Re: [R] ROCR package not installing
try: install.packages(ROCR,dep=T) M.O On to., 2012-11-22 at 16:02 +1000, Philip Robinson wrote: I have tried installing the package (ROCR) with this command: Install.packages(ROCR) And with this command on the command line R CMD INSTALL ROCR_1.0-4.tar.gz But both times I get exactly the same error shown below, I don't understand what is wrong, is this an error in the package code? Thank you Philip probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ R CMD INSTALL ROCR_1.0-4.tar.gz * installing to library '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10' * installing *source* package 'ROCR' ... ** R ** data ** demo ** preparing package for lazy loading Loading required package: gtools Loading required package: gdata gdata: read.xls support for 'XLS' (Excel 97-2004) files ENABLED. gdata: read.xls support for 'XLSX' (Excel 2007+) files ENABLED. Attaching package: 'gdata' The following object(s) are masked from package:utils : object.size Loading required package: caTools Loading required package: bitops Loading required package: grid Loading required package: KernSmooth KernSmooth 2.23 loaded Copyright M. P. Wand 1997-2009 Attaching package: 'gplots' The following object(s) are masked from package:stats : lowess Error in setMethod(plot, signature(x = performance, y = missing), : no existing definition for function plot Error : unable to load R code in package 'ROCR' ERROR: lazy loading failed for package 'ROCR' * removing '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10/ROCR' probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ [[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] cluster analysis in R
It's hard to answer these questions without knowing what the errors are and how they can be reproduced. Best, Ingmar On Thu, Nov 22, 2012 at 1:03 AM, KitKat katherinewri...@trentu.ca wrote: Thanks, I have been trying that site and another one (http://www.statmethods.net/advstats/cluster.html) I don't know if I should be doing mclust or mcclust, but either way, the codes are not working. I am following the guidelines online at: mcclust - http://cran.r-project.org/web/packages/mcclust/mcclust.pdf mclust - http://cran.r-project.org/ I am relatively new to R, but so far I have been able to figure out dfa, manova, pca... I cannot get these codes to work, I keep getting various errors. Are there other resources that have details about what codes to use or what to do when errors result? I have not found anything else helpful Thank you -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-in-R-tp4649635p4650397.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. [[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] ROCR package not installing
My version is: Linux bioinform08 2.6.32-33-generic #70-Ubuntu SMP Thu Jul 7 21:13:52 UTC 2011 x86_64 GNU/Linux Ubuntu 10.04.3 LTS R version 2.10.1 (2009-12-14) Philip -Original Message- From: Pascal Oettli [mailto:kri...@ymail.com] Sent: Thursday, 22 November 2012 4:20 PM To: Philip Robinson Cc: r-help@r-project.org Subject: Re: [R] ROCR package not installing Hello, What is the version of R you use ? Regards, Pascal Le 22/11/2012 15:02, Philip Robinson a écrit : I have tried installing the package (ROCR) with this command: Install.packages(ROCR) And with this command on the command line R CMD INSTALL ROCR_1.0-4.tar.gz But both times I get exactly the same error shown below, I don't understand what is wrong, is this an error in the package code? Thank you Philip probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ R CMD INSTALL ROCR_1.0-4.tar.gz * installing to library '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10' * installing *source* package 'ROCR' ... ** R ** data ** demo ** preparing package for lazy loading Loading required package: gtools Loading required package: gdata gdata: read.xls support for 'XLS' (Excel 97-2004) files ENABLED. gdata: read.xls support for 'XLSX' (Excel 2007+) files ENABLED. Attaching package: 'gdata' The following object(s) are masked from package:utils : object.size Loading required package: caTools Loading required package: bitops Loading required package: grid Loading required package: KernSmooth KernSmooth 2.23 loaded Copyright M. P. Wand 1997-2009 Attaching package: 'gplots' The following object(s) are masked from package:stats : lowess Error in setMethod(plot, signature(x = performance, y = missing), : no existing definition for function plot Error : unable to load R code in package 'ROCR' ERROR: lazy loading failed for package 'ROCR' * removing '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10/ROCR' probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ [[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] ROCR package not installing
Hi Mohammed, This didn't work either: Error in setMethod(plot, signature(x = performance, y = missing), : no existing definition for function plot Error : unable to load R code in package 'ROCR' ERROR: lazy loading failed for package ‘ROCR’ * removing ‘/home/probinson/R/x86_64-pc-linux-gnu-library/2.10/ROCR’ The downloaded packages are in ‘/tmp/Rtmpn1DjHt/downloaded_packages’ Warning message: In install.packages(ROCR, dep = T) : installation of package 'ROCR' had non-zero exit status thanks Philip -Original Message- From: Mohammed Ouassou [mailto:mohammed.ouas...@statkart.no] Sent: Thursday, 22 November 2012 6:02 PM To: Philip Robinson Cc: r-help@r-project.org Subject: Re: [R] ROCR package not installing try: install.packages(ROCR,dep=T) M.O On to., 2012-11-22 at 16:02 +1000, Philip Robinson wrote: I have tried installing the package (ROCR) with this command: Install.packages(ROCR) And with this command on the command line R CMD INSTALL ROCR_1.0-4.tar.gz But both times I get exactly the same error shown below, I don't understand what is wrong, is this an error in the package code? Thank you Philip probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ R CMD INSTALL ROCR_1.0-4.tar.gz * installing to library '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10' * installing *source* package 'ROCR' ... ** R ** data ** demo ** preparing package for lazy loading Loading required package: gtools Loading required package: gdata gdata: read.xls support for 'XLS' (Excel 97-2004) files ENABLED. gdata: read.xls support for 'XLSX' (Excel 2007+) files ENABLED. Attaching package: 'gdata' The following object(s) are masked from package:utils : object.size Loading required package: caTools Loading required package: bitops Loading required package: grid Loading required package: KernSmooth KernSmooth 2.23 loaded Copyright M. P. Wand 1997-2009 Attaching package: 'gplots' The following object(s) are masked from package:stats : lowess Error in setMethod(plot, signature(x = performance, y = missing), : no existing definition for function plot Error : unable to load R code in package 'ROCR' ERROR: lazy loading failed for package 'ROCR' * removing '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10/ROCR' probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ [[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] ROCR package not installing
Very old version. Maybe you should try again after upgrading your version of R (current is 2.15.2). Pascal Le 22/11/2012 18:11, Philip Robinson a écrit : My version is: Linux bioinform08 2.6.32-33-generic #70-Ubuntu SMP Thu Jul 7 21:13:52 UTC 2011 x86_64 GNU/Linux Ubuntu 10.04.3 LTS R version 2.10.1 (2009-12-14) Philip -Original Message- From: Pascal Oettli [mailto:kri...@ymail.com] Sent: Thursday, 22 November 2012 4:20 PM To: Philip Robinson Cc: r-help@r-project.org Subject: Re: [R] ROCR package not installing Hello, What is the version of R you use ? Regards, Pascal Le 22/11/2012 15:02, Philip Robinson a écrit : I have tried installing the package (ROCR) with this command: Install.packages(ROCR) And with this command on the command line R CMD INSTALL ROCR_1.0-4.tar.gz But both times I get exactly the same error shown below, I don't understand what is wrong, is this an error in the package code? Thank you Philip probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ R CMD INSTALL ROCR_1.0-4.tar.gz * installing to library '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10' * installing *source* package 'ROCR' ... ** R ** data ** demo ** preparing package for lazy loading Loading required package: gtools Loading required package: gdata gdata: read.xls support for 'XLS' (Excel 97-2004) files ENABLED. gdata: read.xls support for 'XLSX' (Excel 2007+) files ENABLED. Attaching package: 'gdata' The following object(s) are masked from package:utils : object.size Loading required package: caTools Loading required package: bitops Loading required package: grid Loading required package: KernSmooth KernSmooth 2.23 loaded Copyright M. P. Wand 1997-2009 Attaching package: 'gplots' The following object(s) are masked from package:stats : lowess Error in setMethod(plot, signature(x = performance, y = missing), : no existing definition for function plot Error : unable to load R code in package 'ROCR' ERROR: lazy loading failed for package 'ROCR' * removing '/home/probinson/R/x86_64-pc-linux-gnu-library/2.10/ROCR' probinson@bioinform08:/tmp/RtmpO0rFbx/downloaded_packages$ [[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] error message in Linkcomm package
Dear All, Some of the functions in Linkcomm return an error: unused argument(s) (v = V(x$igraph)) Although I follow the guidance in the manual a number of the functions return this error. The igraph vector is an edgelist. Many thanks for any guidance. Best, Nick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] prediction problem
Hello, I am using the mda package and in particular the fda routine to classify/predict in terms of color to a set of 20 samples for which i don´t know the color. I preformed a flexible discriminant analysis (FDA) using a set of 147 samples for which i know all the information. My script and data follow in attachment. A total of 23 predictors were considered. 20 of the predictors are numeric and 3 are discrete/categorical. The resulting FDA rule was applied to the matrix and for the same predictors in order to predict the color. However a consistent error is occuring: Error in mindist[l] - ndist[l] : **NAs are not allowed in subscripted assignments It is possible the problem being related with the 2 predictors that are discrete/categorical variables? There is already available some rotine to perform a discriminant analysis considering continuous and discrete/categorical variables? If someone can help, would be very grateful. cheers, Catarina n Codecolor GROUP_N Licenca_N type1_W type2_W type3_W type4_W type5_W type6_W type7_W type8_W type9_W type1_V type2_V type3_V type4_V type5_V type7_V type8_V type9_V type10_Vtype11_Vtotal 1 PRT0026940473 verde A8 B7 0,558416119 0 0 0 0,152231706 0,243598713 0 0 0,005176997 0,483991231 0 0 0 0,375554726 0,092127884 0 0 0 0,011391047 714,7 2 PRT0026940480 verde A8 B7 0,431011386 0 0 0 0,090087073 0,389819156 0 0 0 0,27662459 0 0 0 0,329209955 0,254047865 0 0 0 0 597,2 3 PRT0026940500 verde A8 B7 0,349034749 0 0,013127413 0 0,362162162 0,197297297 0 0 0 0,166132511 0 0,005179074 0 0,647152728 0,095422917 0 0 0 0 259 4 PRT0026940507 verde A8 B7 0,38562904 0 0 0 0,235952263 0,245400298 0 0 0 0,195929366 0 0 0 0,561989935 0,149391253 0 0 0 0 402,2 6 PRT0042540249 verde A9 B7 0,160142349 0 0 0,0113879 0 0,098932384 0 0 0 0,071878686 0 0 0,017455771 0 0,127177759 0 0 0 0 140,5 7 PRT0042540358 verde A9 B7 0,258287293 0 0 0 0,04558011 0,165745856 0 0 0 0,308268994 0 0 0 0,147252993 0,073850285 0 0 0 0 72,4 9 PRT0042540498 verde A9 B7 0,535232384 0 0 0 0,230384808 0,143428286 0 0 0 0,220279346 0 0 0 0,50482587 0,113254862 0,089373366 0 0 0 200,1 10 PRT0042540507 verde A9 B7 0,631288766 0 0 0,029634735 0,188146106 0 0 0 0 0,187408732 0 0 0,033825063 0,637267173 0 0 0 0 0 145,1 11 PRT0042540527 verde A9 B7 0,775214835 0 0 0,014925373 0,03256445 0,054726368 0 0 0 0,610476668 0 0 0,031623122 0,225502758 0,110096002 0 0 0 0 221,1 12 PRT0112140260 amarelo A7 B5 0 0 0,009701493 0,664925373 0 0,017164179 0,042537313 0,023880597 0 0 0 0,011007463 0,617276119 0 0,004626866 0 0,120279851 0,012649254 0 134 13 PRT0112140305 azulA7 B5 0 0 0 0 0 0 0,095049505 0 0 0 0 0 0 0 0 0,080972426 0,567469139 0 0 101 14 PRT0112140366 azulA7 B5 0 0 0,014857143 0,102857143 0,040,014857143 0,011428571 0,217142857 0 0 0 0,005893035 0,125398294 0,16692363 0,013019495 0 0,024394422 0,152328091 0 87,5 15 PRT0112140373 azulA7 B5 0 0 0,019488429 0,052375152 0 0,017052375 0,052375152 0 0 0 0 0,020974032 0,041306002 0 0,007062684 0 0,2152573 0 0 82,1 16 PRT0112140382 azulA7 B5 0 0,016591252 0,009049774 0 0 0 0 0 0 0 0,047221686 0,009984565 0 0 0 0 0
Re: [R] Stepwise analysis with fixed variables
Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Stepwise-analysis-with-fixed-variables-tp4650015p4650419.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] problems with RPostgreSQL compilation
Hello, I am trying to install RPostgreSQL, but without success at the moment, getting an errormessage - see below. My OS is Fedora 17, R version 2.15.2, PostgreSQL 9.2.1 with Postgis 2.0.1 spatial extension. Thank You for any help, advice or hint in advance. Regards Tomas install.packages(RPostgreSQL, dependencies=TRUE) trying URL 'http://cran.fyxm.net/src/contrib/RPostgreSQL_0.3-3.tar.gz' Content type 'application/x-gzip' length 474570 bytes (463 Kb) opened URL == downloaded 463 Kb * installing *source* package ‘RPostgreSQL’ ... ** package ‘RPostgreSQL’ successfully unpacked and MD5 sums checked checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed checking build system type... x86_64-unknown-linux-gnu checking host system type... x86_64-unknown-linux-gnu checking target system type... x86_64-unknown-linux-gnu checking for pg_config... /usr/bin/pg_config checking for /usr/include/libpq-fe.h... yes configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-DBI.c -o RS-DBI.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-PQescape.c -o RS-PQescape.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-PostgreSQL.c -o RS-PostgreSQL.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-pgsql-copy.c -o RS-pgsql-copy.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-pgsql-getResult.c -o RS-pgsql-getResult.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-pgsql-pqexec.c -o RS-pgsql-pqexec.o gcc -std=gnu99 -shared -L/usr/local/lib64 -o RPostgreSQL.so RS-DBI.o RS-PQescape.o RS-PostgreSQL.o RS-pgsql-copy.o RS-pgsql-getResult.o RS-pgsql-pqexec.o -L/usr/lib -lpq /usr/bin/ld: cannot find -lpq collect2: error: ld returned 1 exit status make: *** [RPostgreSQL.so] Error 1 ERROR: compilation failed for package ‘RPostgreSQL’ * removing ‘/usr/local/lib64/R/library/RPostgreSQL’ The downloaded source packages are in ‘/tmp/Rtmpv5LkJj/downloaded_packages’ Updating HTML index of packages in '.Library' Making packages.html ... done Warning message: In install.packages(RPostgreSQL, dependencies = TRUE) : installation of package ‘RPostgreSQL’ had non-zero exit status -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] [lattice] Increase distance between tick labels and ticks in wireframe plot (pad)
Hello, I try to increase the distance between tick labels and ticks in a lattice wireframe plot. Here's a minimal example: ## Minimal example x - y - z - c(1,2,3) df - data.frame(x, y, z) wireframe(z ~ x*y, df, scales = list(arrows = FALSE, col = black, font = 1, tck=0.6)) I tried the axis.components option (http://r.789695.n4.nabble.com/Lattice-distance-of-tick-labels-from-axis-line-tp3693014p3693014.html). This works for xyplot, but *not* for wireframe: xyplot(z ~ x, df, scales = list(arrows = FALSE, col = black, font = 1, tck=0.6), par.settings=list(axis.components=list(left=list(pad1=2 wireframe(z ~ x*y, df, scales = list(arrows = FALSE, col = black, font = 1, tck=0.6), par.settings=list(axis.components=list(left=list(pad1=2 Any ideas on how to pad the axis labels in wireframe plots? Thanks, Felix __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 jpeg image with russian font in R?
