Re: How to use "less than" and "greater than" inside a regex for a $variable number

[LanX]

don't really think I understand your task, but

> ...found that I can use an "if-then" type of expression within a regex, but have found no examples for how to do this. How would you do this?

see perlretut#Conditional-expressions

HTH

[Polyglot]

I appreciate the link, but that resource only tells how to work with a known quantity. I need to be able to match a variable number, conditional upon its relative value when compared to another number. In essence, I need to match based on the comparison result.

For example, how would one do something like this?

$string = "I have 5 apples, 6 oranges, and 8 limes.";

#Match the oranges only if they are more than the apples
#and fewer than the limes.
$string =~ m/(\d+)\sapples.*(\d+)\soranges.*(\d+)\slimes(?{if (($1<$2)
+ && ($2<$3))})/g;

If I had something like that I could then plug in $var for $2.

[AnomalousMonk]

The  (*F) operator was introduced with 5.10. Prior to that,  (?!) can be used.

>perl -wMstrict -le
"for my $n (4 .. 9) {
   my $str = qq{I have 5 apples, $n oranges, and 8 limes.};
   print qq{'$str'};
   next unless $str =~ m{
     (\d+) \s+ apples  \D+
     (\d+) \s+ oranges \D+
     (\d+) \s+ limes
     (?(?{ $1 < $2 && $2 < $3 }) | (*F) )
     }xms;
   print qq{'$2'};
   }
"
'I have 5 apples, 4 oranges, and 8 limes.'
'I have 5 apples, 5 oranges, and 8 limes.'
'I have 5 apples, 6 oranges, and 8 limes.'
'6'
'I have 5 apples, 7 oranges, and 8 limes.'
'7'
'I have 5 apples, 8 oranges, and 8 limes.'
'I have 5 apples, 9 oranges, and 8 limes.'

[Polyglot]

I've implemented this approach, as it seems fairly close to the sort of solution I was looking for. Unfortunately, it is still rather slow. I started the process 2.5 days ago now (it's been running over 60 hours) and it is about half-way through the material. So it appears with this method it will take 5 days of 100% CPU on one of four cores of my Dell PowerEdge server. That's a little disappointing. My ugly approach, which may be slightly less thorough, finished after about three days. So it was 40% quicker.

Given the complexity of the regex, I suppose I cannot blame perl or the program itself, it's just the way it is. But without the attempt to narrow the search to finding numbers between their respective forerunners/postrunners, the whole search can complete in less than five minutes.

Anyway, at least I have learned something and I much appreciate your patience in demonstrating this method for me. I may still be able to use this as a final check over a long weekend or something, or perhaps I can limit the amount of material to be checked at a time (~130 books total). Thank you!

[AnomalousMonk]

Actually, I think the problem can be addressed without the need for exotic regex operators or constructs (although this uses the  \K operator introduced with 5.10). Unfortunately, this approach involves the replacement of a substring with the identical substring, an operation that I do not think the regex compiler can optimize away and that therefore may lead to a bit of inefficiency.

>perl -wMstrict -le
"my $book = qq{pg. 1 foo  pg. 2 bar baz  pg. 4 fee fie  pg. 5 foe \n}
          . qq{fum  pg. 6 hoo ha  pg. 9 deedle  pg. 10 \n}
          . qq{blah blah  pg. 14 noddle \n}
          ;
 print qq{[[$book]] \n};
 ;;
 my $pn = qr{ pg[.] \s+ }xms;
 $book =~
   s{ $pn (\d+) \K  (.*?)  (?= $pn (\d+))             }
    { my $m = missing($1, $3);  $m ? qq{$2$m  } : $2; }xmsge;
 print qq{(($book)) \n};
 ;;
 sub missing {
   my ($i, $j) = @_;
   ;;
   return if $j - $i < 2;
   ;;
   my ($ii, $jj) = ($i + 1, $j - 1);
   return $j - $i > 2
        ? qq{(pages $ii - $jj missing)}
        : qq{(page $ii missing)}
        ;
   }
"
[[pg. 1 foo  pg. 2 bar baz  pg. 4 fee fie  pg. 5 foe
fum  pg. 6 hoo ha  pg. 9 deedle  pg. 10
blah blah  pg. 14 noddle
]]

((pg. 1 foo  pg. 2 bar baz  (page 3 missing)  pg. 4 fee fie  pg. 5 foe
fum  pg. 6 hoo ha  (pages 7 - 8 missing)  pg. 9 deedle  pg. 10
blah blah  (pages 11 - 13 missing)  pg. 14 noddle
))