Here is another way to do it. Fairly short and clean, although I ended up writing a complete script as a proof of concept:

#!perl

use strict;

my $numClips=15;
my @types  = ("Menu", "Audio", "Video", "Snacks");
my @answer = ("Yes", "No", "N/A");
my (%file, %counttype, %yes, %no);

## Generate some random data for our test:
for (0..$numClips-1) {
  $file{$_}{FILETYPE}  = @types[rand(@types)];
  $file{$_}{VOICEOVER} = @answer[rand(@answer)];
}

pop(@types); ## Remove "Snacks" for our test
my $types = join('|', @types);
## Or just hardcode the above as: $types = "Menu|Audio" etc...

## The main magic:
for my $clipnum (0..$numClips-1) {
  if ($file{$clipnum}{FILETYPE} =~ /^($types)$/o) {
    $counttype{$1}++;
    $file{$clipnum}{VOICEOVER} eq "No" ? $no{$1}++ : $yes{$1}++;
  }
}

print "Results:\n";
for $a (sort keys %counttype) {
  print "$a: $counttype{$a}\n";
  print "  Yes: $yes{$a}\n";
  print "  No:  $no{$a}\n";
}

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In reply to Re: Keeping a count of matches in a hash that satisfy more than 1 condition by turnstep
in thread Keeping a count of matches in a hash that satisfy more than 1 condition by dmtelf

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