Hi, I m working on R and read one image with russian font using readjpeg() function and write it on pdf (using grid). But russian text is not visible on pdf. How can i fix my pblm? Regards http://r.789695.n4.nabble.com/file/n4650417/pointer.jpg library(JPEG) library(grid) pdf(out.pdf , width = 6.6 ,height = 4.2,family= URWHelvetica, encoding=KOI8-R) a-readJPEG(ArrowImage)# name of arrow grid.newpage() pushViewport(viewport(width=0.9, height=0.9)) pushViewport(viewport(yscale=c(0,0), xscale=c(0,1), x=0, y=0, width=0.15, height=1.0, just=c(left, bottom))) grid.raster(a, y=unit(ypos.img, native)) popViewport() popViewport() dev.off() Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-read-jpeg-image-with-russian-font-in-R-tp4650417.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] Extracting random rows from a dataset
I found this code through internet and I have a one doubt about this example, which is Here you take a 70% random sample considering all the values but what I want is if the selected sample contains p1,p2 then I want all records of p1,p2 and also remaining data in some other data frame...like following output D1 V1 v2 P1 10 P1 3 P1 4 P1 2 P2 30 P2 40 D2 V1 v2 P3 4 P3 1 I hope you understood my doubt.please help me Thanks, Madhu. read.table(textConnection(gsub(\\(|\\), , var) )) #from prior posting V1 V2 1 p1 10 2 p1 3 3 p1 4 4 p2 20 5 p2 30 6 p2 40 7 p3 4 8 p3 1 9 p1 2 ridxs - sample(1:nrow(df),floor(0.7*nrow(df)) ) # the 70% sample row IDs df[ridxs,] V1 V2 5 p2 30 6 p2 40 2 p1 3 7 p3 4 4 p2 20 8 p3 1 df[-ridxs,] V1 V2 1 p1 10 3 p1 4 9 p1 2 [[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] problems with RPostgreSQL compilation
Hello, Do you have the shared libraries required by PostgreSQL installed on your machine (libpq)? Regards, Pascal Le 22/11/2012 18:45, lanc...@t-zones.sk a écrit : Hello, I am trying to install RPostgreSQL, but without success at the moment, getting an errormessage - see below. My OS is Fedora 17, R version 2.15.2, PostgreSQL 9.2.1 with Postgis 2.0.1 spatial extension. Thank You for any help, advice or hint in advance. Regards Tomas install.packages(RPostgreSQL, dependencies=TRUE) trying URL 'http://cran.fyxm.net/src/contrib/RPostgreSQL_0.3-3.tar.gz' Content type 'application/x-gzip' length 474570 bytes (463 Kb) opened URL == downloaded 463 Kb * installing *source* package ‘RPostgreSQL’ ... ** package ‘RPostgreSQL’ successfully unpacked and MD5 sums checked checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed checking build system type... x86_64-unknown-linux-gnu checking host system type... x86_64-unknown-linux-gnu checking target system type... x86_64-unknown-linux-gnu checking for pg_config... /usr/bin/pg_config checking for /usr/include/libpq-fe.h... yes configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-DBI.c -o RS-DBI.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-PQescape.c -o RS-PQescape.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-PostgreSQL.c -o RS-PostgreSQL.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-pgsql-copy.c -o RS-pgsql-copy.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-pgsql-getResult.c -o RS-pgsql-getResult.o gcc -std=gnu99 -I/usr/local/lib64/R/include -DNDEBUG -I/usr/include -I/usr/local/include-fpic -g -O2 -c RS-pgsql-pqexec.c -o RS-pgsql-pqexec.o gcc -std=gnu99 -shared -L/usr/local/lib64 -o RPostgreSQL.so RS-DBI.o RS-PQescape.o RS-PostgreSQL.o RS-pgsql-copy.o RS-pgsql-getResult.o RS-pgsql-pqexec.o -L/usr/lib -lpq /usr/bin/ld: cannot find -lpq collect2: error: ld returned 1 exit status make: *** [RPostgreSQL.so] Error 1 ERROR: compilation failed for package ‘RPostgreSQL’ * removing ‘/usr/local/lib64/R/library/RPostgreSQL’ The downloaded source packages are in ‘/tmp/Rtmpv5LkJj/downloaded_packages’ Updating HTML index of packages in '.Library' Making packages.html ... done Warning message: In install.packages(RPostgreSQL, dependencies = TRUE) : installation of package ‘RPostgreSQL’ had non-zero exit status -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] histogram help
On 11/22/2012 05:15 AM, Rosario Scandurra wrote: Hi, I want to generate an histogram and plot on the y axis the percentage of a categorical variable and on the x axis a nominal variable. I want to move the origin to have 2 categories below 0. Hope somebody could help me. Thanks. Hi Rosario, If you already have your percentages, the barplot function will display them like most people expect an histogram to look by setting space=0. See the names.arg argument to have the names of your nominal variables on the x axis. I'm not sure exactly what you mean by wanting 2 categories below 0. If these are negative percentages, the barplot function will automatically handle them. 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.
Re: [R] Help with loop
Hello, I'm glad it helped. You should have kept this in the list, the odds of getting more answers are bigger. As for your problem with NAs, what do you want to do with them? Should they count as within range? Rui Barradas Em 22-11-2012 10:18, Helene Frigstad escreveu: Hi, yes, that is a very elegant solution. Thank you very much! Do you have any suggestions on how to deal with NA in such a situation? For example, if I have one missing value in the dataset, then the function will only work until it reaches the NA, and then the rest will be NA as well. head (D, 11)Salt time310 35.63511 2004-07-17311 35.62334 2004-07-18312 35.63498 2004-07-19313 35.64032 2004-07-20314 NA 2004-07-21315 35.66930 2004-07-22316 35.65394 2004-07-23317 35.64702 2004-07-24318 35.63810 2004-07-25319 35.63190 2004-07-26320 35.66033 2004-07-27 (change - cumsum(c(FALSE, abs(diff(D$Salt)) 0.05))) [1] 0 0 0 0 NA NA NA NA NA NA NA (split(D, change))$`0`Salt time310 35.63511 2004-07-17311 35.62334 2004-07-18312 35.63498 2004-07-19313 35.64032 2004-07-20 Many thanks, Helene __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] printing difftime summary
On Thu, Nov 22, 2012 at 4:01 AM, Sam Steingold s...@gnu.org wrote:
Hi,
* arun fznegcvax...@lnubb.pbz [2012-11-21 14:04:36 -0800]:
Are you looking for some other function (difftime2string)
ot just remove the quotes from the printed output?
I am wondering what others do when they want to see a summary of difftime.
If it is the latter, then this should do it.
res-do.call(data.frame,lapply(s,difftime2string))
names(res)-names(s)
res
# Min. 1st Qu.Median Mean 3rd Qu. Max.
#1 500.00 ms 17.12 min 99.48 min 8.30 hrs 8.05 hrs 6.98 days
cool, thanks.
I now think that what I want is
--8---cut here---start-8---
difftime.summary - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
r
}
Any reason not summary.difftime to get S3 dispatch?
MW
difftime.summary(infl$delay)
string secs
Min.500.00 ms 0.5
1st Qu. 17.12 min 1027.0
Median 99.48 min 5969.0
Mean 8.30 hrs 29870.0
3rd Qu. 8.05 hrs 28970.0
Max.6.98 days 603100.0
--8---cut here---end---8---
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X
11.0.11103000
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[R] Trick to replace NA
Dear members,
I have a series of values in a vector and some value are missing and
replaced with NA.
For example:
a - c(27, 25, NA, NA, 24, 26, 27, NA, 26)
I would like to replace the NAs with the value taken from the previous
value that is non-NA. The output would be in this case:
27 25 25 25 24 26 27 27 26
Now I do that with a for loop, but I try to eliminate all the loops to
gain in performance
I try this and it works but again I have a while and then I do not
eliminate completely loop:
l - length(a)
while(any(is.na(a))) {a[2:l] - ifelse(is.na(a[2:l]), a[1:(l-1)], a[2:l])}
If someone have another solution, I will be most happy !
Sincerely
Marc Girondot
--
__
Marc Girondot, Pr
Laboratoire Ecologie, Systématique et Evolution
Equipe de Conservation des Populations et des Communautés
CNRS, AgroParisTech et Université Paris-Sud 11 , UMR 8079
Bâtiment 362
91405 Orsay Cedex, France
Tel: 33 1 (0)1.69.15.72.30 Fax: 33 1 (0)1.69.15.73.53
e-mail: marc.giron...@u-psud.fr
Web: http://www.ese.u-psud.fr/epc/conservation/Marc.html
Skype: girondot
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Re: [R] Trick to replace NA
Take a look at
na.locf()
in the zoo package. (LOCF = last observation carried forward)
RMW
On Thu, Nov 22, 2012 at 12:56 PM, Marc Girondot marc_...@yahoo.fr wrote:
Dear members,
I have a series of values in a vector and some value are missing and
replaced with NA.
For example:
a - c(27, 25, NA, NA, 24, 26, 27, NA, 26)
I would like to replace the NAs with the value taken from the previous value
that is non-NA. The output would be in this case:
27 25 25 25 24 26 27 27 26
Now I do that with a for loop, but I try to eliminate all the loops to gain
in performance
I try this and it works but again I have a while and then I do not eliminate
completely loop:
l - length(a)
while(any(is.na(a))) {a[2:l] - ifelse(is.na(a[2:l]), a[1:(l-1)], a[2:l])}
If someone have another solution, I will be most happy !
Sincerely
Marc Girondot
--
__
Marc Girondot, Pr
Laboratoire Ecologie, Systématique et Evolution
Equipe de Conservation des Populations et des Communautés
CNRS, AgroParisTech et Université Paris-Sud 11 , UMR 8079
Bâtiment 362
91405 Orsay Cedex, France
Tel: 33 1 (0)1.69.15.72.30 Fax: 33 1 (0)1.69.15.73.53
e-mail: marc.giron...@u-psud.fr
Web: http://www.ese.u-psud.fr/epc/conservation/Marc.html
Skype: girondot
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Trick to replace NA
Dear Marc,
Try
require(zoo)
na.locf(a)
HTH,
Jorge.-
On Thu, Nov 22, 2012 at 11:56 PM, Marc Girondot wrote:
Dear members,
I have a series of values in a vector and some value are missing and
replaced with NA.
For example:
a - c(27, 25, NA, NA, 24, 26, 27, NA, 26)
I would like to replace the NAs with the value taken from the previous
value that is non-NA. The output would be in this case:
27 25 25 25 24 26 27 27 26
Now I do that with a for loop, but I try to eliminate all the loops to
gain in performance
I try this and it works but again I have a while and then I do not
eliminate completely loop:
l - length(a)
while(any(is.na(a))) {a[2:l] - ifelse(is.na(a[2:l]), a[1:(l-1)], a[2:l])}
If someone have another solution, I will be most happy !
