analysis of proportion (single group summary of prevalence of disease in a population as reported by different study) It ask: 1. PFT: The Freeman-Tukey double arcsine transformed proportion is reported to be equal to 1/2*(asin(sqrt(xi/(ni+1))) + asin(sqrt((xi+1)/(ni+1. Hovewer, i also found the same formula but without 1/2*. 2. how vi, the corresponding (estimated) sampling variance is calculated? (i.e. the formula used to estimate vi) 3. it is possible to return a forest plot with the backtransformed value of proportion, confidenc interval and weight and not with the transformed value after performing rma.uni? Many thanks Mario Petretta Dipartimento di Medicina Clinica Scienze Cardiovascolari e Immunologiche Facoltà di Medicina e Chirurgia Università di Napoli Federico II 081 - 7462233 __ 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] sample() issue
On Mon, Dec 20, 2010 at 01:04:18PM -0600, cory n wrote: length(sample(25000, 25000*(1-.55))) [1] 11249 25000*(1-.55) [1] 11250 Besides FAQ 7.31, look also at http://rwiki.sciviews.org/doku.php?id=misc:r_accuracy for further examples and hints concerning rounding errors in simple situations. Petr Savicky. __ 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] RExcel doesn't get active dataset
Hey everyone When I try to 'get active dataframe' from the context menu in excel using Rcmdr menus in excel, only the header and first row of data gets copied to the excel spread sheet. No clue why this is happening. I followed the instructions from 'R through Excel' pg 25-26. I'm using excel 2010, Rcmdr v 1.6-0 and R 2.11.1. Much appreciate any help Sent via my BlackBerry from Vodacom - let your email find you! __ 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] finding and opening C source called from R functions.
Hi, As will soon be very clear I'm an R novice. I'm trying to better understand the ks.test function in the stats package. When I look at the source code there are several calls to C functions (for example .C(pkstwo, is.integer(length(x[IND])), p=as.double(x[IND]), as.double(tol), PACKAGE = stats)$P). I'm wondering if there is a way to view the source code for the underlying C functions? I googled a post which said to open ks.c to see the source code but I can't find that file anywhere on my hard drive or on the web. Any suggestions would be greatly appreciated. Thanks in advance, Ben ___ This message and any attached documents contain information which may be confidential, subject to privilege or exempt from disclosure under applicable law. These materials are solely for the use of the intended recipient. If you are not the intended recipient of this transmission, you are hereby notified that any distribution, disclosure, printing, copying, storage, modification or the taking of any action in reliance upon this transmission is strictly prohibited. Delivery of this message to any person other than the intended recipient shall not compromise or waive such confidentiality, privilege or exemption from disclosure as to this communication. If you have received this communication in error, please notify the sender immediately and delete this message from your system. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] finding and opening C source called from R functions.
On Mon, Dec 20, 2010 at 12:48 PM, Chiquoine, Ben bchiquo...@tiff.org wrote: Hi, As will soon be very clear I'm an R novice. I'm trying to better understand the ks.test function in the stats package. When I look at the source code there are several calls to C functions (for example .C(pkstwo, is.integer(length(x[IND])), p=as.double(x[IND]), as.double(tol), PACKAGE = stats)$P). I'm wondering if there is a way to view the source code for the underlying C functions? I googled a post which said to open ks.c to see the source code but I can't find that file anywhere on my hard drive or on the web. Any suggestions would be greatly appreciated. Go to cran.r-project.org, find the package stats, download the tar.gz bundle, unpack it (on a linux command line, you can type tar -xvzf stats*.tar.gz), then navigate to the directory stats/src and look at the source code in the directory. __ 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] Sine Regression in R
On Dec 20, 2010, at 3:01 PM, Paolo Rossi wrote: Hi everyone, I am trying to fit a sine function on one year of wind data. I have two questions below. Looking around on the net I managed to get the following: Sine Equation: y = a + b * sin( c + d*x ) b is the amplitude, c is the phase shift, d is something deal with periodicty of data*.* This can be linearised by sin( c+dx ) = cos(c) * sin(dx) + sin(c) * cos(dx). If one calls dx = x1 y = a + b * sin( c + d*x ) y = a + b * [cos(c) * sin(x1) + sin(c) * cos(x1)] y = a + b * cos(c) * sin(x1) + b * sin(c) * cos(x1) y = a + b1 * sin(x1) + b2 * cos(x1) where b1 = b * cos(c) and b2 = b * sin(c) This works fine for me as I am not interested in the value for b and c. By trial and error I also detemined that the sensibel value for d is 1/58. I have 366 days of data and want to fit a single sine onto it. *First question*: 1/ 58 = (2 pi) / 366. I guess 366 is the period of my data, I have 366 days. so *d = (2 pi) / Period*. Is this correct. *Second question*: I reckon that all of this looks very DIY. Is there a better way to do this in R? Maybe through a package I dont know? The package circular might be a candidate. Thanks in advance Paolo 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] finding and opening C source called from R functions.
