[R] Plotting Mean in plotting degree distribution
Hi, Please help! I need to plot a degree distribution. so far (see below) i have the following but the problem: i'm plotting the graph. instead I need to plot the degree distribution. library(igraph) dat- read.graph(file=tf.net, format=pajek) dat- as.undirected(dat, mode = c(each)) is.directed(dat) [1] FALSE degree.distribution(dat, cumulative = TRUE, loops = FALSE) [1] 1.0 0.792857143 0.478571429 0.292857143 0.192857143 0.142857143 0.114285714 0.085714286 [9] 0.078571429 0.071428571 0.064285714 0.05000 0.035714286 0.028571429 0.028571429 0.014285714 [17] 0.014285714 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 [25] 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 [33] 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 [41] 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 0.007142857 plot(dat) the file tf.net can be found at http://fileshare.csb.univie.ac.at/cheat14/INPUT/ any help much appreciated! [[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] Colorspace: bug in HLS conversion?
I just started using the colorspace package, and converting between RGB and HLS is not working as I expect. In particular, converting RGB(0,0,0) to HLS then back to RGB seems broken: # as(as(RGB(0, 0, 0), HLS), RGB) R G B [1,] 1 1 0 That is, converting black to HLS then back again produces yellow! Other values I've tried work as expected (i.e., I get the same RGB value out as I put in). Is this a bug in the code, or in my understanding of colour conversions? Kind Regards, Mike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Error installing extrafonts from local zip file
I am making my first submission to Plos One and am trying to install the extrafonts package to obtain Arial for the figures. extrafonts was archived at the end of last month so I downloaded first the extrafont_0.16.tar.gz file and then the older extrafont_0.15.tar.gz file. I selected Install packages(s) from local zip file from the Packages menu of my installation of R x64 3.0.0 I got: utils:::menuInstallLocal() Error in read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open the connection In addition: Warning messages: 1: In unzip(zipname, exdir = dest) : error 1 in extracting from zip file 2: In read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open compressed file 'extrafont_0.15.tar.gz/DESCRIPTION', probable reason 'No such file or directory' When I unzip these gz files using Winzip I do find a DESCRIPTION file. What am I doing wrong? Thanks Robin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Colorspace: bug in HLS conversion?
Hi, I cannot reproduce your result: R as(as(RGB(0, 0, 0), HLS), RGB) R G B [1,] 0 0 0 You probably should provide the output of sessionInfo(). Regards, Pascal On Thu, Mar 13, 2014 at 7:41 AM, Michael Gauland mikely...@amuri.net wrote: I just started using the colorspace package, and converting between RGB and HLS is not working as I expect. In particular, converting RGB(0,0,0) to HLS then back to RGB seems broken: # as(as(RGB(0, 0, 0), HLS), RGB) R G B [1,] 1 1 0 That is, converting black to HLS then back again produces yellow! Other values I've tried work as expected (i.e., I get the same RGB value out as I put in). Is this a bug in the code, or in my understanding of colour conversions? Kind Regards, Mike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Colorspace: bug in HLS conversion?
On Wed, 12 Mar 2014, Michael Gauland wrote: I just started using the colorspace package, and converting between RGB and HLS is not working as I expect. In particular, converting RGB(0,0,0) to HLS then back to RGB seems broken: # as(as(RGB(0, 0, 0), HLS), RGB) R G B [1,] 1 1 0 That is, converting black to HLS then back again produces yellow! I can't reproduce this. I get 0,0,0 as expected. Other values I've tried work as expected (i.e., I get the same RGB value out as I put in). Is this a bug in the code, or in my understanding of colour conversions? Maybe you are using an old version of colorspace? For version 1.1-1 (from two years ago, version 1.2-4 is current) the NEWS file says: o Bug fix in HLS_to_RGB conversion for s == 0. Kind Regards, Mike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] NAG (NAGFWrapper) or any optimization package recommended?
Hi, Is there any optimization package in R recommended for general purposes (constrained, unconstrained, etc)? Has anyone used the function in NAGFWrapper package http://www.nag.co.uk/numeric/R/r-package Are the optimization functions therein available for free? Thanks, Miao [[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] Error installing extrafonts from local zip file
On 13/03/2014 04:31, robin.thom...@csiro.au wrote: I am making my first submission to Plos One and am trying to install the extrafonts package to obtain Arial for the figures. extrafonts was archived at the end of last month so I downloaded first the extrafont_0.16.tar.gz file and then the older extrafont_0.15.tar.gz file. Those are not zip files. You need to use (at the command line) R CMD INSTALL extrafont_0.16.tar.gz (assuming you have already installed the dependencies). However, I suggest you use R 3.1.0 alpha which has support for Arial built in. I selected Install packages(s) from local zip file from the Packages menu of my installation of R x64 3.0.0 I got: utils:::menuInstallLocal() Error in read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open the connection In addition: Warning messages: 1: In unzip(zipname, exdir = dest) : error 1 in extracting from zip file 2: In read.dcf(file.path(pkgname, DESCRIPTION), c(Package, Type)) : cannot open compressed file 'extrafont_0.15.tar.gz/DESCRIPTION', probable reason 'No such file or directory' When I unzip these gz files using Winzip I do find a DESCRIPTION file. What am I doing wrong? Thanks Robin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] UTF-8 characters in pandoc template variables using knitr
Dear all, Consider the simpel RMarkdown file below. I've saved it as test.md with UTF-8 encoding. Notice that I have embedded a custom pandoc variable 'test' in the file. This variable holds an UTF-8 character ©. %My title %The authorslist !--pandoc format: latex V: test:Copyright notice. © copyright holder -- # First section fgsdfg jsdksdfgsdfg Now I try to compile it from within R using knitr. library(knitr) pandoc(test.md, format = latex) knitr inserts an extra character in the system command to pandoc. I assume it has something to do with the encoding. Any suggestions on how to solve this? executing pandoc -V test:Copyright notice. © copyright holder -f markdown -t latex -o test.pdf test.md I've posted the question last week on StackOverflow (http://stackoverflow.com/questions/22198832/utf-8-characters-in-pandoc-template-variables-using-knitr) but did not got any response. Best regards, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] NAG (NAGFWrapper) or any optimization package recommended?
On 13-03-2014, at 09:50, jpm miao miao...@gmail.com wrote: Hi, Is there any optimization package in R recommended for general purposes (constrained, unconstrained, etc)? See CRAN Task Views, Optimization. http://cran.r-project.org/web/views/ Has anyone used the function in NAGFWrapper package http://www.nag.co.uk/numeric/R/r-package Are the optimization functions therein available for free? Well not from what I could read on that webpage. See the second paragraph in the section labelled “Installation”. The NAG library is not free. Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] UTF-8 characters in pandoc template variables using knitr
And what did the maintainer say? (See the posting guide.) On 13/03/2014 09:11, ONKELINX, Thierry wrote: Dear all, Consider the simpel RMarkdown file below. I've saved it as test.md with UTF-8 encoding. Notice that I have embedded a custom pandoc variable 'test' in the file. This variable holds an UTF-8 character ©. %My title %The authorslist !--pandoc format: latex V: test:Copyright notice. © copyright holder -- # First section fgsdfg jsdksdfgsdfg Now I try to compile it from within R using knitr. library(knitr) pandoc(test.md, format = latex) knitr inserts an extra character in the system command to pandoc. I assume it has something to do with the encoding. Any suggestions on how to solve this? executing pandoc -V test:Copyright notice. © copyright holder -f markdown -t latex -o test.pdf test.md I've posted the question last week on StackOverflow (http://stackoverflow.com/questions/22198832/utf-8-characters-in-pandoc-template-variables-using-knitr) but did not got any response. Best regards, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] replace duplicates with 0
Dear all! Is there a possibility to replace all duplicates values in data frame with 0? Thank you very much! -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[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] replace duplicates with 0
Hello, Dis you at least search? ?duplicated Regards, Pascal On Thu, Mar 13, 2014 at 9:35 PM, catalin roibu catalinro...@gmail.com wrote: Dear all! Is there a possibility to replace all duplicates values in data frame with 0? Thank you very much! -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[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. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] replace duplicates with 0
Hello, See ?duplicated Something like dat1[duplicated(dat1), ] - 0 Hope this helps, Rui Barradas Em 13-03-2014 12:35, catalin roibu escreveu: Dear all! Is there a possibility to replace all duplicates values in data frame with 0? Thank you very much! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] replace duplicates with 0
Hi Catalin, The following should give you some ideas: set.seed(123) x - rpois(50, 2) x idx - duplicated(x) x[idx] - 0 x Best, Jorge.- On Thu, Mar 13, 2014 at 11:35 PM, catalin roibu catalinro...@gmail.comwrote: Dear all! Is there a possibility to replace all duplicates values in data frame with 0? Thank you very much! -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[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] Downloading a RData file and problem with loading it
Hi all, I have downloaded the file 5.R.RData and I am trying to load it into R using load(5.R.RData) But it doesn't work.What's wrong, Many thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Downloading a RData file and problem with loading it
Do you really think we can answer this? Ok, the file is not a valid RData file, or was corrupted in the file transfer. Was that helpful? If not, try being specific (what URL did this file come from? Which version of R? What was the output of session in sessionInfo()?) and posting in plain text as the Posting Guide advises you to (that is a hint... go read it). --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On March 13, 2014 6:22:45 AM PDT, varin sacha varinsa...@yahoo.fr wrote: Hi all, I have downloaded the file 5.R.RData�and I am trying to load it into R using load(5.R.RData)� But it doesn't work.What's wrong, Many thanks� [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] behaviour of rows and colomns suppression in a matrix
Hi List,
while running a script on a set of matrices I came into a case I would not
have guessed to arrive.