Sincerely
Marc Girondot
--
__**
Marc Girondot, Pr
Laboratoire Ecologie, Systématique et Evolution
Equipe de Conservation des Populations et des Communautés
CNRS, AgroParisTech et Université Paris-Sud 11 , UMR 8079
Bâtiment 362
91405 Orsay Cedex, France
Tel: 33 1 (0)1.69.15.72.30 Fax: 33 1 (0)1.69.15.73.53
e-mail: marc.giron...@u-psud.fr
Web:
http://www.ese.u-psud.fr/epc/**conservation/Marc.htmlhttp://www.ese.u-psud.fr/epc/conservation/Marc.html
Skype: girondot
__**
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posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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Re: [R] Help with loop
Hi again, I was thinking that the interval would stop when there was a NA, ignore subsequent NAs and only start a new interval when there was a new value. So in the example below, the interval would stop at row 320, and a new interval started at row 334. head (D, 20) Salt time 319 35.63190 2004-07-26 320 35.66033 2004-07-27 321 NA 2004-07-28 322 NA 2004-07-29 323 NA 2004-07-30 324 NA 2004-07-31 325 NA 2004-08-01 326 NA 2004-08-02 327 NA 2004-08-03 328 NA 2004-08-04 329 NA 2004-08-05 330 NA 2004-08-06 331 NA 2004-08-07 332 NA 2004-08-08 333 NA 2004-08-09 334 35.57781 2004-08-10 335 35.59829 2004-08-11 Many thanks, Helene -- View this message in context: http://r.789695.n4.nabble.com/Help-with-loop-tp4650313p4650432.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] lapack routines cannot be loaded [Help request]
Dear BioConductor and R fellow users I apologize in advance for double posting, but I am not sure which list would actually be best fit for this message. I am experiencing a weird error with my R installation on Ubuntu 10.04.4 (LTS) 64bit: When I run R on the terminal everything goes smoothly: $R R version 2.15.2 (2012-10-26) -- Trick or Treat Copyright (C) 2012 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-pc-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. However, as soon as I try to use limma (or WGCNA... I haven't tried other packages yet) the following mistake pops up ( lapack routines cannot be loaded ) library(limma) fit - lmFit(Data.rma, design) Error in chol2inv(fit$qr$qr, size = fit$qr$rank) : lapack routines cannot be loaded In addition: Warning message: In chol2inv(fit$qr$qr, size = fit$qr$rank) : unable to load shared object '/usr/lib64/R/modules//lapack.so': /usr/lib64/R/modules//lapack.so: undefined symbol: dpstrf_ It is independent of the dataset I am using. I have already tried to recompile the whole BioConductor set of packages, and updated the general packages via CRAN, but nothing changed. Following I am attaching my sessionInfo(), and you will find enclosed to this email the (incriminated) lapack.so file, should you be willing/able to take a look at it. Any insights into what could be going on, and how to address the issue? sessionInfo() R version 2.15.2 (2012-10-26) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8 [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8 [7] LC_PAPER=CLC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] limma_3.12.3 hgu133plus2cdf_2.10.0 AnnotationDbi_1.18.4 [4] affy_1.34.0 Biobase_2.16.0BiocGenerics_0.2.0 loaded via a namespace (and not attached): [1] affyio_1.24.0 BiocInstaller_1.4.9 DBI_0.2-5 [4] IRanges_1.14.4preprocessCore_1.18.0 RSQLite_0.11.2 [7] stats4_2.15.2 tools_2.15.2 zlibbioc_1.2.0 Thank you in advance, Marco -- Dr Marco Manca University of Maastricht Faculty of Health, Medicine and Life Sciences (FHML) Cardiovascular Research Institute (CARIM) Mailing address: PO Box 616, 6200 MD Maastricht (The Netherlands) Visiting address: UNS40 West building - 5th floor Room5.544, Universiteit Singel 40, 6229 HX Maastricht E-mail: m.ma...@maastrichtuniversity.nl Office telephone: +31(0)433884289 Personal mobile: +31(0)626441205 Twitter: @markomanka * This email and any files transmitted with it are confidential and solely for the use of the intended recipient. It may contain material protected by privacy or doctor-patient/consultant-client privilege. If you are not the intended recipient or the person responsible for delivering to the intended recipient, be advised that you have received this email in error and that any use is STRICTLY PROHIBITED. If you have received this email in error please notify us by telephone on +31626441205 Dr Marco MANCA *__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data Extraction
Hello, I would appreciate if someone could help me resolve the following: 1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work 2. Is these message harmful? The following object(s) are masked from 'df1 (position 3)': X1, X2, X3, X4, X5 Thanks, Pradip Muhuri #Reproducible Example set.seed(5) df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5)) attach (df1) #delete rows if any of them NA for X1 df1[!is.na( X1),][,1:5] # This works #delete rows if any of them NA for X1, X2, X3, X4 or X5 df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rapache memory problem
Hello mrzung, it depends on several factors. First of all: do your clients use RApache for computations in separate sessions (possible which computations does not build on the previous ones) or do they need a stateful application? The first would be pretty easy to resolve, just clean the R environment at the end your R scripts (which is a must in long running R sessions anyway). If the users have direct access to R console (or some similar dynamic solution and the app needs previous environments despite the fact that Apache works with children), then e.g. clean the session when they log out. I think opencpu http://opencpu.org/ would handle all this, and with Apparmor (or with another great tool of Jeroen to set the limits dynamically from R: RAppArmor https://github.com/jeroenooms/RAppArmor) you can force (R)Apache not take more then given amount of RAM, and kill the process. Well, not always when the users would want that :) Best, Gergely On Thu, Nov 22, 2012 at 5:56 AM, mrzung mrzun...@gmail.com wrote: Hi all; Now I'm developing web application by using rapache in ubuntu. My problem is that as users execute the application, the server PC cumulates memory. After users close or refresh the application page, the server PC memory is still cumulated. Now I'm renewing apache machine by manually in terminal. Is there any way to kill memory when users refresh or close the application page? Thanks, -- View this message in context: http://r.789695.n4.nabble.com/rapache-memory-problem-tp4650412.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. [[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] Data Extraction
Hi, is.na( X1 | X2 | X3 | X4 | X5) isn't a valid construct. You'd need !(is.na(X1) | is.na(X2) etc ) Or more elegantly df1[apply(df1, 1, function(x)all(!is.na(x))), ] Sarah On Thursday, November 22, 2012, Muhuri, Pradip (SAMHSA/CBHSQ) wrote: Hello, I would appreciate if someone could help me resolve the following: 1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work 2. Is these message harmful? The following object(s) are masked from 'df1 (position 3)': X1, X2, X3, X4, X5 Thanks, Pradip Muhuri #Reproducible Example set.seed(5) df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5)) attach (df1) #delete rows if any of them NA for X1 df1[!is.na( X1),][,1:5] # This works #delete rows if any of them NA for X1, X2, X3, X4 or X5 df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work __ R-help@r-project.org javascript:; mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com http://www.functionaldiversity.org [[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] Data Extraction
Hi do you want this? df1[complete.cases(df1),] X1 X2 X3 X4 X5 2 8 8 3 2 10 6 8 6 7 10 1 11 4 5 5 10 8 12 6 1 7 8 4 17 5 7 3 1 3 18 10 7 3 8 7 19 7 5 3 5 6 20 10 5 2 4 6 Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Muhuri, Pradip (SAMHSA/CBHSQ) Sent: Thursday, November 22, 2012 3:11 PM To: Muhuri, Pradip (SAMHSA/CBHSQ); r-help@r-project.org Subject: [R] Data Extraction Hello, I would appreciate if someone could help me resolve the following: 1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work 2. Is these message harmful? The following object(s) are masked from 'df1 (position 3)': X1, X2, X3, X4, X5 Thanks, Pradip Muhuri #Reproducible Example set.seed(5) df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5)) attach (df1) #delete rows if any of them NA for X1 df1[!is.na( X1),][,1:5] # This works #delete rows if any of them NA for X1, X2, X3, X4 or X5 df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplot2 and the legend
Dear all, i try to plot with ggplot2. Therefor I have an matrix with 3 colums. With cbind I add an additional column called col. I need this column col because in a later step and want to specify here some plot details which I will get from another analysis If I want to plot with this code, I have the problem that the legend is wrong. Blue changed to green and green to blue. Furthermore the colors in the plot seem a little bit strange. Can anybody help me out? All the best and thanks Peter #THE CODE p0 - ggplot(DATA, aes(Freq, Var1)) p1 - p0+ geom_point(aes(Freq, reorder(Var1, rep(score,cl.count)), colour=col)) p2 - p1 +geom_segment(aes(x = 0, xend = Freq, y = Var1, yend = Var1, colour=col)) p3 - p2 + facet_grid(.~Var2) plot(p3) THE DATA # #class(DATA) = data.frame #class(DATA$Var1) = class(DATA$Var2) = factor #class(DATA$Freq) = numeric #class(DATA$col) = character Var1Var2Freqcol 1 208869_s_at JT 10.6802376685463blue 2 203394_s_at JT 10.0160470327077blue 3 201533_at JT 9.58401500697783blue 4 216520_s_at JT 9.07764840614892blue 5 210999_s_at JT 9.00219110130877green 6 212284_x_at JT 8.62193952504277blue 7 203752_s_at JT 8.69193736253539green 8 203574_at JT 8.41492571590263blue 9 219629_at JT 8.19111292436667blue 10 201473_at JT 8.035082989931 green 11 210592_s_at JT 8.00639829720742green 12 200768_s_at JT 7.88012483915651green 13 201565_s_at JT 7.89885432096935green 14 AFFX-r2-P1-cre-3_at JT 7.86490927895639blue 15 202014_at JT 7.69345791769434green 16 212671_s_at JT -7.43350555284605 green 17 203395_s_at JT 7.51127977823914blue 18 208980_s_at JT 7.52175408445448blue 19 207783_x_at JT 7.31954223019292blue 20 215446_s_at JT -6.28236730203077 green 21 203034_s_at JT 7.24582647671987blue 22 211943_x_at JT 7.22247828112639green 23 203974_at JT -6.84217612309489 green 24 201810_s_at JT 7.0727443704914 green 25 201693_s_at JT 7.11350154823655blue 26 213614_x_at JT 7.07586402329325green 27 AFFX-CreX-5_at JT 7.05274590106307blue 28 201798_s_at JT -6.98858491240942 blue 29 212869_x_at JT 6.97833756793094green 30 201531_at JT 6.96660504363297green 31 201170_s_at JT 6.90904695259236green 32 200886_s_at JT -6.92645789612039 blue 33 AFFX-r2-P1-cre-5_at JT 6.85399276205727blue 34 201429_s_at JT 6.68557384216737green 35 212665_at JT 6.75579868986925green 36 208695_s_at JT 6.8087149011011 green 37 211654_x_at JT -6.7426888322764green 38 209927_s_at JT -6.74856557366229 green 39 203107_x_at JT 6.72120597236789blue 40 211296_x_at JT 6.60226581063165green 41 208869_s_at OA -3.73983541027906 green 42 203394_s_at OA -3.29203821581412 green 43 201533_at OA -3.2383756900468blue 44 216520_s_at OA -4.59979375515969 blue 45 210999_s_at OA -5.10853597491871 green 46 212284_x_at OA -5.84925048581651 blue 47 203752_s_at OA -4.54427623838229 blue 48 203574_at OA -2.59882534850623 blue 49 219629_at OA -4.35550869055387 blue 50 201473_at OA -3.15685663855675 blue 51 210592_s_at OA -4.52611461089035 blue 52 200768_s_at OA -2.76833454650619 blue 53 201565_s_at OA -3.80972141262264 green 54 AFFX-r2-P1-cre-3_at OA -3.53240421347696 green 55 202014_at OA -2.6078496058 blue 56 212671_s_at OA 1.68584117381632green 57 203395_s_at OA -2.27910272551108 blue 58 208980_s_at OA -3.66266591800146 green 59 207783_x_at OA -3.93093352561262 blue 60 215446_s_at OA 6.26626194024723green 61 203034_s_at OA -3.84606488854982 blue 62 211943_x_at OA -3.30831181470312 green 63 203974_at OA 1.17068337099947green 64 201810_s_at OA -2.07855331860079 green 65 201693_s_at OA -2.74740330465003 blue 66 213614_x_at OA -3.0385167086
Re: [R] error message in Linkcomm package
Your questions are all but unanswerable. Read the posting guide and http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example --- 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. Nick Duncan nickd...@gmail.com wrote: Dear All, Some of the functions in Linkcomm return an error: unused argument(s) (v = V(x$igraph)) Although I follow the guidance in the manual a number of the functions return this error. The igraph vector is an edgelist. Many thanks for any guidance. Best, Nick __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Data Extraction
On 22-11-2012, at 15:11, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hello,
I would appreciate if someone could help me resolve the following:
1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
2. Is these message harmful? The following object(s) are masked from 'df1
(position 3)':
X1, X2, X3, X4, X5
Thanks,
Pradip Muhuri
#Reproducible Example
set.seed(5)
df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5))
attach (df1)
#delete rows if any of them NA for X1
df1[!is.na( X1),][,1:5] # This works
#delete rows if any of them NA for X1, X2, X3, X4 or X5
df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
Yet another way of doing this is
df1[!is.na(rowSums(df1)),][1:5]
But Petr's solution appears to be quickest.
See this:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),][,1:ncol(df)]}
f2 - function(df) {df[!is.na(rowSums(df)),][1:ncol(df)]}
f3 - function(df) {df[complete.cases(df),][1:ncol(df)]}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.675 13.172 100
2 d2 - f2(df) 0.4011.437 100
3 d3 - f3(df) 0.2791.000 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
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] Data Extraction
Unnecessarily complicated. ?na.omit (linked from ?complete.cases)
df - na.omit(df)
-- Bert
On Thu, Nov 22, 2012 at 6:49 AM, Berend Hasselman b...@xs4all.nl wrote:
On 22-11-2012, at 15:11, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hello,
I would appreciate if someone could help me resolve the following:
1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
2. Is these message harmful? The following object(s) are masked from
'df1 (position 3)':
X1, X2, X3, X4, X5
Thanks,
Pradip Muhuri
#Reproducible Example
set.seed(5)
df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5))
attach (df1)
#delete rows if any of them NA for X1
df1[!is.na( X1),][,1:5] # This works
#delete rows if any of them NA for X1, X2, X3, X4 or X5
df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
Yet another way of doing this is
df1[!is.na(rowSums(df1)),][1:5]
But Petr's solution appears to be quickest.
See this:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na
(x))),][,1:ncol(df)]}
f2 - function(df) {df[!is.na(rowSums(df)),][1:ncol(df)]}
f3 - function(df) {df[complete.cases(df),][1:ncol(df)]}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.675 13.172 100
2 d2 - f2(df) 0.4011.437 100
3 d3 - f3(df) 0.2791.000 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data Extraction
Petr, You have shown a solution that is the simplest. Thanks and regards, Pradip Muhuri Beginner useR From: PIKAL Petr [petr.pi...@precheza.cz] Sent: Thursday, November 22, 2012 9:33 AM To: Muhuri, Pradip (SAMHSA/CBHSQ); r-help@r-project.org Subject: RE: Data Extraction Hi do you want this? df1[complete.cases(df1),] X1 X2 X3 X4 X5 2 8 8 3 2 10 6 8 6 7 10 1 11 4 5 5 10 8 12 6 1 7 8 4 17 5 7 3 1 3 18 10 7 3 8 7 19 7 5 3 5 6 20 10 5 2 4 6 Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Muhuri, Pradip (SAMHSA/CBHSQ) Sent: Thursday, November 22, 2012 3:11 PM To: Muhuri, Pradip (SAMHSA/CBHSQ); r-help@r-project.org Subject: [R] Data Extraction Hello, I would appreciate if someone could help me resolve the following: 1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work 2. Is these message harmful? The following object(s) are masked from 'df1 (position 3)': X1, X2, X3, X4, X5 Thanks, Pradip Muhuri #Reproducible Example set.seed(5) df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5)) attach (df1) #delete rows if any of them NA for X1 df1[!is.na( X1),][,1:5] # This works #delete rows if any of them NA for X1, X2, X3, X4 or X5 df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Data Extraction
Hi Bert,
Your solution is similar to Petr's.
Thanks and regards,
Pradip Muhuri
BeginneR UseR
From: Bert Gunter [gunter.ber...@gene.com]
Sent: Thursday, November 22, 2012 10:20 AM
To: Berend Hasselman
Cc: Muhuri, Pradip (SAMHSA/CBHSQ); r-help@r-project.org
Subject: Re: [R] Data Extraction
Unnecessarily complicated. ?na.omit (linked from ?complete.cases)
df - na.omit(df)
-- Bert
On Thu, Nov 22, 2012 at 6:49 AM, Berend Hasselman
b...@xs4all.nlmailto:b...@xs4all.nl wrote:
On 22-11-2012, at 15:11, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hello,
I would appreciate if someone could help me resolve the following:
1. df1[!is.nahttp://is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not
work
2. Is these message harmful? The following object(s) are masked from 'df1
(position 3)':
X1, X2, X3, X4, X5
Thanks,
Pradip Muhuri
#Reproducible Example
set.seed(5)
df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5))
attach (df1)
#delete rows if any of them NA for X1
df1[!is.nahttp://is.na( X1),][,1:5] # This works
#delete rows if any of them NA for X1, X2, X3, X4 or X5
df1[!is.nahttp://is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
Yet another way of doing this is
df1[!is.nahttp://is.na(rowSums(df1)),][1:5]
But Petr's solution appears to be quickest.
See this:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1,
function(x)all(!is.nahttp://is.na(x))),][,1:ncol(df)]}
f2 - function(df) {df[!is.nahttp://is.na(rowSums(df)),][1:ncol(df)]}
f3 - function(df) {df[complete.cases(df),][1:ncol(df)]}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.675 13.172 100
2 d2 - f2(df) 0.4011.437 100
3 d3 - f3(df) 0.2791.000 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
Berend
__
R-help@r-project.orgmailto:R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data Extraction
Hi Berend,
You have compared all 3 ways. ... very nicely evaluated.
Thanks and regards,
Pradip Muhuri
Beginner UseR
From: Berend Hasselman [b...@xs4all.nl]
Sent: Thursday, November 22, 2012 9:49 AM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: r-help@r-project.org
Subject: Re: [R] Data Extraction
On 22-11-2012, at 15:11, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hello,
I would appreciate if someone could help me resolve the following:
1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
2. Is these message harmful? The following object(s) are masked from 'df1
(position 3)':
X1, X2, X3, X4, X5
Thanks,
Pradip Muhuri
#Reproducible Example
set.seed(5)
df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5))
attach (df1)
#delete rows if any of them NA for X1
df1[!is.na( X1),][,1:5] # This works
#delete rows if any of them NA for X1, X2, X3, X4 or X5
df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work
Yet another way of doing this is
df1[!is.na(rowSums(df1)),][1:5]
But Petr's solution appears to be quickest.
