> txt <- "blah blah start=20080101 end=20090224"
> nums <- sub(".*start=(\\d+).*end=(\\d+).*", "\\1 \\2", txt, perl=TRUE)
> nums <- strsplit(sub(".*start=(\\d+).*end=(\\d+).*", "\\1 \\2", txt,
> perl=TRUE), ' ')
> nums
[[1]]
[1] "20080101" "20090224"
On Tue, Feb 24, 2009 at 7:23 PM, <pie...@demartines.com> wrote:
> 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
>
> ______________________________________________
> R-help@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.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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