Below is a small toy example to illustrate the case.
Of course there is a simple workaround (using a simple test), but why does
this occur, and shouldnt it be corrected ?
More probably I miss a point, but which one ? Is this behavior obtained on
purpose and why ?
Sorry if its a FAQ
I didnt find my way to it.
(And sorry for multiple posting if any : I got a warning from r-bounce but
did not understand it).
Thanks, Olivier
#
# toy example 1 (no problem with this one)
toy.matrix.1-matrix(c(1,0,1,1,0,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.1,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.1,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
# finding the items to remove
unused.rows-which(margin.Rows=thresh) # unused.rows == 2
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM1-toy.matrix.1
TM1-TM1[-unused.rows,]
TM1-TM1[,-unused.cols]
TM1
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1 # OK
##
# toy example 2 (oops, no rows to suppress
)
toy.matrix.2-matrix(c(1,1,1,1,1,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.2,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.2,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
unused.rows-which(margin.Rows=thresh) # unused.rows ==
integer(0)
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM2-toy.matrix.2
TM2-TM2[-unused.rows,]
TM2-TM2[,-unused.cols]
TM2
# [,1] [,2] # empty...
###
# I was expecting :
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1
#[3,] 1 1
# which of course is obtained using :
TM2-toy.matrix.2
if(length(unused.rows) !=0) {TM2-TM2[-unused.rows,]}
if(length(unused.rows) !=0 ){TM2-TM2[,-unused.cols]}
TM2
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] change date format
Dear all! I have a new problem with the date format in a data frame. I have rainfall records extracted from an automatic meteo station. In the same data frame (with 1 records) the date is in 2 formats like that: 4/15/11 11:49 AM 5.6.11 2:51 PM My question is how to modify date format for all records with this format type: 4/15/11 11:49 AM. Thank you! Best regards! Catalin -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[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] behaviour of rows and colomns suppression in a matrix
Hi,
Your basic problem seems to be that you expect R to take
TM2[0, ]
as meaning not to subset anything, rather than to take only row 0,
which doesn't exist:
R TM2[0,]
[,1] [,2] [,3]
There's a hint in
?[
which says:
An index value of 'NULL' is treated as if it were
'integer(0)'.
Here are two alternative formulations of your task:
# safer method 1
used.rows-which(margin.Rows thresh)
used.cols-which(margin.Cols thresh)
TM2[used.rows, used.cols]
# safer method 2
unused.rows- margin.Rows=thresh
unused.cols- margin.Cols=thresh
TM2[!unused.rows, !unused.cols]
Sarah
On Thu, Mar 13, 2014 at 10:02 AM, Olivier ETERRADOSSI
olivier.eterrado...@mines-ales.fr wrote:
Hi List,
while running a script on a set of matrices I came into a case I would not
have guessed to arrive.
Below is a small toy example to illustrate the case.
Of course there is a simple workaround (using a simple test), but why does
this occur, and shouldn't it be corrected ?
More probably I miss a point, but which one ? Is this behavior obtained on
purpose and why ?
Sorry if it's a FAQ... I didn't find my way to it.
(And sorry for multiple posting if any : I got a warning from r-bounce but
did not understand it).
Thanks, Olivier
#
# toy example 1 (no problem with this one)
toy.matrix.1-matrix(c(1,0,1,1,0,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.1,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.1,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
# finding the items to remove
unused.rows-which(margin.Rows=thresh)# unused.rows == 2
unused.cols-which(margin.Cols=thresh)# unused.cols == 3
TM1-toy.matrix.1
TM1-TM1[-unused.rows,]
TM1-TM1[,-unused.cols]
TM1
# [,1] [,2]
#[1,]11
#[2,]11 # OK
##
# toy example 2 (oops, no rows to suppress...)
toy.matrix.2-matrix(c(1,1,1,1,1,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.2,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.2,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
unused.rows-which(margin.Rows=thresh) # unused.rows ==
integer(0)
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM2-toy.matrix.2
TM2-TM2[-unused.rows,]
TM2-TM2[,-unused.cols]
TM2
# [,1] [,2]# empty...
###
# I was expecting :
# [,1] [,2]
#[1,]11
#[2,]11
#[3,]11
# which of course is obtained using :
TM2-toy.matrix.2
if(length(unused.rows) !=0) {TM2-TM2[-unused.rows,]}
if(length(unused.rows) !=0 ){TM2-TM2[,-unused.cols]}
TM2
--
Sarah Goslee
http://www.functionaldiversity.org
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and provide commented, minimal, self-contained, reproducible code.
