similar to: Number Regular Expressions

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

Dear r-help members, I have a number in the form of a string, say: a<-"-01020.909200" I'd like to extract "1020." as well as ".9092" Front<-grep(pattern="[1-9]+[0-9]*\\.", value=TRUE, x=a, fixed=FALSE) End<-grep(pattern="\\.[0-9]*[1-9]+", value=TRUE, x=a, fixed=FALSE) However, both strings give "-01020.909200", exactly

Ist there a possibility to write a R-Script using something like #!/usr/bin/R and then alle the requestet commands? Of course "R CMD BATCH" exists, but there was (for me) no possibility to write something like a HERE-script. regards erkan -- ?ber den grenzen mu? die freiheit wohl wolkenlos sein

Dear Everyone, I try to automatically manipulate the data of a variable (class = factor) like x 220 220a 221 221b B221 Into two variables (class = numeric) like x y 220 0 220 1 221 0 221 1 221 1 y has to carry the information about the class (number or string) of the former x-Variable. I could do it by hand like x[x == "220a"] <- 220

Hello, I have a date in the format Year-Month Name (e.g. 1990-January) and R classes it as a character. I want to convert this character into a date format, but when I try as.Date(1990-January, "%Y-%B"), I get back NA. The function strptime also gives me NA back. Thanks. [[alternative HTML version deleted]]

Is there a way to have an R program send an email? Something like this: address <- 'abc at d.com' text <- 'This is the email body' send.email(address, text) Thanks. FS

Hi, just curious, but does anyone know the source/reason of observing the following error on OSX but not on WinXP and Linux? I've tried with a few different versions of R (v2.5.1, v2.6.1, v2.6.2, v2.7.0devel). The locale does not seem to affect the error, i.e. I've tested a few different and it is still only OSX that gives the error but not the other two. > regexpr("[a-Z]",

Dear All, My query is: Do we always need to use perl = TRUE option when doing ignore.case=TRUE? A small example : my_text = "RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI VS RAMESH GUPTA.\n Dated: 01.03.2016 Item no.01\n Present: Ms. Sonakshi, the proxy counsel for Ms. Usha Singh, the counsel for ARCIL.\n None for the CDs.\n

Hi there, I know, this question is not directly an R-help question but probably someone can give me a hint how to deal with the following problem. I have a vector with file/folder names and want to filter for all entries which have 6 numbers in a row and nothing else. folders <- c("folder1", "f2", "F234562", "12345678", "234567",

Hello all, Let's say I have a character string "Race-ethnicity-----coding information" I want to extract all text before the multiple dashes, including the word "ethnicity." I wrote a handy function to extract the first matched text: grepcut <- function(pattern,x){ start.and.length <- regexpr(pattern,x) substring(x,start.and.length,start.and.length

Zitat von Ashim Kapoor <ashimkapoor at gmail.com>: > Dear All, > > My query is: > > Do we always need to use perl = TRUE option when doing ignore.case=TRUE? > > A small example : > > my_text = > "RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI > VS RAMESH GUPTA.\n Dated: 01.03.2016 Item no.01\n > Present:

Hi, I have a data set that I read in using readLines, but R classes it as a character string. The data set has multiple columns and rows, but I only want the first 2 columns in the form of a matrix. Thanks for the help. [[alternative HTML version deleted]]

Dear Sirs, while > regexpr('(.{1,2})\\1', 'foo') [1] 2 attr(,"match.length") [1] 2 attr(,"useBytes") [1] TRUE yields the correct match, an incremented upper bound in > regexpr('(.{1,3})\\1', 'foo') [1] -1 attr(,"match.length") [1] -1 attr(,"useBytes") [1] TRUE incorrectly yields no match. R versions tested: 2.11.1

I want to make sure this piece of code I wrote is doing what I want it to do. ll<-function(string) { grep(string,dir(),value=T) } subdir = ll("Coverage_[1-9][0-9]$") I basically wrote a little function that would grab all the files of form Coverage_[0-99] The way I wrote it, will it grab Coverage_5 or does it have to have 2 numbers (10-99)? -- View this message in context:

Hello, This may be a trivial question, but I don't find the answer in R online help. Under Windows, I can copy/paste to the clipboard using readClipboard()/writeClipboard(), or something like cat(..., file = "clipboard"). Are there equivalent function for other platforms? Best, Philippe Grosjean ..............................................<??}))><........ ) ) ) ) ) ( (

Hello! Are there any additional packages (besides "regex") I could use to incorporate regular expressions in my R code? I would be grateful for any tips or help you could provide! Yours, Ann Loraine

Hi all, Is there a method for escaping strings to be used regular expressions? i.e. if I have a user supplied string that I'd like to use as a fixed component is there a method that will turn (e.g.) ".$^" into "\\.\\$\\^" ? Thanks, Hadley -- http://had.co.nz/

Dear R-help, Could someone please try to explain this paradox to me? What is more likely to show up first in a string of coin tosses, "Heads then Tails", or "Heads then Heads"? ##generate 2500 strings of random coin flips ht <- replicate(2500, paste(sample(c("H", "T"), 100, replace = TRUE), collapse =

Hi, I am trying to use sub, regexpr on expressions like log(D) ~ log(N)+I(log(N)^2)+log(t) being a model specification. The aim is to produce: "ln D ~ ln N + ln^2 N + ln t" The variable names N, t may change, the number of terms too. I succeded only partially, help on regular expressions is hard to understand for me, examples on my case are rare. The help page on R-help

I'm trying to write a gsub() call that takes a string and escapes all the unescaped quote marks in it. So the string \" would be left unchanged, but \\" would be changed to \\\" because the double backslash doesn't act as an escape for the quote, the first just escapes the second. I have the usual problems of writing regular expressions involving backslashes which