Re: Comparing unordered hierarchical arrays

Update: I misread your post. Sorry.

~~This might be a little quicker as it avoids the sorting~~

#! perl -slw
use strict;

sub cmpAoAs{
    my( $a, $b ) = @_;
    my %h;
    $h{ $_ }++ for map{ @$_ } @$a;
    $h{ $_ }-- for map{ @$_ } @$b;
    return 0 == grep{ $_ } values %h;
}

my $ha = [ [1,2], [5,5,5], [1,2], [3,4] ];
my $hb = [ [4,3], [1,2], [2,1], [5,5,5] ];

print 'Same' if cmpAoAs $ha, $hb;
__END__
P:\test>342841
Same
[download]

However, on the surface of what you've told us, it loooks a little like your using the wrong structure if only the presence and count of a value (rather than it's position) determine equivalence. But I realise that I am not party to the full sp on your application:)

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail

Comment on Re: Comparing unordered hierarchical arrays Download Code

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Re: Re: Comparing unordered hierarchical arrays by Zaxo (Archbishop) on Apr 06, 2004 at 06:47 UTC
I don't think `my $ha = [ [1,2], [5,5,5], [1,2], [3,4] ]; my $hb = [ [1,3], [5,5,5], [1,2], [2,4] ];` [download] are supposed to compare equal, as your function will return. After Compline, Zaxo	[reply] [d/l]
Re: Re: Comparing unordered hierarchical arrays by ant9000 (Monk) on Apr 06, 2004 at 06:51 UTC
Your solution seems flawed to me: consider `my $ha = [ [1,2], [5,5,5], [1,2], [3,4] ]; my $hb = [ [4,3], [1,2], [2,1,5], [5,5] ];` [download] These are clearly different, but your code still outputs "same" since it is discarding the entire original structure (and not only the order of elements).	[reply] [d/l]