On 20/12/2010 3:56 PM, Peter Langfelder wrote: On Mon, Dec 20, 2010 at 12:48 PM, Chiquoine, Benbchiquo...@tiff.org wrote: Hi, As will soon be very clear I'm an R novice. I'm trying to better understand the ks.test function in the stats package. When I look at the source code there are several calls to C functions (for example .C(pkstwo, is.integer(length(x[IND])), p=as.double(x[IND]), as.double(tol), PACKAGE = stats)$P). I'm wondering if there is a way to view the source code for the underlying C functions? I googled a post which said to open ks.c to see the source code but I can't find that file anywhere on my hard drive or on the web. Any suggestions would be greatly appreciated. Go to cran.r-project.org, find the package stats, download the tar.gz bundle, unpack it (on a linux command line, you can type tar -xvzf stats*.tar.gz), then navigate to the directory stats/src and look at the source code in the directory. In many cases that will work, but not in this one: stats is a base package, so there is no tar.gz bundle. See Uwe Ligges' article in R News on how to find source in general. It's in http://cran.r-project.org/doc/Rnews/Rnews_2006-4.pdf and I think it is all still current. 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] R Code and the Pygments Python SyntaxHighlighter
(reviving from the grave) On Thu, Mar 27, 2008 at 5:44 PM, Hans W. Borchers hwborch...@gmail.com wrote: is someone going to write a R/S language lexer for the Pygments Python syntax highlighter http://pygments.org/? As it is used now by Trac, Django, or the Python documentation tool Sphinx, the R community can apply it in Python-based Wikis like Moinmoin and others. I was once again [0] looking for some solution to automatic syntax highlighting (or input/output colouring) in a bare bones R terminal and I've stumbled on both Pygments and this thread. Current Pygemnts fancies [1] support for both S and R. Is anyone aware of any progress in this direction? Perhaps some R interface to Pygments? Regards Liviu [0] http://www.mail-archive.com/r-help@r-project.org/msg86512.html [1] http://pygments.org/languages/ __ 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] .Rd file for S4-method warning
Dear R users,
I want to create a proper .Rd file for the show method for an S4 class.
I am encountering problems in the \usage{} line, I guess. An example:
setClass(testClass,
representation(a=character))
setMethod(show, testClass, function(object){
})
The .Rd file:
\name{show,-method}
\alias{show,testClass-method}
\alias{show}
\title{Show method for testClass...}
\usage{\S4method{show}{testClass}(object)
}
\description{Show method for testClass}
\arguments{\item{testClass}{object}
}
CHECK says:
* checking Rd \usage sections ... WARNING
Undocumented arguments in documentation object 'show,-method'
object
What would be a correct \usage line? Writing R extensions says:
\S4method{generic}{signature_list}(argument_list)
What am I doing wrong?
It works though if I simply delete the \usage line.
Unfortunately I use roxygen and the line is created automatically,
so I need to create it properly.
Thanks in advance,
Mark
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.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.
Re: [R] .Rd file for S4-method warning
On 20/12/2010 5:18 PM, Mark Heckmann wrote:
Dear R users,
I want to create a proper .Rd file for the show method for an S4 class.
I am encountering problems in the \usage{} line, I guess. An example:
setClass(testClass,
representation(a=character))
setMethod(show, testClass, function(object){
})
The .Rd file:
\name{show,-method}
\alias{show,testClass-method}
\alias{show}
\title{Show method for testClass...}
\usage{\S4method{show}{testClass}(object)
}
\description{Show method for testClass}
\arguments{\item{testClass}{object}
}
CHECK says:
* checking Rd \usage sections ... WARNING
Undocumented arguments in documentation object 'show,-method'
object
What would be a correct \usage line? Writing R extensions says:
\S4method{generic}{signature_list}(argument_list)
That's okay, the warning is about the fact that you didn't document
object in the \arguments section.
You had
\item{testClass}{object}
but you should have had
\item{object}{some description of what object is}
As yours was written, it's documentation for the testclass argument,
which doesn't exist.
What am I doing wrong?
It works though if I simply delete the \usage line.
Unfortunately I use roxygen and the line is created automatically,
so I need to create it properly.
Does roxygen also create the argument? Looks like a bug or limitation
(I seem to recall that roxygen doesn't support S4, or didn't in the past...)
