is any of array items contained in a string

Question

I have a keywords list and a blacklist. I want to delete all keywords that contain any of blacklist item. At the moment Im doing it this way:

my @keywords = ( 'some good keyword', 'some other good keyword', 'some bad keyword');
my @blacklist = ( 'bad' );

A: for my $keyword ( @keywords ) {
    B: for my $bl ( @blacklist ) {
        next A if $keyword =~ /$bl/i;      # omitting $keyword
    }
    # some keyword cleaning (for instance: erasing non a-zA-Z0-9 characters, etc)
}

I was wondering is there any fastest way to do this, becouse at the moment I have about 25 milion keywords and couple of hundrets words in blacklist.

Answer 1

The most straightforward option is to join the blacklist entries into a single regular expression, then grep the keyword list for those which don't match that regex:

#!/usr/bin/env perl    

use strict;
use warnings;
use 5.010;

my @keywords = 
  ('some good keyword', 'some other good keyword', 'some bad keyword');
my @blacklist = ('bad');

my $re = join '|', @blacklist;
my @good = grep { $_ !~ /$re/ } @keywords;

say join "\n", @good;

Output:

some good keyword
some other good keyword

Answer 2

Precompiling the search may help my @blacklist = ( qr/bad/i ) if you want to keep the nested loops.

Alternatively, changing from my @blacklist = ( 'bad', 'awful', 'worst' ) to my $blacklist = qr/bad|awful|worst/; and then replacing the inner loop with if ( $keywords[$i] =~ $blacklist ) ....

Answer 3

This should do it:

my @indices;
for my $i (0..$#keywords) {
  for my $bl (@blacklist) {
    if ($keywords[$i] =~ $bl) {
      push(@indices, $i);
      last;
    }
  }
}
for my $i (@indices) {
  @keywords = splice(@keywords, $i);
}