[R] Duplicate of columns when merging two data frames
Dear list users, I have two data frames df1 and df2, where the columns of df1 are Sensor_RM Place_RM Station_RM Y_init_RM M_init_RM D_init_RM Y_fin_RM M_fin_RM D_fin_RM and the columns of df2 are Sensor_RM Station_RM Place_RM Province_RM Region_RM Net_init_RM GaussBoaga_EST_RM GaussBoaga_NORD_RM Gradi_Long_RM Primi_Long_RM Secondi_Long_RM Gradi_Lat_RM Primi_Lat_RM Secondi_Lat_RM Long_Cent_RM Lat_Cent_RM Height_RM When I merge the two data frames through df3 - merge(df1, df2, by=c(Sensor_RM, Station_RM)) I get a new data frame with columns Sensor_RM Station_RM Place_RM.x Y_init_RM M_init_RM D_init_RM Y_fin_RM M_fin_RM D_fin_RM Place_RM.y Province_RM Region_RM Net_init_RM GaussBoaga_EST_RM GaussBoaga_NORD_RM Gradi_Long_RM Primi_Long_RM Secondi_Long_RM Gradi_Lat_RM Primi_Lat_RM Secondi_Lat_RM Long_Cent_RM Lat_Cent_RM Height_RM I am sure that df1$Place_RM and df2$Place_RM are equal. I checked it from the shell using awk and diff. Why then I have a duplicate of Place_RM, i.e. Place_RM.x and Place_RM.y, and only of them? Thank you for your help Stefano AVVISO IMPORTANTE: Questo messaggio di posta elettronica può contenere informazioni confidenziali, pertanto è destinato solo a persone autorizzate alla ricezione. I messaggi di posta elettronica per i client di Regione Marche possono contenere informazioni confidenziali e con privilegi legali. Se non si è il destinatario specificato, non leggere, copiare, inoltrare o archiviare questo messaggio. Se si è ricevuto questo messaggio per errore, inoltrarlo al mittente ed eliminarlo completamente dal sistema del proprio computer. Ai sensi dellâart. 6 della DGR n. 1394/2008 si segnala che, in caso di necessità ed urgenza, la risposta al presente messaggio di posta elettronica può essere visionata da persone estranee al destinatario. IMPORTANT NOTICE: This e-mail message is intended to be received only by persons entitled to receive the confidential information it may contain. E-mail messages to clients of Regione Marche may contain information that is confidential and legally privileged. Please do not read, copy, forward, or store this message unless you are an intended recipient of it. If you have received this message in error, please forward it to the sender and delete it completely from your computer 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] change date format
I would use gsub() to change instances like 5.6.11 to 5/6/11. Hopefully, both versions have month and day in the same order. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 3/13/14 6:57 AM, catalin roibu catalinro...@gmail.com wrote: Dear all! I have a new problem with the date format in a data frame. I have rainfall records extracted from an automatic meteo station. In the same data frame (with 1 records) the date is in 2 formats like that: 4/15/11 11:49 AM 5.6.11 2:51 PM My question is how to modify date format for all records with this format type: 4/15/11 11:49 AM. Thank you! Best regards! Catalin -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[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] Duplicate of columns when merging two data frames
You have duplicates after merging because, although you know they are the same, the R merge() function does not. And the people who wrote that function knew that sometimes two columns with the same name are not identical. Since you know that the columns are identical, drop the column from one of the two data frames before merging. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 3/13/14 8:19 AM, Stefano Sofia stefano.so...@regione.marche.it wrote: Dear list users, I have two data frames df1 and df2, where the columns of df1 are Sensor_RM Place_RM Station_RM Y_init_RM M_init_RM D_init_RM Y_fin_RM M_fin_RM D_fin_RM and the columns of df2 are Sensor_RM Station_RM Place_RM Province_RM Region_RM Net_init_RM GaussBoaga_EST_RM GaussBoaga_NORD_RM Gradi_Long_RM Primi_Long_RM Secondi_Long_RM Gradi_Lat_RM Primi_Lat_RM Secondi_Lat_RM Long_Cent_RM Lat_Cent_RM Height_RM When I merge the two data frames through df3 - merge(df1, df2, by=c(Sensor_RM, Station_RM)) I get a new data frame with columns Sensor_RM Station_RM Place_RM.x Y_init_RM M_init_RM D_init_RM Y_fin_RM M_fin_RM D_fin_RM Place_RM.y Province_RM Region_RM Net_init_RM GaussBoaga_EST_RM GaussBoaga_NORD_RM Gradi_Long_RM Primi_Long_RM Secondi_Long_RM Gradi_Lat_RM Primi_Lat_RM Secondi_Lat_RM Long_Cent_RM Lat_Cent_RM Height_RM I am sure that df1$Place_RM and df2$Place_RM are equal. I checked it from the shell using awk and diff. Why then I have a duplicate of Place_RM, i.e. Place_RM.x and Place_RM.y, and only of them? Thank you for your help Stefano AVVISO IMPORTANTE: Questo messaggio di posta elettronica può contenere informazioni confidenziali, pertanto è destinato solo a persone autorizzate alla ricezione. I messaggi di posta elettronica per i client di Regione Marche possono contenere informazioni confidenziali e con privilegi legali. Se non si è il destinatario specificato, non leggere, copiare, inoltrare o archiviare questo messaggio. Se si è ricevuto questo messaggio per errore, inoltrarlo al mittente ed eliminarlo completamente dal sistema del proprio computer. Ai sensi dell¹art. 6 della DGR n. 1394/2008 si segnala che, in caso di necessità ed urgenza, la risposta al presente messaggio di posta elettronica può essere visionata da persone estranee al destinatario. IMPORTANT NOTICE: This e-mail message is intended to be received only by persons entitled to receive the confidential information it may contain. E-mail messages to clients of Regione Marche may contain information that is confidential and legally privileged. Please do not read, copy, forward, or store this message unless you are an intended recipient of it. If you have received this message in error, please forward it to the sender and delete it completely from your computer 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.
[R] outlier detection in nonlinear regression
Dear useR's, I have several biological data sets from laboratory experiments where a few experiments failed completely due to bacterial contamination. Such cases can easily be seen by eye as outliers. Fitting a Michaelis-Menten kinetics works well with robust nonlinear regression from package robustbase, but this helps me only partly because I need to remove the outliers for a subsequent analysis, so I need a reproducible automatic method. I tried the following two heuristic approaches by either using the robust weights directly (A) or by applying a 3-sigma rule to the residuals (B), see example below. My question: Is there any method available in R that is less heuristic than this? Thank you in advance Anders library(robustbase) ## create data set.seed(375) x - rep(0:10, 3) y - 5 * x / (2 + x) + rnorm(x, sd = 0.2) ## make outliers y[c(3, 5, 10)] - y[c(3, 5, 10)] + c(2, -1, -2) ## robust regression m - nlrob(y ~ a * x /(b + x), data = list(x = x, y = y), start = list(a = 1, b = 1)) summary(m) xnew - list(x = seq(0, 10, 0.1)) plot(x, y) lines(xnew$x, predict(m, newdata = xnew)) ## A) use robust weights to detect outliers ## ?? how to select a critical value? outA - m$rweight 0.5 points(x[outA], y[outA], pch = 16, col = red) ## B) use 3 sigma of residuals to detect outliers eps - residuals(m) outB - abs(eps) (3 * sd(eps)) points(x[outB], y[outB], pch = 3, col = blue) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Duplicate of columns when merging two data frames
On Mar 13, 2014, at 10:19 AM, Stefano Sofia stefano.so...@regione.marche.it wrote: Dear list users, I have two data frames df1 and df2, where the columns of df1 are Sensor_RM Place_RM Station_RM Y_init_RM M_init_RM D_init_RM Y_fin_RM M_fin_RM D_fin_RM and the columns of df2 are Sensor_RM Station_RM Place_RM Province_RM Region_RM Net_init_RM GaussBoaga_EST_RM GaussBoaga_NORD_RM Gradi_Long_RM Primi_Long_RM Secondi_Long_RM Gradi_Lat_RM Primi_Lat_RM Secondi_Lat_RM Long_Cent_RM Lat_Cent_RM Height_RM When I merge the two data frames through df3 - merge(df1, df2, by=c(Sensor_RM, Station_RM)) I get a new data frame with columns Sensor_RM Station_RM Place_RM.x Y_init_RM M_init_RM D_init_RM Y_fin_RM M_fin_RM D_fin_RM Place_RM.y Province_RM Region_RM Net_init_RM GaussBoaga_EST_RM GaussBoaga_NORD_RM Gradi_Long_RM Primi_Long_RM Secondi_Long_RM Gradi_Lat_RM Primi_Lat_RM Secondi_Lat_RM Long_Cent_RM Lat_Cent_RM Height_RM I am sure that df1$Place_RM and df2$Place_RM are equal. I checked it from the shell using awk and diff. Why then I have a duplicate of Place_RM, i.e. Place_RM.x and Place_RM.y, and only of them? Thank you for your help Stefano From the Details section of ?merge: If the columns in the data frames not used in merging have any common names, these have suffixes (.x and .y by default) appended to try to make the names of the result unique. If this is not possible, an error is thrown. If you don't want both columns in the resultant data frame, use them in the 'by' argument or remove one of them prior to merge()ing. If you use them in the 'by' argument, be sure that they will be compared as exactly equal, which can be problematic if they are floating point values. If so, you would be better of subsetting one of the source data frames to remove the column first: df3 - merge(df1, subset(df2, select = -Place_RM), by=c(Sensor_RM, Station_RM)) Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 question about intraindividual variables
I am creating variables representing intraindividual means and standard deviations for longitudinal data with the following code: data$TraitHAPPYmean - with(data, ave(Happy, ID, FUN=function(x) mean(x, na.rm=TRUE) ) ) data$TraitHAPPYsd - with(data, ave(Happy, ID, FUN=function(x) sd(x, na.rm=TRUE) ) ) where 'Happy' is the variable that was collected at multiple timepoint, 'ID' is the grouping variable, and TraitHAPPYMean SD are the new variables. I'd like to also create a variable that just returns the # of cases for each grouping variable (ID). For example, some IDs might have 10 cases of Happy that went into the intraindividual mean, while others have only 5. I realize that this is probably a very basic question, but I just cannot figure it out! Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Estimate LA/AIDS elasticities with Heckman correction in R
Hi, I've been trying to figure out the code to estimate the LA/AIDS model with the inverse mills ratio from the Heckman-2 step to correct for zero consumption in R with no luck. Could someone please direct me on how to do this in R? Thanks, Belinda A. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] change date format
Try: vec1 - c(4/15/11 11:49 AM, 6/12/12 2:30 PM, 5.6.11 2:51 PM) vec2 - gsub([.],/,vec1) A.K. On Thursday, March 13, 2014 9:57 AM, catalin roibu catalinro...@gmail.com wrote: Dear all! I have a new problem with the date format in a data frame. I have rainfall records extracted from an automatic meteo station. In the same data frame (with 1 records) the date is in 2 formats like that: 4/15/11 11:49 AM 5.6.11 2:51 PM My question is how to modify date format for all records with this format type: 4/15/11 11:49 AM. Thank you! Best regards! Catalin -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX: +4 0230 52 16 64 silvic.usv.ro [[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] behaviour of rows and colomns suppression in a matrix
Hi,
You could use:
TM2[!margin.Rows =thresh,!margin.Cols =thresh]
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1
#[3,] 1 1
#For the first case:
TM1[!margin.Rows =thresh,!margin.Cols =thresh]
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1
A.K.
