regexp capturing group in R

mdat <- gregexpr("[[:digit:]]{8}", txt)
dates <- mapply(function(x, y) substr(txt, x, x + y - 1), mdat[[1]],

dates

-----Original Message-----
From: r-help-bounces at r-project.org 
[mailto:r-help-bounces at r-project.org] On Behalf Of 
pierre at demartines.com
Sent: Tuesday, February 24, 2009 7:23 PM
To: r-help at r-project.org
Subject: [R] regexp capturing group in R

Hello,

Newbie question: how do you capture groups in a regexp in R?

Let's say I have txt="blah blah start=20080101 end=20090224".
I'd like to get the two dates start and end.

In Perl, one would say:

my ($start,$end) = ($txt =~ /start=(\d{8}).*end=(\d{8})/);

I've tried:

txt <- "blah blah start=20080101 end=20090224"
m <- regexpr("start=(\\d{8}).*end=(\\d{8})", filename, 
perl=T); dates = substring(filename, m, m+attr(m,"match.length")-1);

but I get the whole matching substring...

Any idea?

~Pierre