See this:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),][,1:ncol(df)]}
f2 - function(df) {df[!is.na(rowSums(df)),][1:ncol(df)]}
f3 - function(df) {df[complete.cases(df),][1:ncol(df)]}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.675 13.172 100
2 d2 - f2(df) 0.4011.437 100
3 d3 - f3(df) 0.2791.000 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
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] selcting a random sample and saving it in a seprate dataframe and also remaining part in other data frame
HI Madhu, I guess you got your solution from Rui: dat1-data.frame(x=c(1,1,2,2,2,3,4,4,4),y=c(23,45,87,46,78,12,87,79,76)) s-sample(unique(dat1[,1]),length(unique(dat1[,1]))*0.8) s #[1] 3 4 2 You can have a list containing both the dataframes list1-list(dat1[dat1$x%in%s,],dat1[!dat1$x%in%s,]) list1 [[1]] # x y #3 2 87 #4 2 46 #5 2 78 #6 3 12 #7 4 87 #8 4 79 #9 4 76 #[[2]] # x y #1 1 23 #2 1 45 is.data.frame(list1[[1]]) #[1] TRUE A.K. - Original Message - From: Madhu Ganganapalli mganganapa...@upstreamsoftware.com To: arun smartpink...@yahoo.com Cc: Sent: Thursday, November 22, 2012 2:53 AM Subject: selcting a random sample and saving it in a seprate dataframe and also remaining part in other data frame ** My question is: I have the following data frame and my distinct values of variable x are 1,2,3,4. data-data.frame(x=c(1,1,2,2,2,3,4,4,4),y=c(23,45,87,46,78,12,87,79));** Here my data has 8 observations but I mentioned that distinct observations are 4 so 80% data means I have to get a random sample from these 4 observations only, in such a way that Suppose while selecting 80% random sample from x I got 1,2, and 3(80% means 80/100*4=3 roughly) so I want a following out put in separate data frame. X y 1 23 1 45 2 87 2 46 2 78 3 12 That means if 1 is in 80% of my random sample then the data corresponding to remaining 1's also should be there in my data frame. One more thing is after creating this data frame, we have only one distinct observations which is 4 in our actual data frame What I mean is we have to get two data sets simultaneously in two different data frames, which is of above output format. In this case second data frame is X y 4 12 4 87 4 79 This will help while building a model, because we use only 80% data for modeling and remaining 20% for validation so that is way I want two datasets simultaneously in two different data frames. Please help me. Thanks, Madhu. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data Extraction
Hi Sarah, I am glad you have precisely caught where I made the mistake. Thank you so much. regards, Pradip Muhuri From: Sarah Goslee [sarah.gos...@gmail.com] Sent: Thursday, November 22, 2012 9:21 AM To: Muhuri, Pradip (SAMHSA/CBHSQ) Cc: r-help@r-project.org Subject: Re: [R] Data Extraction Hi, is.nahttp://is.na/( X1 | X2 | X3 | X4 | X5) isn't a valid construct. You'd need !(is.nahttp://is.na(X1) | is.nahttp://is.na(X2) etc ) Or more elegantly df1[apply(df1, 1, function(x)all(!is.nahttp://is.na(x))), ] Sarah On Thursday, November 22, 2012, Muhuri, Pradip (SAMHSA/CBHSQ) wrote: Hello, I would appreciate if someone could help me resolve the following: 1. df1[!is.nahttp://is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work 2. Is these message harmful? The following object(s) are masked from 'df1 (position 3)': X1, X2, X3, X4, X5 Thanks, Pradip Muhuri #Reproducible Example set.seed(5) df1-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5)) attach (df1) #delete rows if any of them NA for X1 df1[!is.nahttp://is.na( X1),][,1:5] # This works #delete rows if any of them NA for X1, X2, X3, X4 or X5 df1[!is.nahttp://is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work __ R-help@r-project.orgjavascript:; mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data Extraction
On 22-11-2012, at 16:50, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
You have compared all 3 ways. ... very nicely evaluated.
Bert's solution is indeed nice and simple. But Petr's solution is still the
quickest:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),]}
f2 - function(df) {df[!is.na(rowSums(df)),]}
f3 - function(df) {df[complete.cases(df),]}
f4 - function(df) {data.frame(na.omit(df))}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df), d4 - f4(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.588 14.888 100
2 d2 - f2(df) 0.4031.672 100
3 d3 - f3(df) 0.2411.000 100
4 d4 - f4(df) 0.5572.311 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
identical(d1,d4)
[1] TRUE
Berend
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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] ggplot2 and the legend
a) Don't crosspost. Read the posting guide(s) for more cautions. This question is probably more appropriate on the ggplot Google Group than either of these two mailing lists. b) Post reproducible examples. c) Use data frames with ggplot, not matrices. The read.csv or read.table functions automatically create data frames. It really is important for you to show how you create the data objects that give you problems when asking for help, since for all we know you just don't know the difference. See point (a) above. --- 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. Peter Kupfer peter.kup...@me.com wrote: Dear all, i try to plot with ggplot2. Therefor I have an matrix with 3 colums. With cbind I add an additional column called col. I need this column col because in a later step and want to specify here some plot details which I will get from another analysis If I want to plot with this code, I have the problem that the legend is wrong. Blue changed to green and green to blue. Furthermore the colors in the plot seem a little bit strange. Can anybody help me out? All the best and thanks Peter #THE CODE p0 - ggplot(DATA, aes(Freq, Var1)) p1 - p0+ geom_point(aes(Freq, reorder(Var1, rep(score,cl.count)), colour=col)) p2 - p1 +geom_segment(aes(x = 0, xend = Freq, y = Var1, yend = Var1, colour=col)) p3 - p2 + facet_grid(.~Var2) plot(p3) THE DATA # #class(DATA) = data.frame #class(DATA$Var1) = class(DATA$Var2) = factor #class(DATA$Freq) = numeric #class(DATA$col) = character Var1 Var2Freqcol 1 208869_s_at JT 10.6802376685463blue 2 203394_s_at JT 10.0160470327077blue 3 201533_at JT 9.58401500697783blue 4 216520_s_at JT 9.07764840614892blue 5 210999_s_at JT 9.00219110130877green 6 212284_x_at JT 8.62193952504277blue 7 203752_s_at JT 8.69193736253539green 8 203574_at JT 8.41492571590263blue 9 219629_at JT 8.19111292436667blue 10 201473_at JT 8.035082989931 green 11 210592_s_at JT 8.00639829720742green 12 200768_s_at JT 7.88012483915651green 13 201565_s_at JT 7.89885432096935green 14 AFFX-r2-P1-cre-3_at JT 7.86490927895639blue 15 202014_at JT 7.69345791769434green 16 212671_s_at JT -7.43350555284605 green 17 203395_s_at JT 7.51127977823914blue 18 208980_s_at JT 7.52175408445448blue 19 207783_x_at JT 7.31954223019292blue 20 215446_s_at JT -6.28236730203077 green 21 203034_s_at JT 7.24582647671987blue 22 211943_x_at JT 7.22247828112639green 23 203974_at JT -6.84217612309489 green 24 201810_s_at JT 7.0727443704914 green 25 201693_s_at JT 7.11350154823655blue 26 213614_x_at JT 7.07586402329325green 27 AFFX-CreX-5_at JT 7.05274590106307blue 28 201798_s_at JT -6.98858491240942 blue 29 212869_x_at JT 6.97833756793094green 30 201531_at JT 6.96660504363297green 31 201170_s_at JT 6.90904695259236green 32 200886_s_at JT -6.92645789612039 blue 33 AFFX-r2-P1-cre-5_at JT 6.85399276205727blue 34 201429_s_at JT 6.68557384216737green 35 212665_at JT 6.75579868986925green 36 208695_s_at JT 6.8087149011011 green 37 211654_x_at JT -6.7426888322764green 38 209927_s_at JT -6.74856557366229 green 39 203107_x_at JT 6.72120597236789blue 40 211296_x_at JT 6.60226581063165green 41 208869_s_at OA -3.73983541027906 green 42 203394_s_at OA -3.29203821581412 green 43 201533_at OA -3.2383756900468blue 44 216520_s_at OA -4.59979375515969 blue 45 210999_s_at OA -5.10853597491871 green 46 212284_x_at OA -5.84925048581651 blue 47 203752_s_at OA -4.54427623838229 blue 48 203574_at OA -2.59882534850623 blue 49 219629_at OA
Re: [R] Data Extraction - benchmark()
Hi Berend,
I see you are one of the contributors to the rbecnhmark package.
I am sorry that I am bothering you again. I have tried to run your code
(slightly tweaked) involving the benchmark function, and I am getting the
following error message. What am I doing wrong?
Error in benchmark(d1 - s1(df), d2 - s2(df), d3 - s3(df), d4 - s4(df), :
could not find function s1
identical (d1,d2), identical (d1,d3), identical (d1,d4), identical (d1,d5),
identical (d1,d6)
Error: unexpected ',' in identical (d1,d2),
sessionInfo ()
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] rbenchmark_1.0.0
loaded via a namespace (and not attached):
[1] tools_2.15.1
I would appreciate receiving your help if your time permits ..
Thanks and regards,
Pradip Muhuri
# Berend's code extended
N - 10
set.seed(13)
df-data.frame(matrix(sample(c(1:10,NA),N, replace=TRUE),ncol=50))
s1 - df[complete.cases(df),]
s2 - na.omit(df)
s3 - df[apply(df, 1, function(x)all(!is.na(x))), ]
s4 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),][,1:ncol(df)]}
s5 - function(df) {df[!is.na(rowSums(df)),][1:ncol(df)]}
s6 - function(df) {df[complete.cases(df),][1:ncol(df)]}
require(rbenchmark)
benchmark( d1 - s1(df), d2 - s2(df), d3 - s3(df), d4 - s4(df), d5 -
s5(df), d6 - s6(df),
columns=c(test,elapsed, relative, replications) )
identical (d1,d2), identical (d1,d3), identical (d1,d4), identical (d1,d5),
identical (d1,d6)
From: Berend Hasselman [b...@xs4all.nl]
Sent: Thursday, November 22, 2012 11:03 AM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: r-help@r-project.org
Subject: Re: [R] Data Extraction
On 22-11-2012, at 16:50, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
You have compared all 3 ways. ... very nicely evaluated.
Bert's solution is indeed nice and simple. But Petr's solution is still the
quickest:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),]}
f2 - function(df) {df[!is.na(rowSums(df)),]}
f3 - function(df) {df[complete.cases(df),]}
f4 - function(df) {data.frame(na.omit(df))}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df), d4 - f4(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.588 14.888 100
2 d2 - f2(df) 0.4031.672 100
3 d3 - f3(df) 0.2411.000 100
4 d4 - f4(df) 0.5572.311 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
identical(d1,d4)
[1] TRUE
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.
[R] error in replor: Results suspect!
Hello all, I am trying to fit a gee model using repolr and I get a warning message that Results suspect. Warning message: Note: In ogee(formula, data = , id =., : Cgee had an error (code= 104). Results suspect. I checked the source code as Prof. Ripley advice in 2007 but the reason of this warning is not clear yet. Can anybody explain the reason of this error please ? Ana [[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] Data Extraction
Berend et. al:
1, Except you did not use my solution. It is not what you have but
instead:
f4 - function(df)na.omit.df
wrapping the return in data.frame() is both unnecessary and inefficient.
2. But my point is not to be the speediest nor even to show how clever I am
-- using a built-in function is hardly being clever. Rather to emphasize
that the OP requested a standard task for which R ALREADY HAS BUILT IN
FUNCTIONALITY. Reading ?na.omit shows that the returned object contains
information (about the omitted rows) that other R functions (e.g.
predict() ) make use of. That is the point. Rather than roll your own,
one should use that functionality. It is generally there for a good reason,
as in this case.
Cheers,
Bert
On Thu, Nov 22, 2012 at 8:03 AM, Berend Hasselman b...@xs4all.nl wrote:
On 22-11-2012, at 16:50, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
You have compared all 3 ways. ... very nicely evaluated.
Bert's solution is indeed nice and simple. But Petr's solution is still
the quickest:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),]}
f2 - function(df) {df[!is.na(rowSums(df)),]}
f3 - function(df) {df[complete.cases(df),]}
f4 - function(df) {data.frame(na.omit(df))}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df), d4 - f4(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.588 14.888 100
2 d2 - f2(df) 0.4031.672 100
3 d3 - f3(df) 0.2411.000 100
4 d4 - f4(df) 0.5572.311 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
identical(d1,d4)
[1] TRUE
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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data Extraction - benchmark()
On 22-11-2012, at 18:20, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
I see you are one of the contributors to the rbecnhmark package.
I am sorry that I am bothering you again. I have tried to run your code
(slightly tweaked) involving the benchmark function, and I am getting the
following error message. What am I doing wrong?
Error in benchmark(d1 - s1(df), d2 - s2(df), d3 - s3(df), d4 - s4(df), :
could not find function s1
Because you haven't defined a function s1 (or s2, s3, s4 for that matter).
You did s1 - df[complete.cases(df),]
Berend
identical (d1,d2), identical (d1,d3), identical (d1,d4), identical (d1,d5),
identical (d1,d6)
Error: unexpected ',' in identical (d1,d2),
sessionInfo ()
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] rbenchmark_1.0.0
loaded via a namespace (and not attached):
[1] tools_2.15.1
I would appreciate receiving your help if your time permits ..
Thanks and regards,
Pradip Muhuri
# Berend's code extended
N - 10
set.seed(13)
df-data.frame(matrix(sample(c(1:10,NA),N, replace=TRUE),ncol=50))
s1 - df[complete.cases(df),]
s2 - na.omit(df)
s3 - df[apply(df, 1, function(x)all(!is.na(x))), ]
s4 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),][,1:ncol(df)]}
s5 - function(df) {df[!is.na(rowSums(df)),][1:ncol(df)]}
s6 - function(df) {df[complete.cases(df),][1:ncol(df)]}
require(rbenchmark)
benchmark( d1 - s1(df), d2 - s2(df), d3 - s3(df), d4 - s4(df), d5 -
s5(df), d6 - s6(df),
columns=c(test,elapsed, relative, replications) )
identical (d1,d2), identical (d1,d3), identical (d1,d4), identical (d1,d5),
identical (d1,d6)
From: Berend Hasselman [b...@xs4all.nl]
Sent: Thursday, November 22, 2012 11:03 AM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: r-help@r-project.org
Subject: Re: [R] Data Extraction
On 22-11-2012, at 16:50, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
You have compared all 3 ways. ... very nicely evaluated.
Bert's solution is indeed nice and simple. But Petr's solution is still the
quickest:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),]}
f2 - function(df) {df[!is.na(rowSums(df)),]}
f3 - function(df) {df[complete.cases(df),]}
f4 - function(df) {data.frame(na.omit(df))}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df), d4 - f4(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.588 14.888 100
2 d2 - f2(df) 0.4031.672 100
3 d3 - f3(df) 0.2411.000 100
4 d4 - f4(df) 0.5572.311 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
identical(d1,d4)
[1] TRUE
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] printing difftime summary
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-11-22 12:11:55
+]:
I now think that what I want is
--8---cut here---start-8---
difftime.summary - function (v) {
s - summary(as.numeric(v))
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
r[[units(v)]] - s
r
}
Any reason not summary.difftime to get S3 dispatch?
I hoped that someone will ask this :-)
1. because its argument has type vector of difftime, not difftime
(coming from CLOS, I do not expect summary(vector of difftime) to
dispatch to summary.difftime, but to summary.vector.of.difftime or something)
2. because difftime.summary returns a data.frame and not a
Classes 'summaryDefault', 'table' as I assume summary must return.
if these are not valid issues, then I wonder why my function should not
be the system default method.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://memri.org http://honestreporting.com
http://jihadwatch.org http://openvotingconsortium.org http://ffii.org
Sex is like air. It's only a big deal if you can't get any.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to read jpeg image with russian font in R?