On Thursday, March 13, 2014 10:02 AM, Olivier ETERRADOSSI
olivier.eterrado...@mines-ales.fr wrote:
Hi List,
while running a script on a set of matrices I came into a case I would not
have guessed to arrive.
Below is a small toy example to illustrate the case.
Of course there is a simple workaround (using a simple test), but why does
this occur, and shouldn’t it be corrected ?
More probably I miss a point, but which one ? Is this behavior obtained on
purpose and why ?
Sorry if it’s a FAQ… I didn’t find my way to it.
(And sorry for multiple posting if any : I got a warning from r-bounce but
did not understand it).
Thanks, Olivier
#
# toy example 1 (no problem with this one)
toy.matrix.1-matrix(c(1,0,1,1,0,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.1,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.1,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
# finding the items to remove
unused.rows-which(margin.Rows=thresh) # unused.rows == 2
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM1-toy.matrix.1
TM1-TM1[-unused.rows,]
TM1-TM1[,-unused.cols]
TM1
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1 # OK
##
# toy example 2 (oops, no rows to suppress…)
toy.matrix.2-matrix(c(1,1,1,1,1,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.2,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.2,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
unused.rows-which(margin.Rows=thresh) # unused.rows ==
integer(0)
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM2-toy.matrix.2
TM2-TM2[-unused.rows,]
TM2-TM2[,-unused.cols]
TM2
# [,1] [,2] # empty...
###
# I was expecting :
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1
#[3,] 1 1
# which of course is obtained using :
TM2-toy.matrix.2
if(length(unused.rows) !=0) {TM2-TM2[-unused.rows,]}
if(length(unused.rows) !=0 ){TM2-TM2[,-unused.cols]}
TM2
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Intraindividual means, SDs, and n's
Hi, Try: set.seed(42) dat1 - data.frame(Happy=sample(c(NA,1:25),40,replace=TRUE),ID=rep(1:5,each=8)) with(dat1,ave(Happy,ID,FUN=function(x) sum(!is.na(x #or library(plyr) ddply(dat1,.(ID),mutate,mean=mean(Happy,na.rm=TRUE),SD=sd(Happy,na.rm=TRUE),n=sum(!is.na(Happy))) #or you could use: library(psych) with(dat1,describeBy(Happy,ID,mat=TRUE))[,4:6] # n mean sd #11 8 15.75000 7.611082 #12 8 16.25000 6.296257 #13 8 16.12500 9.508455 #14 8 15.0 7.230886 #15 6 15.16667 6.765107 A.K. I am creating variables representing intraindividual means and standard deviations for longitudinal data with the following code: data$TraitHAPPYmean - with(data, ave(Happy, ID, FUN=function(x) mean(x, na.rm=TRUE) ) ) data$TraitHAPPYsd - with(data, ave(Happy, ID, FUN=function(x) sd(x, na.rm=TRUE) ) ) I'd like to also create a variable that just returns the # of cases for each grouping variable (ID). For example, some IDs might have 10 cases of Happy (the target measure here), while others have only 5. I realize that this is probably a very basic question, but I just cannot figure it out! 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] creating table with sequences of numbers based on the table
Hi, Try: Either tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) indx - rep(1:nrow(tab),tab$Freq) tab1 - transform(tab[indx,],ind=ave(seq_along(indx),indx,FUN=seq_along))[,-2] #or tab2 - transform(tab[indx,],ind=unlist(sapply(tab$Freq,seq)))[,-2] identical(tab1,tab2) #[1] TRUE #or tab3 - transform(tab[indx,], ind= with(tab,seq_len(sum(Freq))-rep(cumsum(c(0L,Freq[-length(Freq)])),Freq)))[,-2] identical(tab1,tab3) #[1] TRUE A.K. I have a problem with transfering one table to another automatically. From table like this: tab pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30 I want to use number of individuals (freq) and then in next table just list them with following numbers (depending on total number of individuals) Like this: in pop ind 1 1 1 2 1 3 1 4 . . . . 1 30 2 1 2 2 2 3 2 4 . . 2 25 3 1 3 2 . . . . How can i do it? I think i have to use loops but so far I failed. Thank you in advance, Best, Malgorzata Gazda __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] behaviour of rows and colomns suppression in a matrix
Thank you Sarah,
But no, I was not expecting this. For me integer(0) is not 0.
That's why I finally tested length(unused.rows), which is 0 when
unused.rows is integer(0).
Olivier
-Message d'origine-
De : Sarah Goslee [mailto:sarah.gos...@gmail.com]
Envoyé : jeudi 13 mars 2014 16:12
À : Olivier ETERRADOSSI
Cc : r-help
Objet : Re: [R] behaviour of rows and colomns suppression in a matrix
Hi,
Your basic problem seems to be that you expect R to take
TM2[0, ]
as meaning not to subset anything, rather than to take only row 0, which
doesn't exist:
R TM2[0,]
[,1] [,2] [,3]
There's a hint in
?[
which says:
An index value of 'NULL' is treated as if it were
'integer(0)'.
Here are two alternative formulations of your task:
# safer method 1
used.rows-which(margin.Rows thresh)
used.cols-which(margin.Cols thresh)
TM2[used.rows, used.cols]
# safer method 2
unused.rows- margin.Rows=thresh
unused.cols- margin.Cols=thresh
TM2[!unused.rows, !unused.cols]
Sarah
On Thu, Mar 13, 2014 at 10:02 AM, Olivier ETERRADOSSI
olivier.eterrado...@mines-ales.fr wrote:
Hi List,
while running a script on a set of matrices I came into a case I would
not have guessed to arrive.
Below is a small toy example to illustrate the case.
Of course there is a simple workaround (using a simple test), but why
does this occur, and shouldn't it be corrected ?
More probably I miss a point, but which one ? Is this behavior
obtained on purpose and why ?
Sorry if it's a FAQ... I didn't find my way to it.
(And sorry for multiple posting if any : I got a warning from r-bounce
but did not understand it).
Thanks, Olivier
#
# toy example 1 (no problem with this one)
toy.matrix.1-matrix(c(1,0,1,1,0,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.1,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.1,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
# finding the items to remove
unused.rows-which(margin.Rows=thresh)# unused.rows == 2
unused.cols-which(margin.Cols=thresh)# unused.cols == 3
TM1-toy.matrix.1
TM1-TM1[-unused.rows,]
TM1-TM1[,-unused.cols]
TM1
# [,1] [,2]
#[1,]11
#[2,]11 # OK
##
# toy example 2 (oops, no rows to suppress...)
toy.matrix.2-matrix(c(1,1,1,1,1,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.2,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.2,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
unused.rows-which(margin.Rows=thresh) # unused.rows ==
integer(0)
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM2-toy.matrix.2
TM2-TM2[-unused.rows,]
TM2-TM2[,-unused.cols]
TM2
# [,1] [,2]# empty...
###
# I was expecting :
# [,1] [,2]
#[1,]11
#[2,]11
#[3,]11
# which of course is obtained using :
TM2-toy.matrix.2
if(length(unused.rows) !=0) {TM2-TM2[-unused.rows,]}
if(length(unused.rows) !=0 ){TM2-TM2[,-unused.cols]}
TM2
--
Sarah Goslee
http://www.functionaldiversity.org
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] behaviour of rows and colomns suppression in a matrix
On Thu, Mar 13, 2014 at 11:51 AM, Olivier ETERRADOSSI olivier.eterrado...@mines-ales.fr wrote: Thank you Sarah, But no, I was not expecting this. For me integer(0) is not 0. That's why I finally tested length(unused.rows), which is 0 when unused.rows is integer(0). Also from ?[ i, j, ...: indices specifying elements to extract or replace. Indices are 'numeric' or 'character' vectors or empty (missing) or 'NULL'. Numeric values are coerced to integer as by 'as.integer' (and hence truncated towards zero). Character vectors will be matched to the 'names' of the object (or for matrices/arrays, the 'dimnames'): see 'Character indices' below for further details. So yes, in this case 0 is treated as integer(0). When you have problems that seem odd to you, carefully reading the relevant help is always a good place to start. Sarah __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 get the adjusted R squared from GLS() estimator
Hello,
Although lm() gives a way to get the adjusted R squared by
adjr2 - summary(mdl)$adj.r.squared
I cannot find a way to extract the adjusted R squared from gls(), any hint?
