similar to: perhaps regular expression bug with | sign ??

I am working with Sweave and would like to print out into my latex document the result of the R command version$platform So what I first tried in my .Rnw document was \Sexpr{print(version$platform)}. However, the output from this command is the string "x86_64-apple-darwin10.8.0" (without the quotes). This contains an underscore, which is a special character in tex and so I get an error

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20100208/52a6d080/attachment.pl>

Hello, I have several strings where I am trying to eliminate the period and everything after the period, using a regular expression. However, I am having trouble getting this to work. > x = "wa.w" > gsub(x, "\..*", "", perl=TRUE) [1] "" Warning messages: 1: '\.' is an unrecognized escape in a character string 2: unrecognized escape removed

If I run cvec<-c("test.f", "test.sf", "try.g","try.res", "try.f") print(cvec) indx<-grep('\.f',cvec,perl=TRUE) fset<-cvec[indx] print(fset) I get > cvec<-c("test.f", "test.sf", "try.g","try.res", "try.f") > print(cvec) [1] "test.f" "test.sf"

Full_Name: Allen McIntosh Version: 2.0.0 OS: RedHat 9.0 Submission from: (NULL) (67.80.175.118) The POSIX time printing routine gives strange results when asked to print a time that is exactly midnight: TZ=CST6CDT R -q --no-save > strptime("10/5/2004 00:00:01 CDT", "%m/%d/%Y %H:%M:%S %Z") [1] "2004-10-05 00:00:01" > strptime("10/5/2004 00:00:00

What's really the problem with: > regexpr( '\.odt$', "xxxxYodt", perl=TRUE ) Warning: '\.' is an unrecognized escape in a character string Warning: unrecognized escape removed from "\.odt$" [1] 5 attr(,"match.length") [1] 4 I know that I could use: > regexpr( '[.]odt$', "xxxxYodt", perl=TRUE ) But it seems to me that

Hi i am using a script which involves the following loop. It attempts to reduce a data frame(zz) of 95000 * 41 down to a data frame (averagedreplicates) of 95000 * 21 by averaging the replicate values as you can see in the script below. This script however is very slow (2days). Any suggestions to speed it up. NB I have also tried using rowMeans rather than adding the 2 values and dividing by 2.

Hi I am trying to create a loop which averages replicates in my data. The original data has many rows. and consists of 40 column zz[,2:41] plus row headings in zz[,1] I am trying to average each set of values (i.e. zz[1,2:3] averaged and placed in average_value[1,2] and so on. below is my script but it seems to be stuck in an endless loop Any suggestions?? for (i in 1:length(average_value[,1])) {

Dear readers of the list, I have a problem a comparison of two data from a vector. The comparison yields FALSE but should be TRUE. I have checked for mode(), length() and attributes(). See the following code (R2.10.0): ----------------------------------------------- # data vector of 66 double data X =

--- On Fri, 15/5/09, Amit Patel <amitrhelp at yahoo.co.uk> wrote: > From: Amit Patel <amitrhelp at yahoo.co.uk> > Subject: Help with loops > To: r-help at r-project.org > Date: Friday, 15 May, 2009, 12:17 PM > Hi > I am trying to create a loop which averages replicates in > my data. > The original data has many rows. and consists of 40 column > zz[,2:41]

How do I remove a "$" character from a string sub() and gsub() with "$" or "\$" as pattern do not work. > sub("$","","ABC$DEF") [1] "ABC$DEF" > sub("\$","","ABC$DEF") Error: '\$' is an unrecognized escape in character string starting "\$" >

Hi R users I try to use the lme but I can?t!!!!! My script is (some words in french, sorry!!): rm(list=ls(all=TRUE)) #Efface tous les objets en m?moire pour ?viter des erreurs library(MASS) #Chargement des Librairies library(car) library(Hmisc) library(tkWidgets) library(svDialogs) library(multtest) library(nlme) #Rep <- "C:/Documents and Settings/U3M/Bureau/steph/Scripts

I have installed both 32-bit and 64-bit versions of R2.12.0 (2010-06-15 r52300) on my Ubuntu 10.04 64-bit system. I observe the following behavior when running the examples from base::connections. There appears to be a problem with seek() on a .gz file when using a 32-bit installation of R2.12.0, but the problem doesn't appear in the 64-bit installation. I realize that seek() has been

Hello, I'm trying to use the package ttda, wich is involved in text analysis, for my own data about answers in a company survey. I've installed it, as well as ispell, but when trying to use an example: > zz <- file("stupid.txt", "w") # build a data file > cat("{comment - stupid data file} \n" , file = zz) > cat("<uci=1>

Full_Name: Gavin Simpson Version: 2.1.0-patched (1-Jun-2005) OS: Linux (Fedora Core 3) Submission from: (NULL) (128.40.32.76) axis.Date() insists on labelling tick marks. It could be made more flexible by allowing the user to specify if they want the ticks to be labelled, for example, to add un-labelled minor ticks for "months", added to a plot with "years" labelled. The user

Hi, I am currently writing code to input a few thousand files, run them through the Random Forests package, and then output corresponding results. When I use the code below: zz<-textConnection("ex.lm.out", "w") sink(zz)

Hello, Sorry, I know the topic has been talked about already but I searched the archives and I'm still unsure of the proper way to have dovecot (imap-login) listen to several addresses and/or several ports. 1. how do I translate dovecot-1 protocol imap { listen = xx.xx.xx.xx:143 yy.yy.yy.yy:143 ssl_listen = xx.xx.xx.xx:993 yy.yy.yy.yy:939 } I started by service imap-login {

What is the R way of computing cumulative growth rates given a series of discrete values indexed . For instance, given a matrix of 20 observations for each of 5 series (zz), what is the most straight forward technique in R for computing cumulative growth (zzcum) ? It seems for the solution I'm after might be imbedding the following cum growth rate calc as a function into a function call

I've never had any issues with the way that plot.POSIXct chooses the labels of the date axis before, but in this particular case it's output is a little confusing. plot(seq(as.POSIXct("1997-10-01"),length.out=108,by="month"),rnorm(108)) This command produces a chart with every x tick mark labeled as "Jan 01". I can replicate this chart by adding the format

Can someone explain why this happens when one of the list elements is named 'all'? > zz <- list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) ) > names(zz)<-c('test','bar','foo') > do.call(cbind,zz) test bar foo 1 1 101 201 2 2 102 202 3 3 103 203 4 4 104 204 5 5 105 205 6 6 106 206 7 7 107 207 8 8 108 208 9