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Think about Loose Coupling
 
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But if your keys are just numbers, then the array index can be the number you need. Note that in my example, I used undef as a place holder for array elements that did not have an entry. Unless your indices are not even remotely sequential or you have non-integer indices, an array will be faster, take less memory and will apply a useful constraint on your keys.

If there is a possibility that your arrays or hashes do not have equivalent keys, it is considered best practice to test prior to processing those values. Modifying the previously posted codes:

Hashes:

#!/usr/bin/perl use strict; use warnings; use Data::Dumper; my %hash1 = ("1", "20", "2", "20", "4", "19", "5", "20", "6", "18", "7", "20"); my %hash2 = ("1", "19", "2", "20", "4", "16", "5", "19", "6", "20", "8", "20"); my %differences; for my $key ( sort {$a <=> $b} keys %hash1) { unless (defined $hash2{$key}) { print "$key is in \%hash1 but not \%hash2\n"; next; } $differences{$key} = $hash1{$key} - $hash2{$key}; print "$key ($hash1{$key} - $hash2{$key}) = $differences{$key}\n"; } print "\n", Dumper \%differences; __END__ 1 (20 - 19) = 1 2 (20 - 20) = 0 4 (19 - 16) = 3 5 (20 - 19) = 1 6 (18 - 20) = -2 7 is in %hash1 but not %hash2 $VAR1 = { '6' => -2, '4' => 3, '1' => 1, '2' => 0, '5' => 1 };

Arrays:

#!/usr/bin/perl use strict; use warnings; use Data::Dumper; my @array1 = (undef,20,20,19,20,undef,18,20); my @array2 = (undef,19,20,16,19,undef,20,undef,20); my @differences; my $top = $#array1 > $#array2 ? $#array1 : $#array2; for my $i (1 .. $top) { if (defined $array1[$i] and not defined $array2[$i]) { print "$i is in \@array1 but not \@array2\n"; } if (not defined $array1[$i] and defined $array2[$i]) { print "$i is in \@array2 but not \@array1\n"; } next unless defined $array1[$i] and defined $array2[$i]; $differences[$i] = $array1[$i] - $array2[$i]; print "$i ($array1[$i] - $array2[$i]) = $differences[$i]\n"; } print "\n", Dumper \@differences; __END__ 1 (20 - 19) = 1 2 (20 - 20) = 0 3 (19 - 16) = 3 4 (20 - 19) = 1 6 (18 - 20) = -2 7 is in @array1 but not @array2 8 is in @array2 but not @array1 $VAR1 = [ undef, 1, 0, 3, 1, undef, -2 ];

Note that in the array case, I will not miss that the second set has a value that the first one is missing.


In reply to Re^3: How best to compare hash values? by kennethk
in thread How best to compare hash values? by jimbass

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