Thanks,
Rebecca
--
This message, and any attachments, is for the intended r...{{dropped:5}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] behaviour of rows and colomns suppression in a matrix
Thank you, Arun and Sarah
I was not trying to take row 0 that does not exist, rather trying not to
take (I wrote TM[- unused.rows,]) something that does not exist.
So what I understand from Arun's answer is that I was badly using MINUS with
a vector instead of NOT with a logical.
Olivier
-Message d'origine-
De : arun [mailto:smartpink...@yahoo.com]
Envoyé : jeudi 13 mars 2014 16:13
À : r-help@r-project.org
Cc : Olivier ETERRADOSSI
Objet : Re: [R] behaviour of rows and colomns suppression in a matrix
Hi,
You could use:
TM2[!margin.Rows =thresh,!margin.Cols =thresh] # [,1] [,2] #[1,]1
1 #[2,]11 #[3,]11
#For the first case:
TM1[!margin.Rows =thresh,!margin.Cols =thresh] # [,1] [,2] #[1,]1
1 #[2,]11
A.K.
On Thursday, March 13, 2014 10:02 AM, Olivier ETERRADOSSI
olivier.eterrado...@mines-ales.fr wrote:
Hi List,
while running a script on a set of matrices I came into a case I would not
have guessed to arrive.
Below is a small toy example to illustrate the case.
Of course there is a simple workaround (using a simple test), but why does
this occur, and shouldn’t it be corrected ?
More probably I miss a point, but which one ? Is this behavior obtained on
purpose and why ?
Sorry if it’s a FAQ… I didn’t find my way to it.
(And sorry for multiple posting if any : I got a warning from r-bounce but
did not understand it).
Thanks, Olivier
#
# toy example 1 (no problem with this one)
toy.matrix.1-matrix(c(1,0,1,1,0,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.1,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.1,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
# finding the items to remove
unused.rows-which(margin.Rows=thresh)# unused.rows == 2
unused.cols-which(margin.Cols=thresh)# unused.cols == 3
TM1-toy.matrix.1
TM1-TM1[-unused.rows,]
TM1-TM1[,-unused.cols]
TM1
# [,1] [,2]
#[1,]11
#[2,]11 # OK
##
# toy example 2 (oops, no rows to suppress…)
toy.matrix.2-matrix(c(1,1,1,1,1,1,0,0,0),3,3)
# getting the marginal sums
margin.Rows- apply(toy.matrix.2,MARGIN=1,FUN=sum)
margin.Cols- apply(toy.matrix.2,MARGIN=2,FUN=sum)
#giving a threshold for lines and columns suppression
thresh-0
unused.rows-which(margin.Rows=thresh) # unused.rows ==
integer(0)
unused.cols-which(margin.Cols=thresh) # unused.cols == 3
TM2-toy.matrix.2
TM2-TM2[-unused.rows,]
TM2-TM2[,-unused.cols]
TM2
# [,1] [,2]# empty...
###
# I was expecting :
# [,1] [,2]
#[1,]11
#[2,]11
#[3,]11
# which of course is obtained using :
TM2-toy.matrix.2
if(length(unused.rows) !=0) {TM2-TM2[-unused.rows,]}
if(length(unused.rows) !=0 ){TM2-TM2[,-unused.cols]}
TM2
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Re: [R] behaviour of rows and colomns suppression in a matrix
Yes, that's usually what I'm doing... Here I didn't guess that the answer was in [ ! I already noticed that it's sometimes easier to locate something very technical or statistically smart than obvious things (because we think we know, or think we have already read 10 times the basics...) ;-) O. -Message d'origine- De : Sarah Goslee [mailto:sarah.gos...@gmail.com] Envoyé : jeudi 13 mars 2014 16:59 À : Olivier ETERRADOSSI Cc : r-help Objet : Re: [R] behaviour of rows and colomns suppression in a matrix On Thu, Mar 13, 2014 at 11:51 AM, Olivier ETERRADOSSI olivier.eterrado...@mines-ales.fr wrote: Thank you Sarah, But no, I was not expecting this. For me integer(0) is not 0. That's why I finally tested length(unused.rows), which is 0 when unused.rows is integer(0). Also from ?[ i, j, ...: indices specifying elements to extract or replace. Indices are 'numeric' or 'character' vectors or empty (missing) or 'NULL'. Numeric values are coerced to integer as by 'as.integer' (and hence truncated towards zero). Character vectors will be matched to the 'names' of the object (or for matrices/arrays, the 'dimnames'): see 'Character indices' below for further details. So yes, in this case 0 is treated as integer(0). When you have problems that seem odd to you, carefully reading the relevant help is always a good place to start. Sarah __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] SL 6.5 R from source; should I update texlive? if yes, how to point configure to new binary.
Hello all, I would like to install R from source on a SL 6.5 box. I have everything I need except for inconsolata.sty and zi4.sty not found. I am thinking of installing latex 2013 from https://www.tug.org/texlive/. The tex binary will be installed in a different folder. 1. How do I tell ./configure to look to this directory for the tex binary? or 2. Is there something easier that I am overlooking? I appreciate all of the help. --Stephen Sefick ** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama 36849 ** sas0...@auburn.edu http://www.auburn.edu/~sas0025 ** Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman [[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] create dataframe using structure ()
dear all, I have tried to create a dataframe using the structure() function, but it did not really work. In column 1 i have the row number, in column 2 and 3 factors (8 and 4 levels respectively) and in column 4 the actual data. any clue on what went wrong? best regards luigi my.data-structure(list( column_1 = 1:32, column_2 = structure(c(1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4), .Label = c(bd, 2m, 1m, 0.5m), class = factor), column_4 = c(71.62097178, 2.892223468, 144.8618621, 40.90079262, 37.0516856, 32.78206822, 72.44424152, 1512.516293, 3164.748333, 1092.634557, 28733.20269, 1207.87783, 729.6090973, 151.8706088, 241.2466141, 9600.963594, 682.865522, 5869.375226, 554.8924896, 2982.759858, 7690.028092, 40.37298719, 942.3549279, 3403.679967, 103.9862389, 35.28562613, 321.5985377, 0.274424607, 3352.874906, 93.76421187, 21.3709382, 4193.183281), .Names = c(row, stimulation, type, copy), class = data.frame)) attach(my.data) [[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] centroid of diamond
Hello R users,
I am trying to make a baricentric diagram like the ternary plot, but with 4
edges. I want to know how to calculate the centroid of the diamond. The 4
edges are A, B, C, D. If value of A=B=C=D, then the point should be at the
centre of the diamond. If AB and B=C=D=0, Then the point should be at the
corner of A.
