R help - Nov 2012 - 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

On Fri, Nov 2, 2012 at 6:02 PM, Alexander Shenkin <ashenkin at ufl.edu>
wrote:
> 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!
>
Try this:
> library(gsubfn)
> strapplyc(temp, "abcd(1234)")
[[1]]
[1] "1234" "1234"

-- 
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GKX Group, GKX Associates Inc.
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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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