[R] backreferences in gregexpr

[arun]

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

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