Combinatorial regexing

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Hello monks, Code:

$string="abc de fgh ijkl";

while($string =~ /([a-z]+ [a-z]+)/g)
{
    print "$1 $1\n";
}
[download]

Outputs:

abc de
fgh ijkl
[download]

I would like it to output:

abc de
de fgh
fgh ijkl
[download]

I have tried doing:

$string =~ /([a-z]+) ([a-z]+)\G(?: [a-z])*?/g
[download]

and using lookarounds but I haven't been able to find the right combination of \G, lookarounds and matching to come out with what I want. I am also aware that I could just do:

my @words =split $string;
for($i=0;$i<$#words;$i++)
{
   print "$words[$i] $words[$i+1]\n";
}
[download]

I trying to stick to regex's. Thanks, Alex

Comment on Combinatorial regexing Select or Download Code

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Re: Combinatorial regexing by blokhead (Monsignor) on May 04, 2007 at 18:36 UTC
Try something like this: `while ($string =~ / (?=([a-z]+[ ][a-z]+)) [a-z]+ /gx) {` [download] This says: look ahead to capture the next two words, without advancing pos. Then with that stuff captured, advance pos as far as the first word. However, I don't see much wrong with using @words like you've done in your last example... blokhead	[reply] [d/l]
Re: Combinatorial regexing by johngg (Canon) on May 04, 2007 at 22:27 UTC
This uses a look-ahead but in a slightly different way to blokhead's solution. The regex looks for and captures a letter sequence then uses a look-ahead with a capture to find another sequence. `use strict; use warnings; my $string = q{abc de fgh ijkl}; my $rxComb = qr {(?x) (?: ([a-z]+) \s+ (?=([a-z]+)) ) }; while ( $string =~ m{$rxComb}g ) { print qq{$1 $2\n}; }` [download] Here's the output. `abc de de fgh fgh ijkl` [download] I hope this is of use. Cheers, JohnGG	[reply] [d/l] [select]
Re: Combinatorial regexing by shandor (Monk) on May 04, 2007 at 19:15 UTC
How about this? `use warnings; use strict; my $string = "abc de fgh ijkl"; my @array = split /\s+/, $string; my $count = 0; while ($count < $#array) { print $"array[$count] $array[$count+1]\n"; $count++; }` [download] It's not a regexp, but, it displays the output you want. UPDATE: when you read the entire OP, you find funny things... like the fact that the OP-er listed this solution. Oh, well, made me not have to do real work.	[reply] [d/l]