On Nov 21, 2012, at 11:21 PM, Manish Gupta wrote: Hi, I m working on R and read one image with russian font using readjpeg() function and write it on pdf (using grid). But russian text is not visible on pdf. How can i fix my pblm? You are working with two different plotting paradigms. pdf and ps are vector formats and do encode the text as individual characters. But jpeg is a lossy bitmap format, so text is not separately encoded as characters. Regards http://r.789695.n4.nabble.com/file/n4650417/pointer.jpg library(JPEG) There is a package named 'jpeg so I'm wondering if you are unaware the case in naming objects in R is enforced? -- David. library(grid) pdf(out.pdf , width = 6.6 ,height = 4.2,family= URWHelvetica, encoding=KOI8-R) a-readJPEG(ArrowImage)# name of arrow grid.newpage() pushViewport(viewport(width=0.9, height=0.9)) pushViewport(viewport(yscale=c(0,0), xscale=c(0,1), x=0, y=0, width=0.15, height=1.0, just=c(left, bottom))) grid.raster(a, y=unit(ypos.img, native)) popViewport() popViewport() dev.off() Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-read-jpeg-image-with-russian-font-in-R-tp4650417.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. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] [BioC] lapack routines cannot be loaded [Help request]
On 11/22/2012 03:23 AM, Manca Marco (PATH) wrote: Dear BioConductor and R fellow users I apologize in advance for double posting, but I am not sure which list would actually be best fit for this message. It helps to narrow the problem down. It looks like the problem occurs when chol2inv is invoked, and in a plain R session R --vanilla the following would fail example(chol2inv) hence a problem with your installed R (since chol2inv is in the 'base' package), rather than Bioconductor. How did you install your R, especially how was it configured to use LAPACK? Have you modified your system's LAPACK after installing R? Martin I am experiencing a weird error with my R installation on Ubuntu 10.04.4 (LTS) 64bit: When I run R on the terminal everything goes smoothly: $R R version 2.15.2 (2012-10-26) -- Trick or Treat Copyright (C) 2012 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-pc-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. However, as soon as I try to use limma (or WGCNA... I haven't tried other packages yet) the following mistake pops up ( lapack routines cannot be loaded ) library(limma) fit - lmFit(Data.rma, design) Error in chol2inv(fit$qr$qr, size = fit$qr$rank) : lapack routines cannot be loaded In addition: Warning message: In chol2inv(fit$qr$qr, size = fit$qr$rank) : unable to load shared object '/usr/lib64/R/modules//lapack.so': /usr/lib64/R/modules//lapack.so: undefined symbol: dpstrf_ It is independent of the dataset I am using. I have already tried to recompile the whole BioConductor set of packages, and updated the general packages via CRAN, but nothing changed. Following I am attaching my sessionInfo(), and you will find enclosed to this email the (incriminated) lapack.so file, should you be willing/able to take a look at it. Any insights into what could be going on, and how to address the issue? sessionInfo() R version 2.15.2 (2012-10-26) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8 [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8 [7] LC_PAPER=CLC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] limma_3.12.3 hgu133plus2cdf_2.10.0 AnnotationDbi_1.18.4 [4] affy_1.34.0 Biobase_2.16.0BiocGenerics_0.2.0 loaded via a namespace (and not attached): [1] affyio_1.24.0 BiocInstaller_1.4.9 DBI_0.2-5 [4] IRanges_1.14.4preprocessCore_1.18.0 RSQLite_0.11.2 [7] stats4_2.15.2 tools_2.15.2 zlibbioc_1.2.0 Thank you in advance, Marco -- Dr Marco Manca University of Maastricht Faculty of Health, Medicine and Life Sciences (FHML) Cardiovascular Research Institute (CARIM) Mailing address: PO Box 616, 6200 MD Maastricht (The Netherlands) Visiting address: UNS40 West building - 5th floor Room5.544, Universiteit Singel 40, 6229 HX Maastricht E-mail: m.ma...@maastrichtuniversity.nl Office telephone: +31(0)433884289 Personal mobile: +31(0)626441205 Twitter: @markomanka * This email and any files transmitted with it are confidential and solely for the use of the intended recipient. It may contain material protected by privacy or doctor-patient/consultant-client privilege. If you are not the intended recipient or the person responsible for delivering to the intended recipient, be advised that you have received this email in error and that any use is STRICTLY PROHIBITED. If you have received this email in error please notify us by telephone on +31626441205 Dr Marco MANCA * ___ Bioconductor mailing list bioconduc...@r-project.org https://stat.ethz.ch/mailman/listinfo/bioconductor Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor -- Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793
[R] Using cumsum with 'group by' ?
Hi, First post here. Grateful for any help you can give. I have data which looks like this: idtimex 1 12:015 1 12:02 14 1 12:03 6 1 12:04 3 2 12:01 98 2 12:02 23 2 12:03 1 2 12:04 4 3 12:01 5 3 12:02 65 3 12:03 23 3 12:04 23 But I want to add a column which is the cumulative sum of X, but only by id. I've used cumsum before, but not in this way. So the result should be something like: id time xcumsum 1 12:01 55 1 12:02 14 19 1 12:03 6 25 1 12:04 3 28 2 12:01 98 98 2 12:02 23 121 2 12:03 1 122 2 12:04 4 126 3 12:01 5 5 3 12:02 65 70 3 12:03 23 93 3 12:04 23 116 Any ideas please? -- View this message in context: http://r.789695.n4.nabble.com/Using-cumsum-with-group-by-tp4650457.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with loop
HI, If it is possible to remove those NA values: dat$time - as.Date(dat$time,format=%Y-%m-%d) newdat-dat[complete.cases(dat),] change-cumsum(c(FALSE,abs(diff(newdat$Salt))0.05)) #Rui's solution split(newdat,change) A.K. - Original Message - From: Hefri helenefrigs...@hotmail.com To: r-help@r-project.org Cc: Sent: Thursday, November 22, 2012 6:43 AM Subject: Re: [R] Help with loop Hi again, I was thinking that the interval would stop when there was a NA, ignore subsequent NAs and only start a new interval when there was a new value. So in the example below, the interval would stop at row 320, and a new interval started at row 334. head (D, 20) Salt time 319 35.63190 2004-07-26 320 35.66033 2004-07-27 321 NA 2004-07-28 322 NA 2004-07-29 323 NA 2004-07-30 324 NA 2004-07-31 325 NA 2004-08-01 326 NA 2004-08-02 327 NA 2004-08-03 328 NA 2004-08-04 329 NA 2004-08-05 330 NA 2004-08-06 331 NA 2004-08-07 332 NA 2004-08-08 333 NA 2004-08-09 334 35.57781 2004-08-10 335 35.59829 2004-08-11 Many thanks, Helene -- View this message in context: http://r.789695.n4.nabble.com/Help-with-loop-tp4650313p4650432.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] How to control plotting formula above lm diagnostic plots?
Dear All, I am trying to plot four diagnostic plots for my lm in one window. Here is some random data for an example: a = rnorm(20, mean=2, sd=0.2) b = rnorm(20, mean=1, sd=0.4) model=lm(a~b) When I set the page as: par(mfrow=c(1,1), oma=c(1,0,2,0)) and plot my model: plot(model) above all four plots the formula lm(a~b) is printed. What is interesting if I set the page to have two plots per page (still plotting all four using plot(model)), the formula is plotted only above the last two. When I set the page to have one plot per page (still plotting all four using plot(model)) the formula is not printed above any of the plots. Is there any way to control plotting the formula above the diagnostic plots if I have multiple per page? To control the text size, font, position, or just not to have it plotted it at all? Playing with the option 'main' or 'cex.main' inside the 'plot' function does not change anything. Many thanks Kasia -- View this message in context: http://r.789695.n4.nabble.com/How-to-control-plotting-formula-above-lm-diagnostic-plots-tp4650465.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] Partial dependence plot in randomForest package (all flat responses)
Hi, I'm trying to make a partial plot with package randomForest in R. After I perform my random forest object I type partialPlot(data.rforest, pred.data=act2, x.var=centroid, C) where data.rforest is my randomforest object, act2 is the original dataset, centroid is one of the predictor and C is one of the classes in my response variable. Whatever predictor or response class I try I always get a plot with a straight line (a completely flat response). Similarly, If I set a categorical variable as predictor, I get a barplot with all the bar with the same height. I suppose I'm doing something wrong here because all other analysis on the same rforest object seem correct (e.g. varImp or MDSplot). Is it possible it is related to some option set in random forest object? Can somebody see the problem here? Thanks for your time -- View this message in context: http://r.789695.n4.nabble.com/Partial-dependence-plot-in-randomForest-package-all-flat-responses-tp4650470.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] Mixed models and learning curves
My name is Giovanna and I am a PhD student in Norway. I am a beginner with statistics and R, hence my ignorance. Apologies from now. I have been collecting data on time performances of 5 subjects using a 1:3 scale tower yarder. The task was consisting in yarding 5 small logs placed on permanently marked course. Four subjects had different previous experiences (None, Some) and the fifth was a trainer (Control). Each cycle time per each log was registered, the sum of the 5 logs' cycle time was giving the replication time. We had 6 replication per subject . I would like to predict the time necessary to perform the task. I have been modelling the time to perform the task (prod.time)versus the replication number (Trial-in the dataset), the previous experience (factor) and their interaction. As random effect I have been using the subjects. ma-lme(prod.time~Trial+Previous.experience+Trial*Previous.experience, data= Data27_04, random=~1|Student, method=ML) summary(ma) Linear mixed-effects model fit by maximum likelihood Data: Data27_04 AIC BIClogLik 1517.445 1541.259 -750.7226 Random effects: Formula: ~1 | Student (Intercept) Residual StdDev:7.337648 42.42332 Fixed effects: prod.time ~ Trial + Previous.experience + Trial * Previous.experience Value Std.Error DF t-value p-value (Intercept)102.44173 9.561987 137 10.713435 0. Trial -6.48494 2.252271 137 -2.879291 0.0046 Previous.experience1 -37.36173 14.786033 2 -2.526826 0.1274 Previous.experience247.22627 12.451072 2 3.792948 0.0630 Trial:Previous.experience1 6.55351 3.496401 137 1.874360 0.0630 Trial:Previous.experience2 -7.55163 2.940879 137 -2.567813 0.0113 Correlation: (Intr) Trial Prvs.1 Prvs.2 Tr:P.1 Trial -0.841 Previous.experience10.253 -0.208 Previous.experience2 -0.234 0.199 -0.540 Trial:Previous.experience1 -0.207 0.264 -0.835 0.447 Trial:Previous.experience2 0.199 -0.226 0.447 -0.836 -0.550 Standardized Within-Group Residuals: Min Q1Med Q3Max -2.3519731 -0.6903211 -0.1031114 0.6503216 4.6699702 Number of Observations: 145 Number of Groups: 5 Do you think this is good enough to demonstrate a learning effect. Learning curves are exponential. I have been trying to log transform the response variable but then p-values are saying that previous experience has no significance. mb-lme(log.prodtime~Trial+Previous.experience+Trial*Previous.experience, data= Data27_04, random=~1|Student, method=ML) summary(mb) Linear mixed-effects model fit by maximum likelihood Data: Data27_04 AIC BIClogLik 225.1042 248.9181 -104.5521 Random effects: Formula: ~1 | Student (Intercept) Residual StdDev: 0.04484554 0.495812 Fixed effects: log.prodtime ~ Trial + Previous.experience + Trial * Previous.experience Value Std.Error DF t-value p-value (Intercept) 4.448206 0.10593072 137 41.99165 0. Trial -0.060150 0.02629765 137 -2.28726 0.0237 Previous.experience1 -0.333664 0.16351518 2 -2.04057 0.1781 Previous.experience20.368358 0.13776525 2 2.67381 0.1160 Trial:Previous.experience1 0.051714 0.04084708 137 1.26604 0.2076 Trial:Previous.experience2 -0.043036 0.03435150 137 -1.25282 0.2124 Correlation: (Intr) Trial Prvs.1 Prvs.2 Tr:P.1 Trial -0.886 Previous.experience10.248 -0.221 Previous.experience2 -0.237 0.209 -0.535 Trial:Previous.experience1 -0.220 0.266 -0.881 0.473 Trial:Previous.experience2 0.208 -0.225 0.474 -0.883 -0.551 Standardized Within-Group Residuals: Min Q1Med Q3Max -2.7119095 -0.8005032 0.1127388 0.8621127 2.1988560 Number of Observations: 145 Number of Groups: 5 The model is surely better (AIC, BIC) also the residuals are looking better but then should I reduce the model leaving only the Trial number? How would you present the results in a clear way? I am still struggling to figure it out. The concept of mixed models is clear in my head but it is hard to present it. How should I then plot the learning curve? I have been plotting the data I have adding a smooth line. Is this good enough? Looking forward for your response Best regards Giovanna Giovanna Ottaviani Aalmo Stipendiat/Ph..D. Student --- Norsk institutt for skog og landskap Pb 115, NO-1431 Ås T (+47) 64 94 9094 M(+47) 980 30 422 F(+47) 64 94 90 80 --- www.skogoglandskap.nohttp://www.skogoglandskap.no/ --- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list
Re: [R] cluster analysis in R
These are the errors I've been having. I have been trying 3 different things 1- Mclust: This is the example I have been following: # Model Based Clustering library(mclust) fit - Mclust(mydata) plot(fit, mydata) # plot results print(fit) # display the best model What I have done: fit - Mclust(mydat) plot(fit, mydat) #plot results Error in match.arg(what, c(BIC, classification, uncertainty, density), : 'arg' must be NULL or a character vector 2- Mclust using different website (cran-r) instructions This is the example: mydatMclust - Mclust(mydat) summary(mydatMclust) summary(mydatMclust, parameters = TRUE) plot(mydatMclust) There are a couple other steps but the plot is the problem. I get two plots, there should be four. One should be plotting all my individuals but it's plotting my variables instead. It's also taking a very long time. R script at this point says: Waiting to confirm page change… 3. Mcclust Instructions from cran-r: data(cls.draw2) # sample of 500 clusterings from a Bayesian cluster model tru.class - rep(1:8,each=50) # the true grouping of the observations psm2 - comp.psm(cls.draw2) # posterior similarity matrix # optimize criteria based on PSM mbind2 - minbinder(psm2) mpear2 - maxpear(psm2) # Relabelling k - apply(cls.draw2,1, function(cl) length(table(cl))) max.k - as.numeric(names(table(k))[which.max(table(k))]) relab2 - relabel(cls.draw2[k==max.k,]) # compare clusterings found by different methods with true grouping arandi(mpear2$cl, tru.class) arandi(mbind2$cl, tru.class) arandi(relab2$cl, tru.class) I called my data: mydat so I changed that where appropriate. I cannot get past one early step, psm2 - comp.psm(cls.draw2).. the error reads: Error: could not find function comp.psm I think I have all appropriate packages installed. I don't know what more to do on these three errors. Any help would be great! Thank you -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-in-R-tp4649635p4650466.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] ROC Curve: negative AUC
Hi all, does anyone know why the area under the curve (AUC) is negative? I'm using ROC function with a logistic regression, package Epi. First time it happens... Thanks a lot! Bruno -- View this message in context: http://r.789695.n4.nabble.com/ROC-Curve-negative-AUC-tp4650469.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] Using cumsum with 'group by' ?
transform(d, cumsumXById = ave(x, id, FUN=cumsum)) id time x cumsumXById 1 1 12:01 5 5 2 1 12:02 14 19 3 1 12:03 6 25 4 1 12:04 3 28 5 2 12:01 98 98 6 2 12:02 23 121 7 2 12:03 1 122 8 2 12:04 4 126 9 3 12:01 5 5 10 3 12:02 65 70 11 3 12:03 23 93 12 3 12:04 23 116 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of TheRealJimShady Sent: Thursday, November 22, 2012 9:09 AM To: r-help@r-project.org Subject: [R] Using cumsum with 'group by' ? Hi, First post here. Grateful for any help you can give. I have data which looks like this: idtimex 1 12:015 1 12:02 14 1 12:03 6 1 12:04 3 2 12:01 98 2 12:02 23 2 12:03 1 2 12:04 4 3 12:01 5 3 12:02 65 3 12:03 23 3 12:04 23 But I want to add a column which is the cumulative sum of X, but only by id. I've used cumsum before, but not in this way. So the result should be something like: id time xcumsum 1 12:01 55 1 12:02 14 19 1 12:03 6 25 1 12:04 3 28 2 12:01 98 98 2 12:02 23 121 2 12:03 1 122 2 12:04 4 126 3 12:01 5 5 3 12:02 65 70 3 12:03 23 93 3 12:04 23 116 Any ideas please? -- View this message in context: http://r.789695.n4.nabble.com/Using-cumsum-with- group-by-tp4650457.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] Data Extraction - benchmark()
Hi Berend,
Thank you very much for pointing out the mistake and for your patience. I have
corrected the the script, which has worked fine.
regards,
Pradip Muhuri
From: Berend Hasselman [b...@xs4all.nl]
Sent: Thursday, November 22, 2012 12:42 PM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: r-help@r-project.org
Subject: Re: [R] Data Extraction - benchmark()
On 22-11-2012, at 18:20, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
I see you are one of the contributors to the rbecnhmark package.