For diamond, how to convert the value of A,B,C,D into cartesian
co-ordinates ?. if x1,x2,y1,y2 are A,B,C,D, then someone suggested:
new_point - function(x1, x2, y1, y2, grad=1.73206){
b1 - y1-(grad*x1)
b2 - y2-(-grad*x2)
M - matrix(c(grad, -grad, -1,-1), ncol=2)
intercepts - as.matrix(c(b1,b2))
t_mat - -solve(M) %*% intercepts
data.frame(x=t_mat[1,1], y=t_mat[2,1])
}
But this is not working. Please do suggest some help.
thanks and best regards,
Alaguraj.V
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[R] R bug on survival library Windows 7
Dear All, I have updated my OS - Windows 7, and I found a situation where R stops working with reported error R for Windows GUI front-end has stopped working. My script worked before (on windows XP). The error occured when I perform this: coxph(Surv(time,event)~F11_rs2036914+frailty(n), data=base) from survival library. If I delete the specification frailty(n) this works! But I have clustered data and need this specification. Here is my system information: R version 3.0.2 (2013-09-25) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=French_France.1252 LC_CTYPE=French_France.1252 [3] LC_MONETARY=French_France.1252 LC_NUMERIC=C [5] LC_TIME=French_France.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base Do someone meet this problem? Anyone has some idea how to avoid this crash in Windows 7 system? Note that I can't go back on windows XP because my institution do the choice of windows 7. Sincerely, Christel Castelli, PhD Ingénieur biostatisticien Responsable de la cellule médico-économie Département BESPIM CHU de Nîmes __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] create dataframe using structure ()
I have tried to create a dataframe using the structure() function, Why do you want to do this? but it did not really work. Use str() or dput() to compare this to a data.frame made in the usual way. You have .Names and class being arguments to list() instead of to structure() and you didn't supply any row.names in the call to structure. Also, why call your columns column_n in one place and row, stimulation, ... in another? Is that why you didn't want to call data.frame in the first place, so you could make the column names variable? Bill Dunlap TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Luigi Marongiu Sent: Thursday, March 13, 2014 11:59 AM To: r-help@r-project.org Subject: [R] create dataframe using structure () dear all, I have tried to create a dataframe using the structure() function, but it did not really work. In column 1 i have the row number, in column 2 and 3 factors (8 and 4 levels respectively) and in column 4 the actual data. any clue on what went wrong? best regards luigi my.data-structure(list( column_1 = 1:32, column_2 = structure(c(1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4), .Label = c(bd, 2m, 1m, 0.5m), class = factor), column_4 = c(71.62097178, 2.892223468, 144.8618621, 40.90079262, 37.0516856, 32.78206822, 72.44424152, 1512.516293, 3164.748333, 1092.634557, 28733.20269, 1207.87783, 729.6090973, 151.8706088, 241.2466141, 9600.963594, 682.865522, 5869.375226, 554.8924896, 2982.759858, 7690.028092, 40.37298719, 942.3549279, 3403.679967, 103.9862389, 35.28562613, 321.5985377, 0.274424607, 3352.874906, 93.76421187, 21.3709382, 4193.183281), .Names = c(row, stimulation, type, copy), class = data.frame)) attach(my.data) [[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] create dataframe using structure ()
Hello, You're misplacing one close parenthesis. See the commented lines below. structure(list( column_1 = 1:32, column_2 = structure(c(1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4), .Label = c(bd, 2m, 1m, 0.5m), class = factor), column_4 = c(71.62097178, 2.892223468, 144.8618621, 40.90079262, 37.0516856, 32.78206822, 72.44424152, 1512.516293, 3164.748333, 1092.634557, 28733.20269, 1207.87783, 729.6090973, 151.8706088, 241.2466141, 9600.963594, 682.865522, 5869.375226, 554.8924896, 2982.759858, 7690.028092, 40.37298719, 942.3549279, 3403.679967, 103.9862389, 35.28562613, 321.5985377, 0.274424607, 3352.874906, 93.76421187, 21.3709382, 4193.183281)), # close 2 )) here .Names = c(row, stimulation, type, copy), row.names = c(NA, -32L), class = data.frame) # and only one ) after df Hope this helps, Rui Barradas Em 13-03-2014 18:58, Luigi Marongiu escreveu: dear all, I have tried to create a dataframe using the structure() function, but it did not really work. In column 1 i have the row number, in column 2 and 3 factors (8 and 4 levels respectively) and in column 4 the actual data. any clue on what went wrong? best regards luigi my.data-structure(list( column_1 = 1:32, column_2 = structure(c(1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4), .Label = c(bd, 2m, 1m, 0.5m), class = factor), column_4 = c(71.62097178, 2.892223468, 144.8618621, 40.90079262, 37.0516856, 32.78206822, 72.44424152, 1512.516293, 3164.748333, 1092.634557, 28733.20269, 1207.87783, 729.6090973, 151.8706088, 241.2466141, 9600.963594, 682.865522, 5869.375226, 554.8924896, 2982.759858, 7690.028092, 40.37298719, 942.3549279, 3403.679967, 103.9862389, 35.28562613, 321.5985377, 0.274424607, 3352.874906, 93.76421187, 21.3709382, 4193.183281), .Names = c(row, stimulation, type, copy), class = data.frame)) attach(my.data) [[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 bug on survival library Windows 7
CASTELLI christel christel.CASTELLI at chu-nimes.fr writes: Dear All, I have updated my OS - Windows 7, and I found a situation where R stops working with reported error R for Windows GUI front-end has stopped working. My script worked before (on windows XP). The error occured when I perform this: coxph(Surv(time,event)~F11_rs2036914+frailty(n), data=base) from survival library. If I delete the specification frailty(n) this works! But I have clustered data and need this specification. Here is my system information: R version 3.0.2 (2013-09-25) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=French_France.1252 LC_CTYPE=French_France.1252 [3] LC_MONETARY=French_France.1252 LC_NUMERIC=C [5] LC_TIME=French_France.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base Do someone meet this problem? Anyone has some idea how to avoid this crash in Windows 7 system? Note that I can't go back on windows XP because my institution do the choice of windows 7. Sincerely, Christel Castelli, PhD Ingénieur biostatisticien Responsable de la cellule médico-économie Département BESPIM CHU de Nîmes You may get lucky and discover that someone else has seen a very similar symptom, but it is very most likely that you will have to produce a reproducible example http://tinyurl.com/reproducible-000. That means that (1) you provide data, or simulated data, that (2) reliably produces this crash in a clean session on your machine (for some kinds of bugs the best you can do may be if I run this 1000 times in succession it almost always crashes) (3) can produce this crash on someone else's machine (same version of R, same OS, perhaps the same locale). Googling for this problem suggests it may be a memory exhaustion issue: http://r.789695.n4.nabble.com/ r-for-windows-gui-front-end-has-stopped-working-td459.html (broken URL) How much memory do you have, how big is your data set? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] creating table with sequences of numbers based on the table
Less coding with plyr: tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) # Function to do the work on each row f - function(pop, Freq) data.frame(ind = seq_len(Freq)) library(plyr) u - mdply(tab, f)[, -2] Dennis On Thu, Mar 13, 2014 at 8:01 AM, arun smartpink...@yahoo.com wrote: Hi, Try: Either tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) indx - rep(1:nrow(tab),tab$Freq) tab1 - transform(tab[indx,],ind=ave(seq_along(indx),indx,FUN=seq_along))[,-2] #or tab2 - transform(tab[indx,],ind=unlist(sapply(tab$Freq,seq)))[,-2] identical(tab1,tab2) #[1] TRUE #or tab3 - transform(tab[indx,], ind= with(tab,seq_len(sum(Freq))-rep(cumsum(c(0L,Freq[-length(Freq)])),Freq)))[,-2] identical(tab1,tab3) #[1] TRUE A.K. I have a problem with transfering one table to another automatically. From table like this: tab pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30 I want to use number of individuals (freq) and then in next table just list them with following numbers (depending on total number of individuals) Like this: in popind 1 1 1 2 1 3 1 4 . . . . 1 30 2 1 2 2 2 3 2 4 . . 2 25 3 1 3 2 . . . . How can i do it? I think i have to use loops but so far I failed. Thank you in advance, Best, Malgorzata Gazda __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating table with sequences of numbers based on the table
I think this'll be way simpler and also faster: ans - data.frame(pop = rep.int(tab$pop, tab$Freq), ind=sequence(tab$Freq)) Arun From:Â Dennis Murphy djmu...@gmail.com Reply:Â Dennis Murphy djmu...@gmail.com Date:Â March 13, 2014 at 9:57:20 PM To:Â arun smartpink...@yahoo.com Cc:Â R help r-help@r-project.org Subject:Â Re: [R] creating table with sequences of numbers based on the table Less coding with plyr: tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) # Function to do the work on each row f - function(pop, Freq) data.frame(ind = seq_len(Freq)) library(plyr) u - mdply(tab, f)[, -2] Dennis On Thu, Mar 13, 2014 at 8:01 AM, arun smartpink...@yahoo.com wrote: Hi, Try: Either tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) indx - rep(1:nrow(tab),tab$Freq) tab1 - transform(tab[indx,],ind=ave(seq_along(indx),indx,FUN=seq_along))[,-2] #or tab2 - transform(tab[indx,],ind=unlist(sapply(tab$Freq,seq)))[,-2] identical(tab1,tab2) #[1] TRUE #or tab3 - transform(tab[indx,], ind= with(tab,seq_len(sum(Freq))-rep(cumsum(c(0L,Freq[-length(Freq)])),Freq)))[,-2] identical(tab1,tab3) #[1] TRUE A.K. I have a problem with transfering one table to another automatically. From table like this: tab pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30 I want to use number of individuals (freq) and then in next table just list them with following numbers (depending on total number of individuals) Like this: in pop ind 1 1 1 2 1 3 1 4 . . . . 1 30 2 1 2 2 2 3 2 4 . . 2 25 3 1 3 2 . . . . How can i do it? I think i have to use loops but so far I failed. Thank you in advance, Best, Malgorzata Gazda __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. [[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 get the adjusted R squared from GLS() estimator
Well if I had it and you asked nicely, then I would be happy to give
it to you. Oh, you mean the gls function, not GLS as my initials (my
parents are OLS and WLS, perhaps I was destined to regress), sorry.
