Search and replace dynamically in perl regex

Question

$string = "On 0 8 February 2 0 1 4 , he visited the fair";

I want to replace the spaces between the numbers, so it becomes "On 08 February 2014, he visited the fair"

How can I do this with regex? I can do a for loop index by index, but given large amounts of text, it will be slow.

This is an idea of what I'm trying to achieve:

$string =~ s/([0-9]\s)+/substr($string,$-[0],$+[0]-$-[0])/g;

but it won't work since substr is not treated as a function within the regex. Any ideas?

mpapec · Accepted Answer · 2014-09-03 07:31:57Z

3

You can use look behind, and look ahead to filter such spaces,

$string =~ s/[0-9]\K [ ]+ (?=[,0-9])//xg;

answered Sep 3, 2014 at 7:31

mpapec

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1 Comment

I did this $string =~ s/[0-9]\K [ ]+ (?=[[:punct:]0-9])//xg;

Jens · Accepted Answer · 2014-09-03 07:32:29Z

0

Try this code:

do{
    $string =~s/(\d+)\s+(\d+)/$1$2/g;
} while($string =~/(\d+)\s+(\d+)/);

answered Sep 3, 2014 at 7:32

Jens

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vishnukg · Accepted Answer · 2014-09-03 10:52:10Z

0

You can also try the following regex $string =~ s/(?<=\d)\s+(?![a-z])//ig;

answered Sep 3, 2014 at 10:52

vishnukg

1