I am sorry that I am bothering you again. I have tried to run your code
(slightly tweaked) involving the benchmark function, and I am getting the
following error message. What am I doing wrong?
Error in benchmark(d1 - s1(df), d2 - s2(df), d3 - s3(df), d4 - s4(df), :
could not find function s1
Because you haven't defined a function s1 (or s2, s3, s4 for that matter).
You did s1 - df[complete.cases(df),]
Berend
identical (d1,d2), identical (d1,d3), identical (d1,d4), identical (d1,d5),
identical (d1,d6)
Error: unexpected ',' in identical (d1,d2),
sessionInfo ()
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] rbenchmark_1.0.0
loaded via a namespace (and not attached):
[1] tools_2.15.1
I would appreciate receiving your help if your time permits ..
Thanks and regards,
Pradip Muhuri
# Berend's code extended
N - 10
set.seed(13)
df-data.frame(matrix(sample(c(1:10,NA),N, replace=TRUE),ncol=50))
s1 - df[complete.cases(df),]
s2 - na.omit(df)
s3 - df[apply(df, 1, function(x)all(!is.na(x))), ]
s4 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),][,1:ncol(df)]}
s5 - function(df) {df[!is.na(rowSums(df)),][1:ncol(df)]}
s6 - function(df) {df[complete.cases(df),][1:ncol(df)]}
require(rbenchmark)
benchmark( d1 - s1(df), d2 - s2(df), d3 - s3(df), d4 - s4(df), d5 -
s5(df), d6 - s6(df),
columns=c(test,elapsed, relative, replications) )
identical (d1,d2), identical (d1,d3), identical (d1,d4), identical (d1,d5),
identical (d1,d6)
From: Berend Hasselman [b...@xs4all.nl]
Sent: Thursday, November 22, 2012 11:03 AM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: r-help@r-project.org
Subject: Re: [R] Data Extraction
On 22-11-2012, at 16:50, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:
Hi Berend,
You have compared all 3 ways. ... very nicely evaluated.
Bert's solution is indeed nice and simple. But Petr's solution is still the
quickest:
N - 10
set.seed(13)
df - data.frame(matrix(sample(c(1:10,NA),N,replace=TRUE),ncol=50))
library(rbenchmark)
f1 - function(df) {df[apply(df, 1, function(x)all(!is.na(x))),]}
f2 - function(df) {df[!is.na(rowSums(df)),]}
f3 - function(df) {df[complete.cases(df),]}
f4 - function(df) {data.frame(na.omit(df))}
benchmark(d1 - f1(df), d2 - f2(df), d3 - f3(df), d4 - f4(df),
columns=c(test,elapsed, relative, replications))
test elapsed relative replications
1 d1 - f1(df) 3.588 14.888 100
2 d2 - f2(df) 0.4031.672 100
3 d3 - f3(df) 0.2411.000 100
4 d4 - f4(df) 0.5572.311 100
identical(d1,d2)
[1] TRUE
identical(d1,d3)
[1] TRUE
identical(d1,d4)
[1] TRUE
Berend
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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] Using cumsum with 'group by' ?
On 22-11-2012, at 18:08, TheRealJimShady wrote: Hi, First post here. Grateful for any help you can give. I have data which looks like this: idtimex 1 12:015 1 12:02 14 1 12:03 6 1 12:04 3 2 12:01 98 2 12:02 23 2 12:03 1 2 12:04 4 3 12:01 5 3 12:02 65 3 12:03 23 3 12:04 23 But I want to add a column which is the cumulative sum of X, but only by id. I've used cumsum before, but not in this way. So the result should be something like: id time xcumsum 1 12:01 55 1 12:02 14 19 1 12:03 6 25 1 12:04 3 28 2 12:01 98 98 2 12:02 23 121 2 12:03 1 122 2 12:04 4 126 3 12:01 5 5 3 12:02 65 70 3 12:03 23 93 3 12:04 23 116 Any ideas please? Assuming your data are in a dataframe named df this should do what you want df[,cumsum] - ave(df$x,by=df$id, FUN=cumsum) 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] ROC Curve: negative AUC
wrong direction for ranking On Thu, Nov 22, 2012 at 1:58 PM, brunosm brunos...@gmail.com wrote: the area under the curve (AUC) is negative? I'm using ROC function with a logistic regression, package Epi. First time it happens... Thanks a lot! Bruno -- == WenSui Liu Credit Risk Manager, 53 Bancorp wensui@53.com 513-295-4370 == [[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] ROC Curve: negative AUC
On Fri, Nov 23, 2012 at 7:58 AM, brunosm brunos...@gmail.com wrote: Hi all, does anyone know why the area under the curve (AUC) is negative? I'm using ROC function with a logistic regression, package Epi. First time it happens... Have you looked at the ROC curve? That should tell you whether there's some strange non-convexity going on or whether the variable is just being put into the calculations backwards. -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland [[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] Plotting specific points with type='l'
I have a dataframe (x) and I'm plotting the 5th column vs the index. I also have a vector (v) with a few select points that I want to emphasize with a dot for those points head(x) period AP EU LA NA 1 Jan 2007 0.18 0.45 0.19 3.19 2 Feb 2007 0.14 0.48 0.36 3.55 3 Mar 2007 0.14 0.42 0.46 2.61 4 Apr 2007 0.24 0.73 0.32 4.32 5 May 2007 0.19 0.60 0.32 4.40 6 Jun 2007 0.14 0.38 0.32 1.09 v -c(2,4,7) plot(x[,5], type='l') How do I put a solid dot at just the points I want to highlight ? In other words, a solid dot a the 2nd, 4th, and 7th point on the plot. All other points according the the type='l'. I tried ... points(x[v,5], pch=19) ... but the points didn't plot in the right spot. -- View this message in context: http://r.789695.n4.nabble.com/Plotting-specific-points-with-type-l-tp4650475.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] Simulating a VEC
Hello everyone, I have estimated a VEC using functions of package urca. Now I need to simulate several trajectories of the variables of the model based in the information of the model. To do that first I converted the VEC to a VAR using vec2var (from package vars) and then I converted de VAR to an ARMA using ARMA function of package dse. The idea is to use the function simulate (from package dse), to simulate the ARMA from the VEC using the initial values of the variables of the model, the exogenous variables in the simulation period and the matrix of covariances of the residuals of the VEC model, to model the residuals in each simulation. Then I made several iterations of this simulation, to obtain different trajectories and conclude about the model; but sometimes the plot of the simulations shows values of the variables on the order of +-1e137 (a really big number) and I don't know why it is happening. Does anyone of you know what can be happening? Thanks in advance, Laura Echeverri -- View this message in context: http://r.789695.n4.nabble.com/Simulating-a-VEC-tp4650463.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] ifelse + numeric
Thanks, it worked without quotation marks. -- View this message in context: http://r.789695.n4.nabble.com/ifelse-numeric-tp4650390p4650482.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] Plotting specific points with type='l'
try this # you need to provide sample data x - runif(10) v - c(2, 4, 7) plot(x, type = 'l') # highlight points points(v, x[v], pch = 19) On Thu, Nov 22, 2012 at 2:22 PM, eric ericst...@aol.com wrote: I have a dataframe (x) and I'm plotting the 5th column vs the index. I also have a vector (v) with a few select points that I want to emphasize with a dot for those points head(x) period AP EU LA NA 1 Jan 2007 0.18 0.45 0.19 3.19 2 Feb 2007 0.14 0.48 0.36 3.55 3 Mar 2007 0.14 0.42 0.46 2.61 4 Apr 2007 0.24 0.73 0.32 4.32 5 May 2007 0.19 0.60 0.32 4.40 6 Jun 2007 0.14 0.38 0.32 1.09 v -c(2,4,7) plot(x[,5], type='l') How do I put a solid dot at just the points I want to highlight ? In other words, a solid dot a the 2nd, 4th, and 7th point on the plot. All other points according the the type='l'. I tried ... points(x[v,5], pch=19) ... but the points didn't plot in the right spot. -- View this message in context: http://r.789695.n4.nabble.com/Plotting-specific-points-with-type-l-tp4650475.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] SEM raw moment matrix
Hello, I estimated a model using SEM package in R, which was fit to a raw moment matrix, and includes an intercept term. The only goodness of fit statistics that are output are Model Chisquare, AIC, AICc, BIC, CAIC, and normalized residuals. How can I get the other goodness of fit statistics, like adjusted goodness of fit, RMSEA, and R-squared? And how can I get the final value of the log-likelihood of the model? Thanks, Maya [[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] plotting 1000 simulations, error message: plot.new has not been called yet
Why are you using jpeg to create files with a .png extension? wouldn't it
make more sense to use the png function?
There are a couple of options, you could plot using the matplot function
(with different subsets in the loop).
Or you can start an initial empty plot and use lines to add to it, then use
dev.copy to copy the current version to a file.
Or you can look at the animation package.
Or you could do nested loops, but hopefully one of the above is better than
that approach.
On Wed, Nov 21, 2012 at 11:07 AM, Maximilian Lklweryc maxlklwe...@gmail.com
wrote:
Hi,
I know this is not a mailing list for r, but I posted my question on
several help pages and did not get any help. I really don't know how
to solve my problem, maybe you could help me?
want to simulate stock paths. I have simulated 1000 paths with 22
trading days (1 starting value). Now I want to include it into my
presentation, but animated, so I need the png files.
I want to create 1000 png files, starting with the first stock path,
then the second and so on.
So I start with the first path, add a second to the plot, add the
third and so on, so at the end I have a plot with 1000 simulations,
here is my code:
for(i in 1:1000){
#jpeg(paste(1000s,i,.png,sep=))
plot(c(1:23),matrix[,1],type=l,ylim=c(17,24))
lines(c(1:23),matrix[,i],type=l,col=i)
#dev.off()
}
Here is the problem, that each additional part disappears when the
loop gets to the next value, so I tried:
plot(0,0 , xlim=c(1,23),ylim=c(17,24),xlab=,ylab=)
for(i in 1:1000){
jpeg(paste(1000s,i,.png,sep=))
lines(c(1:23),matrix[,i],type=l,col=i)
dev.off()
}
(I know this is not a working example, but my problem is just a
logical one with the loop) I get the following error message when I
the last code: plot.new has not been called yet.
The matrix has 1000 columns and 23 row entries, this should be 1000
simulations of stock pathes for 22 trading days.
How can I change that the error does not appear anymore? Thanks!
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538...@gmail.com
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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] Summary statistics for matrix columns
frespider wrote Hi, is there a way I can calculate a summary statistics for a columns matrix let say we have this matrix x - matrix(sample(1:8000),nrow=100) colnames(x)- paste(Col,1:ncol(x),sep=) if I used summary summary(x) i get the output for each column but I need the output to be in matrix with rownames and all the columns beside it this how I want it Col76 Col77 Min. :739 1st Qu. :1846 1630 Median : 3631 3376 Mean: 3804 3617 Sd : 3rd Qu.:5772 5544 IQR: Max. :79527779 Is there an easy way? Thanks How about ... x - matrix(sample(1:8000),nrow=100) colnames(x)- paste(Col,1:ncol(x),sep=) apply(x,2,function(x) c(summary(x), sd=sd(x), IQR=IQR(x))) HTH Pete -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650490.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] Data frame manipulation
Hello, I have a table that was constructed in a wrong way (dput data on bottom - wrong data-frame): Local Mês Dia Colonia X6h X7h X8h X9hX10hX11h X12hX13hX14hX15hX16hX17h 1 Conceição Junho 1 3 2.16137 2.20412 2.08991 1.72428 1.69897 1.62325 1.44716 1.51851 1.43136 1.47712 1.51851 1.04139 2 Conceição Junho 2 3 2.46538 2.13672 2.06819 1.97772 2.0 1.80618 1.64345 1.20412 1.62325 1.36173 1.69020 1.57978 3 Conceição Junho 3 3 2.53275 2.52504 2.49276 2.3 2.12710 2.26007 2.24551 1.95424 2.09342 1.04139 1.53148 1.17609 4 Conceição Junho 1 4 1.65321 2.16435 1.91381 1.75587 1.74036 1.17609 1.66276 1.51851 1.39794 1.04139 1.11394 1.04139 5 Conceição Junho 2 4 2.30320 1.71600 2.02531 2.05690 1.86332 1.66276 1.17609 1.04139 1.30103 1.27875 1.3 1.3 6 Conceição Junho 3 4 2.71012 2.30320 2.53403 1.80618 2.24551 2.20683 2.02531 1.07918 1.36173 1.39794 1.11394 1.93450 7 Conceição Junho 1 5 2.21748 1.99564 2.26007 2.28103 2.10380 1.41497 0.47712 1.07918 0.90309 1.04139 1.49136 1.23045 8 Conceição Junho 2 5 2.10721 2.16435 2.05308 2.38561 2.14613 1.61278 1.27875 0.47712 1.61278 1.0 1.44716 1.07918 9 Conceição Junho 3 5 1.62325 1.93450 2.33041 2.24797 2.29885 2.48001 2.29003 1.43136 1.49136 1.17609 1.41497 1.14613 10 Conceição julho 1 3 2.20952 2.01284 1.79239 1.59106 1.62325 1.51851 1.41497 1.38021 1.66276 1.46240 1.53148 1.66276 I have to create a new column (hour) and transpose just the last 12 columns, and first four columns have to be copied 12 time, like this (dput data on bottom - correct data-frame): Local Mês Dia Colonia HoraN 1 Conceição Junho 1 3 6h 2.161370 2 Conceição Junho 1 3 7h 2.204120 3 Conceição Junho 1 3 8h 2.089910 4 Conceição Junho 1 3 9h 1.724280 5 Conceição Junho 1 3 10h 1.698970 6 Conceição Junho 1 3 11h 1.623250 7 Conceição Junho 1 3 12h 1.447160 8 Conceição Junho 1 3 13h 1.518510 9 Conceição Junho 1 3 14h 1.431360 10 Conceição Junho 1 3 15h 1.477120 11 Conceição Junho 1 3 16h 1.518510 12 Conceição Junho 1 3 17h 1.041390 13 Conceição Junho 2 3 6h 2.465383 Some one could give me some ideas? I don't even know how to start... Thanks in advanced, -- Raoni Rosa Rodrigues Research Associate of Fish Transposition Center CTPeixes Universidade Federal de Minas Gerais - UFMG Brasil rodrigues.ra...@gmail.com wrong data frame: structure(list(Local = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = Conceição, class = factor), Mês = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L), .Label = c(julho, Junho ), class = factor), Dia = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L), Colonia = c(3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 3L), X6h = c(2.16137, 2.46538, 2.53275, 1.65321, 2.3032, 2.71012, 2.21748, 2.10721, 1.62325, 2.20952), X7h = c(2.20412, 2.13672, 2.52504, 2.16435, 1.716, 2.3032, 1.99564, 2.16435, 1.9345, 2.01284), X8h = c(2.08991, 2.06819, 2.49276, 1.91381, 2.02531, 2.53403, 2.26007, 2.05308, 2.33041, 1.79239), X9h = c(1.72428, 1.97772, 2.3, 1.75587, 2.0569, 1.80618, 2.28103, 2.38561, 2.24797, 1.59106), X10h = c(1.69897, 2, 2.1271, 1.74036, 1.86332, 2.24551, 2.1038, 2.14613, 2.29885, 1.62325), X11h = c(1.62325, 1.80618, 2.26007, 1.17609, 1.66276, 2.20683, 1.41497, 1.61278, 2.48001, 1.51851), X12h = c(1.44716, 1.64345, 2.24551, 1.66276, 1.17609, 2.02531, 0.47712, 1.27875, 2.29003, 1.41497), X13h = c(1.51851, 1.20412, 1.95424, 1.51851, 1.04139, 1.07918, 1.07918, 0.47712, 1.43136, 1.38021), X14h = c(1.43136, 1.62325, 2.09342, 1.39794, 1.30103, 1.36173, 0.90309, 1.61278, 1.49136, 1.66276), X15h = c(1.47712, 1.36173, 1.04139, 1.04139, 1.27875, 1.39794, 1.04139, 1, 1.17609, 1.4624), X16h = c(1.51851, 1.6902, 1.53148, 1.11394, 1.3, 1.11394, 1.49136, 1.44716, 1.41497, 1.53148), X17h = c(1.04139, 1.57978, 1.17609, 1.04139, 1.3, 1.9345, 1.23045, 1.07918, 1.14613, 1.66276)), .Names = c(Local, Mês, Dia, Colonia, X6h, X7h, X8h, X9h, X10h, X11h, X12h, X13h, X14h, X15h, X16h, X17h), row.names = c(NA, 10L), class = data.frame) Correct data frame: structure(list(Local = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = Conceição, class = factor), Mês = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(julho, Junho ), class = factor), Dia = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Colonia = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), Hora = structure(1:10, .Label = c(6h, 7h, 8h, 9h, 10h, 11h, 12h, 13h, 14h, 15h, 16h, 17h), class = factor), N = c(2.16137, 2.20412, 2.08991, 1.72428, 1.69897, 1.62325, 1.44716, 1.51851, 1.43136, 1.47712)), .Names = c(Local, Mês, Dia, Colonia, Hora, N), row.names = c(NA, 10L), class = data.frame) [[alternative HTML version deleted]]
Re: [R] [lattice] Increase distance between tick labels and ticks in wireframe plot (pad)
On 2012-11-22 01:16, Felix Schönbrodt wrote: Hello, I try to increase the distance between tick labels and ticks in a lattice wireframe plot. Here's a minimal example: ## Minimal example x - y - z - c(1,2,3) df - data.frame(x, y, z) wireframe(z ~ x*y, df, scales = list(arrows = FALSE, col = black, font = 1, tck=0.6)) I tried the axis.components option (http://r.789695.n4.nabble.com/Lattice-distance-of-tick-labels-from-axis-line-tp3693014p3693014.html). This works for xyplot, but *not* for wireframe: xyplot(z ~ x, df, scales = list(arrows = FALSE, col = black, font = 1, tck=0.6), par.settings=list(axis.components=list(left=list(pad1=2 wireframe(z ~ x*y, df, scales = list(arrows = FALSE, col = black, font = 1, tck=0.6), par.settings=list(axis.components=list(left=list(pad1=2 Any ideas on how to pad the axis labels in wireframe plots? Thanks, Felix I think that you want the 'distance=' argument to scales (which is documented in ?wireframe). Try this: wireframe(z ~ x*y, df, scales = list(arrows = FALSE, col = black, font = 1, tck = c(0.8, 0.6, 0.4), distance =c(2, 5, 8))) Adjust as you prefer. Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Adding a function with default parameters into the Rcmdr menu
Hi everyone,
I made some tests with Rcmdr, to add a function with default parameters :
For example (very simple):
myfunction-function(var=314){
print(hello)
print(var)
}
if I run myfunction() directly i see :
myfunction()
[1] hello
[1] 314
it's ok.