The gls function in the nlme package (is that the one that you are
asking about? or is there another gls function?) fits using maximum
likelihood (or restricted maximum likelihood) rather than looking at
sums of squares, so an adjusted r-squared is not a direct result like
in ordinary least squares. The idea of r-squared does not really
translate well to models beyond ordinary least squares (see
fortune(252), fortune(253), and fortune(254)), so adjusted r-squared
would not either.
There are other measures of overall model fit that penalize or adjust
for the number of terms in the model, e.g. AIC and BIC, perhaps one of
those would be better for what you are trying to accomplish.
One possibility would be to square the correlation between the
original y-values and the predicted y-values (y-hats) as an estimate
of r-squared, then apply the same adjustment
(http://en.wikipedia.org/wiki/Adjusted_R-squared#Adjusted_R2), but
there is no guarantee that it has the same effect for the generalized
model (might be an interesting project for a student to look at this
by simulation). I would suggest looking into AIC or BIC instead.
On Thu, Mar 13, 2014 at 9:59 AM, Yuan, Rebecca
rebecca.y...@bankofamerica.com wrote:
Hello,
Although lm() gives a way to get the adjusted R squared by
adjr2 - summary(mdl)$adj.r.squared
I cannot find a way to extract the adjusted R squared from gls(), any hint?
Thanks,
Rebecca
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating table with sequences of numbers based on the table
Or use sequence() and rep(), as in u1 - with(tab, data.frame(pop=rep(pop,Freq), ind=sequence(Freq))) I think that u1 is the same as your u. Bill Dunlap TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dennis Murphy Sent: Thursday, March 13, 2014 1:55 PM To: arun Cc: R help Subject: Re: [R] creating table with sequences of numbers based on the table Less coding with plyr: tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) # Function to do the work on each row f - function(pop, Freq) data.frame(ind = seq_len(Freq)) library(plyr) u - mdply(tab, f)[, -2] Dennis On Thu, Mar 13, 2014 at 8:01 AM, arun smartpink...@yahoo.com wrote: Hi, Try: Either tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) indx - rep(1:nrow(tab),tab$Freq) tab1 - transform(tab[indx,],ind=ave(seq_along(indx),indx,FUN=seq_along))[,-2] #or tab2 - transform(tab[indx,],ind=unlist(sapply(tab$Freq,seq)))[,-2] identical(tab1,tab2) #[1] TRUE #or tab3 - transform(tab[indx,], ind= with(tab,seq_len(sum(Freq))-rep(cumsum(c(0L,Freq[- length(Freq)])),Freq)))[,-2] identical(tab1,tab3) #[1] TRUE A.K. I have a problem with transfering one table to another automatically. From table like this: tab pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30 I want to use number of individuals (freq) and then in next table just list them with following numbers (depending on total number of individuals) Like this: in popind 1 1 1 2 1 3 1 4 . . . . 1 30 2 1 2 2 2 3 2 4 . . 2 25 3 1 3 2 . . . . How can i do it? I think i have to use loops but so far I failed. Thank you in advance, Best, Malgorzata Gazda __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating table with sequences of numbers based on the table
I think we're down to counting the number of characters in each solution! Arun's 3 two-line versus your two-line solution (not counting loading plyr). How about three short lines? pop - rep(1:nrow(tab), tab$Freq) ind - unlist(sapply(tab$Freq, seq_len)) tab2 - data.frame(pop, ind) - David L Carlson Department of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dennis Murphy Sent: Thursday, March 13, 2014 3:55 PM To: arun Cc: R help Subject: Re: [R] creating table with sequences of numbers based on the table Less coding with plyr: tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) # Function to do the work on each row f - function(pop, Freq) data.frame(ind = seq_len(Freq)) library(plyr) u - mdply(tab, f)[, -2] Dennis On Thu, Mar 13, 2014 at 8:01 AM, arun smartpink...@yahoo.com wrote: Hi, Try: Either tab - read.table(text=pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30,sep=,header=TRUE) indx - rep(1:nrow(tab),tab$Freq) tab1 - transform(tab[indx,],ind=ave(seq_along(indx),indx,FUN=seq_along) )[,-2] #or tab2 - transform(tab[indx,],ind=unlist(sapply(tab$Freq,seq)))[,-2] identical(tab1,tab2) #[1] TRUE #or tab3 - transform(tab[indx,], ind= with(tab,seq_len(sum(Freq))-rep(cumsum(c(0L,Freq[-length(Freq)]) ),Freq)))[,-2] identical(tab1,tab3) #[1] TRUE A.K. I have a problem with transfering one table to another automatically. From table like this: tab pop Freq 1 1 30 2 2 25 3 3 30 4 4 30 5 5 30 6 6 30 7 7 30 I want to use number of individuals (freq) and then in next table just list them with following numbers (depending on total number of individuals) Like this: in popind 1 1 1 2 1 3 1 4 . . . . 1 30 2 1 2 2 2 3 2 4 . . 2 25 3 1 3 2 . . . . How can i do it? I think i have to use loops but so far I failed. Thank you in advance, Best, Malgorzata Gazda __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] centroid of diamond
Your question makes absolutely no sense at all. See inline below.
On 14/03/14 08:03, al Vel wrote:
Hello R users,
I am trying to make a baricentric diagram like the ternary plot, but with 4
edges. I want to know how to calculate the centroid of the diamond.
Which centroid? A diamond (convex quadrilateral?) has 3 well defined
centroids, in general all different.
The 4 edges are A, B, C, D.
Are A, B, C, and D the *lengths* of the edges? Or are they just labels
for the edges?
If value of A=B=C=D, then the point should be at the
centre of the diamond. If AB and B=C=D=0,
Since we are apparently talking about numerical values here it would
seem that A, B, C, and D are the lengths of the sides. How can you have
a diamond (convex quadrilateral?) with 3 sides of length 0 and the
other non-zero?
Then the point should be at the corner of A.
What ***on earth*** does the corner of A mean?
For diamond, how to convert the value of A,B,C,D into cartesian
co-ordinates ?. if x1,x2,y1,y2 are A,B,C,D, then someone suggested:
new_point - function(x1, x2, y1, y2, grad=1.73206){
b1 - y1-(grad*x1)
b2 - y2-(-grad*x2)
M - matrix(c(grad, -grad, -1,-1), ncol=2)
intercepts - as.matrix(c(b1,b2))
t_mat - -solve(M) %*% intercepts
data.frame(x=t_mat[1,1], y=t_mat[2,1])
}
But this is not working. Please do suggest some help.
