backreferences in gregexpr
HI,
I am not sure whether this helps:
temp1<-regmatches(temp, gregexpr("(?:abcd)(1234)", temp)) #your code
substr(unlist(temp1),5,8)
#[1] "1234" "1234"
A.K.
----- Original Message -----
From: Alexander Shenkin <ashenkin at ufl.edu>
To: r-help at r-project.org
Cc:
Sent: Friday, November 2, 2012 6:02 PM
Subject: [R] backreferences in gregexpr
Hi Folks,
I'm trying to extract just the backreferences from a regex.
temp = "abcd1234abcd1234"
regmatches(temp, gregexpr("(?:abcd)(1234)", temp))
[[1]] [1] "abcd1234" "abcd1234" What I would like is: [1] "1234" "1234" Note: I know I can just match 1234 here, but the actual example is complicated enough that I have to match a larger string, and just want to pass out the backreferenced portion. Any help greatly appreciated! Thanks, Allie ______________________________________________ R-help at 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.