Duncan Murdoch
Thanks in advance,
Mark
–––
Mark Heckmann
Blog: www.markheckmann.de
R-Blog: http://ryouready.wordpress.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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R.matlab memory use
Hi, The data set is indeed highly redundant, mostly zeros (which means I could use sparse matrices, but that's another issue). I have posted an octave ascii file here: http://dl.dropbox.com/u/4000198/freq.txt and a .mat file here: http://dl.dropbox.com/u/4000198/freq.mat A few more questions about what's going on: it makes sense that your ASCII format file is about 75 M (i.e., 10x larger than mine); presumably your data set is highly redundant so that it can be compressed to 300Kb? Any chance you can post your data somewhere, or figure out a way to generate a sample data set that is similar (i.e. starts at about 75 M and compresses to 300Kb when saved appropriately in octave)? Ben Bolker -- Stefano Ghirlanda www.intercult.su.se/~stefano - drghirlanda.wordpress.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] Plot a matrix recursively
for data.frame: for(j in grep('Laser_', names(m)) lines(m[,j])
for matrix: for(j in grep('Laser_', colnames(m)) lines(m[,j])
or
for(j in 2:4) lines(m[,j])
Shorter could be worse if you insert additional columns later.
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-a-matrix-recursively-tp3067283p3106041.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.
Re: [R] plot more plots from one matrix
See library(reshape) -- View this message in context: http://r.789695.n4.nabble.com/plot-more-plots-from-one-matrix-tp3069545p3107761.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] source file for rnorm
Dear Sir/Madam: I am sorry if this is a simple question. I have downloaded the source code for R but could not find the source file that has the rnorm function definition. Could you please let me know which file has the function? Also, in general, how do I find the source file of a given function? Thanks. Yue Zhang, CFA Cohen Steers Capital Management, Inc. 280 Park Ave., 10th Floor New York, NY 10017 Tel: (212)796-9370 Fax: (212)319-8238 Email: yzh...@cohenandsteers.com The information transmitted is only for the intended recipient and may contain confidential, privileged, copyrighted, or otherwise restricted material. It may not be reproduced or retransmitted without permission. If you have received this in error, please contact the sender, delete this transmission and dispose of any printed material as your review, retransmission, dissemination or other use of, or taking any action in reliance upon, this information is not authorized and may be prohibited by law. Cohen Steers may review and retain this message and any addressed to its sender. Opinions or views herein do not necessarily reflect those of Cohen Steers, Inc., its subsidiaries, or affiliates. [[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] Plot a matrix recursively
for data.frame: for(j in grep('Laser_', names(m)) lines(m[,j])
for matrix: for(j in grep('Laser_', colnames(m)) lines(m[,j])
or: for(j in 2:4) lines(m[,j])
Shorter could be worse if you insert additional columns later.
alcesgabbo wrote:
Hi,
I have the following matrix (named m):
key
sensor_date Laser_1 Laser_2
Laser_3
2010-09-30T15:00:12+020063
1
2010-10-31T15:05:07+010054
2
2011-09-30T15:00:12+020063
1
2011-10-31T15:05:07+010054
2
I plot the first column with the following function:
plot(m[,1],type=o, xaxt=n,ylim=c(min(m[,1:length(colnames(m))])-1,
max(m[,1:length(colnames(m))])+1))
for the other columns I use there functions:
lines(m[,2],type=\o\)
lines(m[,3],type=\o\)
ok, it works.
But is there a way to do this prodcedures recursively??
for example:
for each columns {
lines(m[,column],type=\o\)
}
I try with :
lines(m[,2:length(colnames(m))],type=\o\)
but it doesn't work.
Thanks .
Alberto
--
View this message in context:
http://r.789695.n4.nabble.com/Plot-a-matrix-recursively-tp3067283p3106392.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] Filter data
You missed the first, the most important line of str() output. What type is your data? -- View this message in context: http://r.789695.n4.nabble.com/Filter-data-tp3070069p3108011.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] logistic regression or not?
Hi, I have a dataset where the response for each person on one of the 2
treatments was a proportion (percentage of certain number of markers being
positive), I also have the number of positive negative markers available for
each person. what is the best way to analyze this kind of data?
I can think of analyzing this data using glm() with the attached dataset:
test-read.table('test.txt',sep='\t')
fit-glm(cbind(positive,total-positive)~treatment,test,family=binomial)
summary(fit)
anova(fit, test='Chisq')
First, is this still called logistic regression or something else? I thought
with logistic regression, the response variable is a binary factor?
Second, then summary(fit) and anova(fit, test='Chisq') gave me different p
values, why is that? which one should I use?
Third, is there an equivalent model where I can use variable percentage
instead of positive total?
Finally, what is the best way to analyze this kind of dataset where it's almost
the same as ANOVA except that the response variable is a proportion (or success
and failure)?