Try using Google. Wikipedia has a good article on quadrilaterals and
outlines a procedure for finding the area centroid (I presume that's
what you actually want) of a convex quadrilateral.
cheers,
Rolf Turner
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Re: [R] create dataframe using structure ()
Hi, Try: my.data - structure(list(column_1 = 1:32, column_2 = structure(c(1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3= structure(c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4), .Label = c(bd, 2m, 1m, 0.5m), class = factor), column_4 = structure(c(71.62097178, 2.892223468, 144.8618621, 40.90079262, 37.0516856, 32.78206822, 72.44424152, 1512.516293, 3164.748333, 1092.634557, 28733.20269, 1207.87783, 729.6090973, 151.8706088, 241.2466141, 9600.963594, 682.865522, 5869.375226, 554.8924896, 2982.759858, 7690.028092, 40.37298719, 942.3549279, 3403.679967, 103.9862389, 35.28562613, 321.5985377, 0.274424607, 3352.874906, 93.76421187, 21.3709382, 4193.183281))), .Names = c(row, stimulation, type, copy), row.names= c(NA,-32L), class = data.frame) str(my.data) 'data.frame': 32 obs. of 4 variables: $ row : int 1 2 3 4 5 6 7 8 9 10 ... $ stimulation: Factor w/ 8 levels Unstimulated,..: 1 2 3 4 5 6 7 8 1 2 ... $ type : Factor w/ 4 levels bd,2m,1m,..: 1 1 1 1 1 1 1 1 2 2 ... $ copy : num 71.62 2.89 144.86 40.9 37.05 ... A.K. dear all, I have tried to create a dataframe using the structure() function, but it did not really work. In column 1 i have the row number, in column 2 and 3 factors (8 and 4 levels respectively) and in column 4 the actual data. any clue on what went wrong? best regards luigi my.data-structure(list( column_1 = 1:32, column_2 = structure(c(1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8), .Label = c(Unstimulated, ESAT6, CFP10, Rv3615c, Rv2654, Rv3879, Rv3873, PHA), class = factor), column_3 = structure(c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4), .Label = c(bd, 2m, 1m, 0.5m), class = factor), column_4 = c(71.62097178, 2.892223468, 144.8618621, 40.90079262, 37.0516856, 32.78206822, 72.44424152, 1512.516293, 3164.748333, 1092.634557, 28733.20269, 1207.87783, 729.6090973, 151.8706088, 241.2466141, 9600.963594, 682.865522, 5869.375226, 554.8924896, 2982.759858, 7690.028092, 40.37298719, 942.3549279, 3403.679967, 103.9862389, 35.28562613, 321.5985377, 0.274424607, 3352.874906, 93.76421187, 21.3709382, 4193.183281), .Names = c(row, stimulation, type, copy), class = data.frame)) attach(my.data) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] boxplot with x-axis time
Dear R-users, I want to plot boxplots of a single variable collected a few times during almost one year and I would like the x-axis to recognize the date-class of the variable. I found some topics in the archive but: - some questions were poorly posed (http://r.789695.n4.nabble.com/Boxplot-with-dates-td896401.html) and so with no answer; - others, only address the topic of have the date on the x-axes ordered (http://r.789695.n4.nabble.com/Using-boxplot-in-a-daily-time-series-td843060.html#a843061); - finally, another topic (http://r.789695.n4.nabble.com/Boxplot-position-on-X-axis-relative-to-it-s-value-td2196020.html)is quite close to what I need but it is not exactly the same. I attached a simulated data.frame: it 2-column, the1-st are date and the 2-nd is the variable. Here the code I use: pippo- read.csv(pippo.csv) pippo$date- as.Date(pippo$date, format=%Y-%m-%d) boxplot(V ~ date, data=pippo) I would like that the x-axis look like in this plot plot(V ~ date, data=pippo) where the thick reproduce the right time of the year . Thank you for your help__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Creating a box plot with whiskers when I only have mean, sd and n
Dear all, I want to create a boxplot with whiskers. I want to compare several studies. For each study I have 1. mean 2. standard deviation (sd) 3. name 4. number of observations (n) How can i do this in R ? normally I would type bxp or boxplot but this is not allowed when I only know mean and sd -- Frederik Borup MD, Ph.D.-stud. Accident Analysis Group (UAG) The Department of Orthopaedic Surgery, Odense University Hospital (OUH) The Department of Orthopaedic Surgery, SLB Kolding University of Southern Denmark (SDU) [[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] Negative binomial models and censored observations
Hi, I am working with hurdle models in the pscl package to model zero inflated overdispersed count data and want to incorporate censored observations into the equation. 33% of the observed positive count data is right censored, i.e. subject lost to follow up during the duration of the study. Can this be accounted for in the hurdle() function? Thanks, Tim [[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] Creating a box plot with whiskers when I only have mean, sd and n
On 14/03/14 08:52, Frederik Borup wrote: Dear all, I want to create a boxplot with whiskers. I want to compare several studies. For each study I have 1. mean 2. standard deviation (sd) 3. name 4. number of observations (n) How can i do this in R ? normally I would type bxp or boxplot but this is not allowed when I only know mean and sd. (1) Read the help for bxp. (2) Think about what you want your box and whiskers to *represent* given that you only know the summary statistics. (Is it *really* meaningful to create a boxplot in such circumstances?) (3) If you can figure that out, then create an appropriate argument (a list) for bxp() and call bxp() with that argument. cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] GLM vs GAM
Dear R-list, I am wondering whether anyone could explain what'd be the difference between running a 'generalized additive regression' versus 'generalized linear regression' with splines. Are they same models theoretically? My apologies if this is a silly question. Any comments or direction to references will be highly appreciated. Thanks in advance, Ehsan # set.seed(545) require(mgcv) n - 200 x1 - c(rnorm(n), 1+rnorm(n)) x2 - sqrt(c(rnorm(n,4),rnorm(n,6))) y - c(rep(0,n), rep(1,n)) # # GAM version # r1 - gam(y~s(x1, bs = cr)+s(x2, bs = cr),family=binomial) pr1 - predict(r1, type='response') summary(pr1) hist(pr1) # # GLM version # r2 - glm(y~ns(x1)+ns(x2),family=binomial) pr2 - predict(r2, type='response') summary(pr2) hist(pr2) # # Results # summary(pr1) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.394 0.0550200 0.5027000 0.500 0.9322000 1.000 summary(pr2) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.403 0.0573300 0.5229000 0.500 0.9159000 0.9992000 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Colorspace: bug in HLS conversion?
Achim Zeileis Achim.Zeileis at uibk.ac.at writes: Maybe you are using an old version of colorspace? For version 1.1-1 (from two years ago, version 1.2-4 is current) the NEWS file says: ... That was it--thanks! As a developer, I should know enough to check the release notes... --Mike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] centroid of diamond
On 03/14/2014 06:03 AM, al Vel wrote:
Hello R users,
I am trying to make a baricentric diagram like the ternary plot, but with 4
edges. I want to know how to calculate the centroid of the diamond. The 4
edges are A, B, C, D. If value of A=B=C=D, then the point should be at the
centre of the diamond. If AB and B=C=D=0, Then the point should be at the
corner of A.
For diamond, how to convert the value of A,B,C,D into cartesian
co-ordinates ?. if x1,x2,y1,y2 are A,B,C,D, then someone suggested:
new_point- function(x1, x2, y1, y2, grad=1.73206){
b1- y1-(grad*x1)
b2- y2-(-grad*x2)
M- matrix(c(grad, -grad, -1,-1), ncol=2)
intercepts- as.matrix(c(b1,b2))
t_mat- -solve(M) %*% intercepts
data.frame(x=t_mat[1,1], y=t_mat[2,1])
}
But this is not working. Please do suggest some help.
thanks and best regards,
Alaguraj.V
Hi Alaguraj,
A lot depends upon what you mean by diamond. A rhombus is out because
you say that the lengths of the sides can be different (and as you note,
the answer is easy). If you mean a parallelogram (opposite sides are
equal) the answer is also easy, the intersection of the lines joining
opposite vertices. If you mean a kite (sides of equal length are
adjacent) it is a matter of finding the point along the line joining the
two vertices that join the two sets of equal sides. I suspect that what
you have to calculate is the centroid of a quadrilateral with arbitrary
length sides. We can probably assume that it is convex, as Rolf noted,
as the centroid may be outside quadrilaterals that are very concave
(think boomerang, sport). So if you could tell us a bit more about what
constraints you wish to place on your quadrilateral, somebody may be
able to help.
Jim
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Re: [R] Creating a box plot with whiskers when I only have mean, sd and n
On 03/14/2014 06:52 AM, Frederik Borup wrote: Dear all, I want to create a boxplot with whiskers. I want to compare several studies. For each study I have 1. mean 2. standard deviation (sd) 3. name 4. number of observations (n) How can i do this in R ? normally I would type bxp or boxplot but this is not allowed when I only know mean and sd Hi Frederik, You may want box.heresy in the plotrix package. 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.