But if i edit de Rcmdr-menu.txt (in
C:\Users\myname\Documents\R\win-library\2.15\Rcmdr\etc)
and add :
menuMyMenutopMenu
itemtopMenu cascade
MyTestMyMenu
itemMyMenu command Test
myfunction
and I put myfonction into a file Rcmdr-test.R in the same folder
I have a new button with a cascade menu, and myfunction is corectly
sourced.. but :
that's what append :
library(Rcmdr)
Loading required package: tcltk
Loading Tcl/Tk interface ... done
Loading required package: car
Loading required package: MASS
Loading required package: nnet
Sourced: Rcmdr-test.r
Rcmdr Version 1.9-2
If i use the Menu :
[1] hello
[1] %var
-there are %var instead off 314.
myfunction
function(var=314){
print(hello)
print(var)
}
I think that I made a mistake but I dont know were. How can I use Rcmdr
menu AND a default parameter ?
Can you help me?
Thx a lot.
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Re: [R] Data frame manipulation
The 'reshape2' package is your friend: require(reshape2) x - melt(wrong, id = c(Local, Mês, Dia, Colonia), variable.name = Hora) # remove X from Hora x$Hora - as.character(substring(x$Hora, 2)) head(x) # not in the right order Local Mês Dia Colonia Hora value 1 Conceição Junho 1 3 6h 2.16137 2 Conceição Junho 2 3 6h 2.46538 3 Conceição Junho 3 3 6h 2.53275 4 Conceição Junho 1 4 6h 1.65321 5 Conceição Junho 2 4 6h 2.30320 6 Conceição Junho 3 4 6h 2.71012 # sort, but first add blank on Hora for less that 10h for sorting x$Hora - ifelse(nchar(x$Hora) == 2, paste0( , x$Hora), x$Hora) x - x[order(x$Local, x$Mês, x$Dia, x$Colonia, x$Hora), ] head(x,20) Local Mês Dia Colonia Hora value 10 Conceição julho 1 3 6h 2.20952 20 Conceição julho 1 3 7h 2.01284 30 Conceição julho 1 3 8h 1.79239 40 Conceição julho 1 3 9h 1.59106 50 Conceição julho 1 3 10h 1.62325 60 Conceição julho 1 3 11h 1.51851 70 Conceição julho 1 3 12h 1.41497 80 Conceição julho 1 3 13h 1.38021 90 Conceição julho 1 3 14h 1.66276 100 Conceição julho 1 3 15h 1.46240 110 Conceição julho 1 3 16h 1.53148 120 Conceição julho 1 3 17h 1.66276 1 Conceição Junho 1 3 6h 2.16137 11 Conceição Junho 1 3 7h 2.20412 21 Conceição Junho 1 3 8h 2.08991 31 Conceição Junho 1 3 9h 1.72428 41 Conceição Junho 1 3 10h 1.69897 51 Conceição Junho 1 3 11h 1.62325 61 Conceição Junho 1 3 12h 1.44716 71 Conceição Junho 1 3 13h 1.51851 On Thu, Nov 22, 2012 at 8:53 PM, Raoni Rodrigues caciquesamu...@gmail.com wrote: structure(list(Local = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = Conceição, class = factor), Mês = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(julho, Junho ), class = factor), Dia = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Colonia = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), Hora = structure(1:10, .Label = c(6h, 7h, 8h, 9h, 10h, 11h, 12h, 13h, 14h, 15h, 16h, 17h), class = factor), N = c(2.16137, 2.20412, 2.08991, 1.72428, 1.69897, 1.62325, 1.44716, 1.51851, 1.43136, 1.47712)), .Names = c(Local, Mês, Dia, Colonia, Hora, N), row.names = c(NA, 10L), class = data.frame) -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Constant (= wrong) historical quotes via get.hist.quote() from yahoo.finance
Dear expeRts, I would like to download a time series of historical data from the ticker with symbol ROG.VX. Interestingly, I obtain constant values (138.3 for each day in the chosen period) although the yahoo.finance website tells me that the time series is not at all constant. What's wrong? Cheers, Marius require(tseries) hq - get.hist.quote(instrument=ROG.VX, start=2011-09-09, end=2012-03-28, quote=Close, provider=yahoo, drop=TRUE) plot(hq) # = constant stopifnot(hq==138.3) # = constant 138.3 ## However, under http://finance.yahoo.com/q/hp?s=ROG.VXa=08b=09c=2011d=02e=28f=2012g=dz=66y=132 ## the historical prices are not all equal to 138.3 pgpxmc2kGwwm9.pgp Description: PGP signature __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Summary statistics for matrix columns
I also don't like to use split function because I have like around 800 columns Date: Thu, 22 Nov 2012 18:08:54 -0800 From: ml-node+s789695n4650496...@n4.nabble.com To: frespi...@hotmail.com Subject: RE: Summary statistics for matrix columns Hi, How about this: res-do.call(cbind,lapply(split(x,col(x)),function(x) c(summary(x),sd=sd(x),IQR=IQR(x colnames(res)-colnames(x) is.matrix(res) [1] TRUE res Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Min.10.0 1.0 17.0 3.0 18.0 11.0 13.0 15.0 1st Qu. 24.75000 29.5 26.0 7.75000 40.0 17.25000 27.5 34.75000 Median 34.0 46.0 42.5 35.5 49.5 23.5 51.5 51.5 Mean42.5 42.75000 41.75000 35.75000 44.88000 26.88000 44.75000 50.12000 3rd Qu. 67.75000 58.5 50.0 63.25000 54.25000 30.25000 56.25000 70.5 Max.74.0 77.0 76.0 70.0 65.0 63.0 79.0 80.0 sd 25.05993 27.77846 19.57221 28.40397 16.39196 16.60841 21.97239 25.51995 IQR 43.0 29.0 24.0 55.5 14.25000 13.0 28.75000 35.75000 Col9Col10 Min. 2.0 6.0 1st Qu. 24.5 12.5 Median 33.5 48.0 Mean34.88000 40.75000 3rd Qu. 45.25000 63.0 Max.71.0 72.0 sd 24.39811 28.21727 IQR 20.75000 50.5 A.K. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650496.html This email was sent by arun kirshna (via Nabble) -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650498.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.
[R] Summary statistics for matrix columns
Hi, is there a way I can calculate a summary statistics for a columns matrix let say we have this matrix x - matrix(sample(1:8000),nrow=100) colnames(x)- paste(Col,1:ncol(x),sep=) if I used summary summary(x) i get the output for each column but I need the output to be in matrix with rownames and all the columns beside it this how I want it Col76 Col77 Min. :739 1st Qu. :1846 1630 Median : 3631 3376 Mean: 3804 3617 Sd : 3rd Qu.:5772 5544 IQR: Max. :79527779 Is there an easy way? Thanks -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489.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] Summary statistics for matrix columns
Hi peter, but this doesn't give me them in the order I want. Is there a better approach Thanks -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650492.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] Summary statistics for matrix columns
There is still missing some statistics, like sd and IQR and I prefer the output to be matrix Thanks Date: Thu, 22 Nov 2012 18:00:20 -0800 From: ml-node+s789695n4650493...@n4.nabble.com To: frespi...@hotmail.com Subject: Re: Summary statistics for matrix columns HI, You could try this: set.seed(125) x - matrix(sample(1:80),nrow=8) colnames(x)- paste(Col,1:ncol(x),sep=) sapply(data.frame(x),function(x) summary(x)) # Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 #Min.10.00 1.00 17.00 3.00 18.00 11.00 13.00 15.00 2.00 6.00 #1st Qu. 24.75 29.50 26.00 7.75 40.00 17.25 27.50 34.75 24.50 12.50 #Median 34.00 46.00 42.50 35.50 49.50 23.50 51.50 51.50 33.50 48.00 #Mean42.50 42.75 41.75 35.75 44.88 26.88 44.75 50.12 34.88 40.75 #3rd Qu. 67.75 58.50 50.00 63.25 54.25 30.25 56.25 70.50 45.25 63.00 #Max.74.00 77.00 76.00 70.00 65.00 63.00 79.00 80.00 71.00 72.00 A.K. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650493.html To unsubscribe from Summary statistics for matrix columns, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650494.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] Summary statistics for matrix columns
HI, but Sd and IQR not in the order I want , Thanks Date: Thu, 22 Nov 2012 18:08:57 -0800 From: ml-node+s789695n4650496...@n4.nabble.com To: frespi...@hotmail.com Subject: RE: Summary statistics for matrix columns Hi, How about this: res-do.call(cbind,lapply(split(x,col(x)),function(x) c(summary(x),sd=sd(x),IQR=IQR(x colnames(res)-colnames(x) is.matrix(res) [1] TRUE res Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Min.10.0 1.0 17.0 3.0 18.0 11.0 13.0 15.0 1st Qu. 24.75000 29.5 26.0 7.75000 40.0 17.25000 27.5 34.75000 Median 34.0 46.0 42.5 35.5 49.5 23.5 51.5 51.5 Mean42.5 42.75000 41.75000 35.75000 44.88000 26.88000 44.75000 50.12000 3rd Qu. 67.75000 58.5 50.0 63.25000 54.25000 30.25000 56.25000 70.5 Max.74.0 77.0 76.0 70.0 65.0 63.0 79.0 80.0 sd 25.05993 27.77846 19.57221 28.40397 16.39196 16.60841 21.97239 25.51995 IQR 43.0 29.0 24.0 55.5 14.25000 13.0 28.75000 35.75000 Col9Col10 Min. 2.0 6.0 1st Qu. 24.5 12.5 Median 33.5 48.0 Mean34.88000 40.75000 3rd Qu. 45.25000 63.0 Max.71.0 72.0 sd 24.39811 28.21727 IQR 20.75000 50.5 A.K. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650496.html To unsubscribe from Summary statistics for matrix columns, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650497.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] Summary statistics for matrix columns
Hi, it is possible. but don't you think it will slow the code if you convert to data.frame? Thanks Date: Thu, 22 Nov 2012 18:31:35 -0800 From: ml-node+s789695n4650500...@n4.nabble.com To: frespi...@hotmail.com Subject: RE: Summary statistics for matrix columns HI, Is it possible to use as.matrix()? res-sapply(data.frame(x),function(x) c(summary(x),sd=sd(x),IQR=IQR(x))) res1-as.matrix(res) is.matrix(res1) #[1] TRUE res1[c(1:4,7,5,8,6),] #Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 #Min.10.0 1.0 17.0 3.0 18.0 11.0 13.0 15.0 #1st Qu. 24.75000 29.5 26.0 7.75000 40.0 17.25000 27.5 34.75000 #Median 34.0 46.0 42.5 35.5 49.5 23.5 51.5 51.5 #Mean42.5 42.75000 41.75000 35.75000 44.88000 26.88000 44.75000 50.12000 #sd 25.05993 27.77846 19.57221 28.40397 16.39196 16.60841 21.97239 25.51995 #3rd Qu. 67.75000 58.5 50.0 63.25000 54.25000 30.25000 56.25000 70.5 #IQR 43.0 29.0 24.0 55.5 14.25000 13.0 28.75000 35.75000 #Max.74.0 77.0 76.0 70.0 65.0 63.0 79.0 80.0 # Col9Col10 #Min. 2.0 6.0 #1st Qu. 24.5 12.5 #Median 33.5 48.0 #Mean34.88000 40.75000 #sd 24.39811 28.21727 #3rd Qu. 45.25000 63.0 #IQR 20.75000 50.5 #Max.71.0 72.0 Solves the order and the matrix output! A.K. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650500.html To unsubscribe from Summary statistics for matrix columns, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/Summary-statistics-for-matrix-columns-tp4650489p4650501.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] Plotting specific points with type='l'
Hi, If you wanted to emphasize the select points from the 5th column: You could try: plot(x[,5], type='l') points(dat2[c(2,4,7),5],pch=19) A.K. - Original Message - From: eric ericst...@aol.com To: r-help@r-project.org Cc: Sent: Thursday, November 22, 2012 2:22 PM Subject: [R] Plotting specific points with type='l' I have a dataframe (x) and I'm plotting the 5th column vs the index. I also have a vector (v) with a few select points that I want to emphasize with a dot for those points head(x) period AP EU LA NA 1 Jan 2007 0.18 0.45 0.19 3.19 2 Feb 2007 0.14 0.48 0.36 3.55 3 Mar 2007 0.14 0.42 0.46 2.61 4 Apr 2007 0.24 0.73 0.32 4.32 5 May 2007 0.19 0.60 0.32 4.40 6 Jun 2007 0.14 0.38 0.32 1.09 v -c(2,4,7) plot(x[,5], type='l') How do I put a solid dot at just the points I want to highlight ? In other words, a solid dot a the 2nd, 4th, and 7th point on the plot. All other points according the the type='l'. I tried ... points(x[v,5], pch=19) ... but the points didn't plot in the right spot. -- View this message in context: http://r.789695.n4.nabble.com/Plotting-specific-points-with-type-l-tp4650475.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] Factor function for coded numerical values
I have s data set where 2 of the columns give the coded versions of the factors A and B. Factor A is coded with 1, 2, 3. Factor B is coded with 1,2. How do use the factor function to convert these variables into factors, and also use the labels= command to give them a more informative name? -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-for-coded-numerical-values-tp4650483.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] Using cumsum with 'group by' ?