Thanks
John
treatment total positive percentage
1 exposed 11 4 0.363636363636364
2 exposed 10 4 0.4
3 exposed 9 4 0.444
4 exposed 7 4 0.571428571428571
5 exposed 7 4 0.571428571428571
6 exposed 6 5 0.833
8 exposed 12 7 0.583
9 exposed 8 5 0.625
10exposed 13 12 0.923076923076923
11exposed 10 5 0.5
12control 10 1 0.1
13control 11 2 0.181818181818182
14control 8 0 0
16control 12 1 0.0833
15control 8 0 0
17control 10 1 0.1
18control 10 1 0.1
19control 8 1 0.125
20control 8 0 0
21control 9 1 0.111
22control 10 1 0.1
__
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] package arules - 'transpose' of the transactions
Just found 'ngCMatrix' class, it must be the way to go?
-
Message: 47Date: Mon, 20 Dec 2010 17:41:20 +1100
From: Kohleth Chia kohl...@gmail.com
To: r-help@r-project.org
Subject: [R] package arules - 'transpose' of the transactions
Message-ID:
aanlktimmhuj+gnthvjvt8yomf3c5sqorkxcvgb3ut...@mail.gmail.comaanlktimmhuj%2bgnthvjvt8yomf3c5sqorkxcvgb3ut...@mail.gmail.com
Content-Type: text/plain
Suppose this is my list of transactions:
set.seed(200)
3)
tran=random.transactions(100,
inspect(tran)
itemstransactionID
1 {item80}trans1
2 {item8,
item20}trans2
3 {item28}trans3
I want to get the 'transpose' of the data, i.e.
transactionID items
1 {trans2}item8
2 {trans2}item20
3 {trans3}item28
4 {trans1}item80
I tried converting tran into a matrix, then transpose it, then convert it
back to transactions. But my dataset is actually very very large, so I
wonder if there is any faster method?
Thanks
--
KC
[[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] install.packages() - old version deleted, new version did not install
I've just tentatively put code to fix this in place in R-devel. By default it is not enabled, because if it goes wrong it could really mess things up. To enable it, set options(install.lock=TRUE) before installing or updating binary packages in Windows. It will slow down all installs, so it may be that the default should stay no-locking as it is now: only people running multiple instances of R really need this. Since it appealed to my sense of symmetry, you can also say options(install.lock=FALSE) which will mean that source installs done by install.packages() will default to act like R CMD INSTALL --unsafe, i.e. go ahead without locking. The latter change affects all platforms, not just Windows. You should be able to download R-devel builds containing this change by tomorrow; look for revision r53875 or newer. Duncan Murdoch On 17/12/2010 11:13 AM, Jon Olav Skoien wrote: Dear list, (R 2.12.0, Windows 7, 64bit) I recently tried to install a new package (spacetime), that depends on sp among others. I already had the last one installed, but there was probably a newer version on CRAN, so the command install.packages(spacetime) also gave me: also installing the dependencies ‘sp’, ‘zoo’, ‘xts’ sp was already loaded in this session, so installation failed: package 'sp' successfully unpacked and MD5 sums checked Warning: cannot remove prior installation of package 'sp' Unfortunately, the warning should rather say: cannot completely remove prior installation of package 'sp' R managed to remove most of the prior installation of sp, except for the .dll. I could go on using sp in the existing sessions, but not load the package in a new session or open the help pages. This has happened to me several times, and the only solution I have found to this is to close all R-sessions and install the package again. This is normally ok, but this time I had some long-time computations running in another R-session that I did not want to interrupt. For the next time, is there a way to reinstall a package without interrupting running R-sessions? For me it seems like the cause of the problem could have been solved by checking if the .dll can be removed before removing the rest of the package, by adding something like the following in utils:::unpackPkgZip? if (unlink(paste(instPath,/libs/x64/sp.dll, sep = )) != 0) warning(cannot remove...) before ret- unlink(instPath, recursive = TRUE) (line 95) x64 in the path would have to be changed to something architecture dependent... Best wishes, Jon __ 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] source file for rnorm
On Dec 20, 2010, at 5:40 PM, Yue Zhang wrote: Dear Sir/Madam: I am sorry if this is a simple question. I have downloaded the source code for R but could not find the source file that has the rnorm function definition. Could you please let me know which file has the function? Also, in general, how do I find the source file of a given function? Uwe Ligges. R Help Desk: Accessing the sources. R News, 6(4):43-45, October 2006 http://cran.r-project.org/doc/Rnews/Rnews_2006-4.pdf Thanks. Yue Zhang, CFA Cohen Steers Capital Management, Inc. 280 Park Ave., 10th Floor 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.
