similar to: regex question

\b is word boundary. But, unexpectedly, strsplit("dia ma", "\\b") splits character by character. > strsplit("dia ma", "\\b") [[1]] [1] "d" "i" "a" " " "m" "a" > strsplit("dia ma", "\\b", perl=TRUE) [[1]] [1] "d" "i" "a" " "

Dear list! > gregexpr("a+(b+)", "abcdaabbc") [[1]] [1] 1 5 attr(,"match.length") [1] 2 4 What I want is the offsets of the matches for the group (b+), i.e. 2 and 7, not the offsets of the complete matches. Is there a way in R to get that? I know about gsubgn and strapply, but they only give me the strings matched by groups not their offsets. I could write

Gregexpr - extract results with lapply Hello, I need to extract sequences of three upper case letters in a string. In other words, in this string: str <-c("ABC", "this WOUld be gOOD") The result I'm looking for is ABC WOU OOD. With gregexpr, I can get the position and length of the sequences gregexpr('[A-Z]{3}',str,perl=TRUE) [[1]] [1] 1

I need to capture matching words in a string, any ideas ? I tried using gregexpr, but it was no help. In this example, I need to capture ID23423424 and ID324234325 > s <- "sID23423424 apple pID324234325 orange"> gregexpr("ID[0-9]+", s)[[1]][1] 2 20attr(,"match.length")[1] 10 11 _________________________________________________________________

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>

Hi, I have a vector where an integer is followed by a randomly varying number of NA elements. I want to replace the NA's with the preceding integer. Let me explain with a simple?example : avec<- c(1,NA,NA,5,NA,3,4,NA,NA,NA,NA,9,NA,0,NA,NA) I want to get the following vector from a : bvec<-c(1,1,1,5,5,3,4,4,4,4,4,9,9,0,0,0) How can I do this with R? Will appreciate all the help that I

I'm new to R and very excited about its possibilities. But I'm struggling with some very simple things, probably because I haven't found the correct documentation. Here's a simple example which illustrates several of my problems. Suppose I want to have a regexp match against a string, and return all the matching substrings in a vector of strings. regexp <-

For regular expression afficianados, I'd like a cleverer solution to the following problem (my solution works just fine for my needs; I'm just trying to improve my regex skills): Given the string (entered, say, at a readline prompt): "1 2 -5, 3- 6 4 8 5-7 10" ## only integers will be entered parse it to produce the numeric vector: c(1, 2, 3, 4, 5, 3, 4, 5, 6, 8, 5, 6,

The interpretation of regular expressions with repetition quantifiers in the 'gregexpr' function seems to have changed between R Version 2.2.0 and 2.3.0. The 'gsub' function, however, gives the same results in R Versions 2.2.0 and 2.3.0. Below is an example that demonstrates the version differences of the 'gregexpr' function. I am not sure whether this new behavior is

hi, is there an easy way to extract numbers from a string? for example I have; "this Item costs 3.32 Dollars" is there an easy way to extract the 3.32 as a number? thanks!

Hi, I have a silly question regarding the usage of two commands: read.table and gregexpr： For read.table, if I read a matrix and set header = T, I found that all the dash ("-") becomes dots (".") A = read.table("Matrix.txt", sep = "\t", header = F) A[1,1] # "A-B-C-D". A = read.table("Matrix.txt", sep = "\t", header = T)

Dear list, I am trying to use gregexpr to see if entries in a dataframe have either of two possible values for a string. here's an example text<-c("fat", "rat", "cat", "dog", "log", "fish") If I just wanted to find if any one of the elements in text match the pattern "at" I would do gregexpr("\\at", text)

completion is semi-broken in today's r-devel, and the reason seems to be some regular expression changes: > sessionInfo() R version 2.6.0 Under development (unstable) (2007-05-22 r41673) i686-pc-linux-gnu locale: [...] attached base packages: [1] "stats" "graphics" "grDevices" "utils" "datasets" "methods" [7]

Hi all, Several people have noticed that gregexpr is very slow for large subject strings when perl=TRUE is specified. - https://stackoverflow.com/questions/31216299/r-faster-gregexpr-for-very-large-strings - http://r.789695.n4.nabble.com/strsplit-perl-TRUE-gregexpr-perl-TRUE-very-slow-for-long-strings-td4727902.html - https://stat.ethz.ch/pipermail/r-help/2008-October/178451.html I figured out

Hi I have this problem where I need to find if there is any numbers in a string, this is no problem if theres only one number per string. I would then simply use the regexpr() funtion togheter with the substring function to extract the number. But regexpr only picks one number per string either from the beginning or the end, but not multiple. Can this be done? And how for example My string <-

Full_Name: Peter Dolan Version: 2.5.1 OS: Windows Submission from: (NULL) (128.193.227.43) gregexpr does not find all matching substrings if the substrings overlap: > gregexpr("abab","ababab") [[1]] [1] 1 attr(,"match.length") [1] 4 It does work correctly in Version 2.3.1 under linux.

What is the simplest way to extract a matched subexpression? Eg. in perl you can do "hello world" =~ m/hello (.*)/ which would return 1(true) and set $1 to the matched subexpression "world". -- View this message in context: http://n4.nabble.com/regular-expression-submatch-tp1459146p1459146.html Sent from the R help mailing list archive at Nabble.com.

Hi R-users and R-experts, I am having a hard time in figuring out how to tackle regex questions where the "backslash" character is an integral part of the string. Let me explain how I?came across?this problem : I wanted to clearly see all the components in the windows environmental path variable. This is a long string.?For easy readability, I wanted to split up this string so that each

Hi, From R 2.6, I would like to update to R2.7. I would like to have some tips on the recommended method of installing the latest versions of an entire list of packages in R2.7 - i.e. all the packages that I have presently installed in R2.6. I am hoping that there is an easier method than fetching the packages individually as I did, to begin with, for R2.6. Additionally, I would like to install

I've been trying to use the VMAuthenticate function in 1.2+. This function is supposed to "behave[s] the same way as the Authenticate application, but the passwords are taken from voicemail.conf." The problem is that it always gives the "comedian mail" prompt and requests the mailbox number, even though I provide the mailbox number already. The upshot is that