Hi, No problem. One more method if you wanted to try: library(data.table) dat2-data.table(dat1) dat2[,list(x,time,Cumsum=cumsum(x)),list(id)] # id x time Cumsum #1: 1 5 12:01 5 #2: 1 14 12:02 19 #3: 1 6 12:03 25 #4: 1 3 12:04 28 #5: 2 98 12:01 98 #6: 2 23 12:02 121 #7: 2 1 12:03 122 #8: 2 4 12:04 126 #9: 3 5 12:01 5 #10: 3 65 12:02 70 #11: 3 23 12:03 93 #12: 3 23 12:04 116 A.K. - Original Message - From: TheRealJimShady james.david.sm...@gmail.com To: r-help@r-project.org Cc: Sent: Thursday, November 22, 2012 12:27 PM Subject: Re: [R] Using cumsum with 'group by' ? Thank you very much, I will try these tomorrow morning. On 22 November 2012 17:25, arun kirshna [via R] ml-node+s789695n4650459...@n4.nabble.com wrote: HI, You can do this in many ways: dat1-read.table(text= id time x 1 12:01 5 1 12:02 14 1 12:03 6 1 12:04 3 2 12:01 98 2 12:02 23 2 12:03 1 2 12:04 4 3 12:01 5 3 12:02 65 3 12:03 23 3 12:04 23 ,sep=,header=TRUE,stringsAsFactors=FALSE) dat1$Cumsum-ave(dat1$x,dat1$id,FUN=cumsum) #or unlist(tapply(dat1$x,dat1$id,FUN=cumsum),use.names=FALSE) # [1] 5 19 25 28 98 121 122 126 5 70 93 116 #or library(plyr) ddply(dat1,.(id),function(x) cumsum(x[3]))[,2] # [1] 5 19 25 28 98 121 122 126 5 70 93 116 head(dat1) # id time x Cumsum #1 1 12:01 5 5 #2 1 12:02 14 19 #3 1 12:03 6 25 #4 1 12:04 3 28 #5 2 12:01 98 98 #6 2 12:02 23 121 A.K. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Using-cumsum-with-group-by-tp4650457p4650459.html To unsubscribe from Using cumsum with 'group by' ?, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/Using-cumsum-with-group-by-tp4650457p4650461.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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Optimizing nested function with nlminb()
I am trying to optimize custom likelyhood with nlminb()
Arguments h and f are meant to be fixed.
example.R:
compute.hyper.log.likelyhood - function(a, h, f) {
a1 - a[1]
a2 - a[2]
l - 0.0
for (j in 1:length(f)) {
l - l + lbeta(a1 + f[j], a2 + h - f[j]) - lbeta(a1, a2)
}
return(l)
}
compute.optimal.hyper.params - function(start, limits, h_, f_) {
result - nlminb(start,
compute.hyper.log.likelyhood,
h=h_,
f=f_,
scale = -1,
lower = c(limits[1],limits[1]),
upper = c(limits[2],limits[2]))
return (result[[1]])
}
Console launch:
source('~/Desktop/Dropbox/example.R')
h - 1000
start - c(3,3)
limits - c(0.01,100)
f - c(40,30,50)
compute.optimal.hyper.params(start,limits,h,f)
produces the following:
Error in a2 + h : 'h' is missing
Could you please explain me why?
R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i686-pc-linux-gnu (32-bit)
--
Best regards,
Dmitry
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and provide commented, minimal, self-contained, reproducible code.
[R] help in M-estimator by R
hi guys and gals ... How are you all ... i have to do something in robust regression by R programm , and i have some problems as following: *the first :* suppose w(r) =1/(1 r^2) and r - c(7.01,2.07,7.061,5.607,8.502,54.909,12.222) and i want to exclude some values from r so that (abs(r)4.9 )... after ,i want to used (w) to get on coefficients beta0 and beta1 (B1 - (sum(w*y*x)-sum(w*y)*sum(w*x))/(sum(w*(x^2))-(sum(w*x))^2/sum(w)) B0 - (sum(w*y)-B1*sum(x*w))/sum(w). this process must to repeat it until (w) be 0.001 ( tolerance - 0.001) How can i get on this in one program? - i tried to do it by if -else , repeat and while - *the second:* in simple robust regression we have two way to fit the model 1- rlm(formula, data, weights, ..., subset, na.action, method = c(M, MM, model.frame), wt.method = c(inv.var, case), model = TRUE, x.ret = TRUE, y.ret = FALSE, contrasts = NULL) 2- rlm(x, y, weights, ..., w = rep(1, nrow(x)), init = ls, psi = psi.huber, scale.est = c(MAD, Huber, proposal 2), k2 = 1.345, method = c(M, MM), wt.method = c(inv.var, case), maxit = 20, acc = 1e-4, test.vec = resid, lqs.control = NULL) psi.huber(u, k = 1.345, deriv = 0) psi.hampel(u, a = 2, b = 4, c = 8, deriv = 0) psi.bisquare(u, c = 4.685, deriv = 0) and all these ways works by default function as psi.huber and i want to add new psi ( psi.a=r/(1+(r/k)^2)) ,k=1.345 and get on new coefficients to the model, and, Is all the arguments are necessary ? i want short code again ... *how can i do this ... * thanks in advance ... /my best wishes.../ - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-in-M-estimator-by-R-tp4650485.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] (no subject)
__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data frame manipulation
Hi, May be this helps: library(reshape) dat1 # data that needs to be converted res-melt(dat1,id=c(Local,Mês,Dia,Colonia)) names(res)[5:6]-c(Hora,N) res1-res[order(res$Dia),] row.names(res1)-1:nrow(res1) res1$Hora-gsub([X],,res1$Hora) head(res1) # Local Mês Dia Colonia Hora N #1 Conceição Junho 1 3 6h 2.16137 #2 Conceição Junho 1 4 6h 1.65321 #3 Conceição Junho 1 5 6h 2.21748 #4 Conceição julho 1 3 6h 2.20952 #5 Conceição Junho 1 3 7h 2.20412 #6 Conceição Junho 1 4 7h 2.16435 A.K. To: r-help@r-project.org Cc: Sent: Thursday, November 22, 2012 8:53 PM Subject: [R] Data frame manipulation Hello, I have a table that was constructed in a wrong way (dput data on bottom - wrong data-frame): Local Mês Dia Colonia X6h X7h X8h X9h X10h X11h X12h X13h X14h X15h X16h X17h 1 Conceição Junho 1 3 2.16137 2.20412 2.08991 1.72428 1.69897 1.62325 1.44716 1.51851 1.43136 1.47712 1.51851 1.04139 2 Conceição Junho 2 3 2.46538 2.13672 2.06819 1.97772 2.0 1.80618 1.64345 1.20412 1.62325 1.36173 1.69020 1.57978 3 Conceição Junho 3 3 2.53275 2.52504 2.49276 2.3 2.12710 2.26007 2.24551 1.95424 2.09342 1.04139 1.53148 1.17609 4 Conceição Junho 1 4 1.65321 2.16435 1.91381 1.75587 1.74036 1.17609 1.66276 1.51851 1.39794 1.04139 1.11394 1.04139 5 Conceição Junho 2 4 2.30320 1.71600 2.02531 2.05690 1.86332 1.66276 1.17609 1.04139 1.30103 1.27875 1.3 1.3 6 Conceição Junho 3 4 2.71012 2.30320 2.53403 1.80618 2.24551 2.20683 2.02531 1.07918 1.36173 1.39794 1.11394 1.93450 7 Conceição Junho 1 5 2.21748 1.99564 2.26007 2.28103 2.10380 1.41497 0.47712 1.07918 0.90309 1.04139 1.49136 1.23045 8 Conceição Junho 2 5 2.10721 2.16435 2.05308 2.38561 2.14613 1.61278 1.27875 0.47712 1.61278 1.0 1.44716 1.07918 9 Conceição Junho 3 5 1.62325 1.93450 2.33041 2.24797 2.29885 2.48001 2.29003 1.43136 1.49136 1.17609 1.41497 1.14613 10 Conceição julho 1 3 2.20952 2.01284 1.79239 1.59106 1.62325 1.51851 1.41497 1.38021 1.66276 1.46240 1.53148 1.66276 I have to create a new column (hour) and transpose just the last 12 columns, and first four columns have to be copied 12 time, like this (dput data on bottom - correct data-frame): Local Mês Dia Colonia Hora N 1 Conceição Junho 1 3 6h 2.161370 2 Conceição Junho 1 3 7h 2.204120 3 Conceição Junho 1 3 8h 2.089910 4 Conceição Junho 1 3 9h 1.724280 5 Conceição Junho 1 3 10h 1.698970 6 Conceição Junho 1 3 11h 1.623250 7 Conceição Junho 1 3 12h 1.447160 8 Conceição Junho 1 3 13h 1.518510 9 Conceição Junho 1 3 14h 1.431360 10 Conceição Junho 1 3 15h 1.477120 11 Conceição Junho 1 3 16h 1.518510 12 Conceição Junho 1 3 17h 1.041390 13 Conceição Junho 2 3 6h 2.465383 Some one could give me some ideas? I don't even know how to start... Thanks in advanced, -- Raoni Rosa Rodrigues Research Associate of Fish Transposition Center CTPeixes Universidade Federal de Minas Gerais - UFMG Brasil rodrigues.ra...@gmail.com wrong data frame: structure(list(Local = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = Conceição, class = factor), Mês = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L), .Label = c(julho, Junho ), class = factor), Dia = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L), Colonia = c(3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 3L), X6h = c(2.16137, 2.46538, 2.53275, 1.65321, 2.3032, 2.71012, 2.21748, 2.10721, 1.62325, 2.20952), X7h = c(2.20412, 2.13672, 2.52504, 2.16435, 1.716, 2.3032, 1.99564, 2.16435, 1.9345, 2.01284), X8h = c(2.08991, 2.06819, 2.49276, 1.91381, 2.02531, 2.53403, 2.26007, 2.05308, 2.33041, 1.79239), X9h = c(1.72428, 1.97772, 2.3, 1.75587, 2.0569, 1.80618, 2.28103, 2.38561, 2.24797, 1.59106), X10h = c(1.69897, 2, 2.1271, 1.74036, 1.86332, 2.24551, 2.1038, 2.14613, 2.29885, 1.62325), X11h = c(1.62325, 1.80618, 2.26007, 1.17609, 1.66276, 2.20683, 1.41497, 1.61278, 2.48001, 1.51851), X12h = c(1.44716, 1.64345, 2.24551, 1.66276, 1.17609, 2.02531, 0.47712, 1.27875, 2.29003, 1.41497), X13h = c(1.51851, 1.20412, 1.95424, 1.51851, 1.04139, 1.07918, 1.07918, 0.47712, 1.43136, 1.38021), X14h = c(1.43136, 1.62325, 2.09342, 1.39794, 1.30103, 1.36173, 0.90309, 1.61278, 1.49136, 1.66276), X15h = c(1.47712, 1.36173, 1.04139, 1.04139, 1.27875, 1.39794, 1.04139, 1, 1.17609, 1.4624), X16h = c(1.51851, 1.6902, 1.53148, 1.11394, 1.3, 1.11394, 1.49136, 1.44716, 1.41497, 1.53148), X17h = c(1.04139, 1.57978, 1.17609, 1.04139, 1.3, 1.9345, 1.23045, 1.07918, 1.14613, 1.66276)), .Names = c(Local, Mês, Dia, Colonia, X6h, X7h, X8h, X9h, X10h, X11h, X12h, X13h, X14h, X15h, X16h, X17h), row.names = c(NA, 10L), class = data.frame) Correct data frame:
[R] What is the . in formula ~. syntax?
I know if I have a dataframe with columns y, x1, x2 and I wish to have y as my y value and x1 and x2 as x values I can do: y ~ x1 + x2 or y ~. but can someone explain what . actually is or what its transposed into? I searched for this with no success, reading the formula manual pages. Brian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to override gWidgets:gvarbrowser_classes
How do I get gvarbrowser to display the contents of data.frame named, say atab1? That is modify the Filter by: entries to only show data.frame Also, how do I turn off the selection pull down box, given that I know the name of the data.frame variable? Basically, I will like to browse the variable name in a data.frame gdf(atab1, container =gwindow(Object browser), expand=TRUE) displays the data values, I an interest in showing just the names in a GUI environment similar to gvarbrowser I tried the following age=18:29; height=1:12; atab1 - data.frame(age=age,height=height) mydefaultclasses - list(Data sets1=c(data.frame) ) v - gvarbrowser( container =gwindow(Object broser), gvarbrowser_classes=mydefaultclasses ) This did not seem to work, displayed default selection entries I also tried: options(gWidgets:gvarbrowser_classes=mydefaultclasses) v - gvarbrowser( container =gwindow(Object broser)) -- View this message in context: http://r.789695.n4.nabble.com/How-to-override-gWidgets-gvarbrowser-classes-tp4650509.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] Odp: Object Browser
How do I get gvarbrowser to display only data.frame named, say atab1 or atab2 or atab*? Also, how do I turn off the selection pull down box? #I tried: mydefaultclasses - list(Data sets1=c(data.frame) ) # # Then v - gvarbrowser( container =gwindow(Object broser), gWidgets:gvarbrowser_classes=mydefaultclasses ) #-- # but it does not seem to work #I also tried: options(gWidgets:gvarbrowser_classes=mydefaultclasses) #then v - gvarbrowser( container =gwindow(Object broser)) #this too did not work #I would simply like to display the variable browser window for data.frame atab1 (say). Thank you -- View this message in context: http://r.789695.n4.nabble.com/Object-Browser-tp2594912p4650511.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] Barplot with lines
Hi, I'm trying to plot stacked barplot with lines on it. Here is the data. emp days val1 val2 score 1 21 1 0 1200 2 35 1 1 na 3 42 na na 3000 4 53 2 1 2100 5 64 1 0 na 6 73 na na 1400 My X-axis is days. I'm looking to plot val1,val2 as stacked bars and score as lines with different y-axis. I could get the bar plot and lines on it but the problem is the bars and lines are not aligning well for same point on X-axis. Note, that there might not be a score for some when there is val1 and val2 or vice versa. Any help in this regard is highly appreciated. Thanks in Advance! SK [[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.
