in reply to Is one array a subset of the other

As soon as you're thinking of "sets" of unique items and not particularly interested in their order, use hashes.

Untested. 

my %a1; @a1{qw(a b c d e f g)} = ();
my %a2; @a2{qw(b g)} = ();

my @a1only = grep { not exists $a2{$_} } keys %a1;
my @a2only = grep { not exists $a1{$_} } keys %a2;
my @both = grep { exists $a2{$_} } keys %a1;
my %u; my @either = grep { not $u{$_}++ } (keys %a1, keys %a2);

# your question:
print "subset" if not @a2only;

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Re: Re: Is one array a subset of the other
by pzbagel (Chaplain) on May 19, 2003 at 20:10 UTC

    Why all the hashes? Array lookups are faster, no? Plus this way you have no extra arrays hanging around. First argument to issubset() is a ref to the big set, second argument is a ref the alleged subset: 

    #!/usr/bin/perl

use strict;

my @a1=qw(a b c d e f g);
my @a2=qw(b g);
my @a3=qw(c d e f g h);

sub issubset
{
   ($a,$b)=@_;
   for my $x (@$b)
      {
      return ("not a subset") unless grep {$x eq $_} @$a;
      }
   return ("a subset");
}

print "a2 is ",issubset(\@a1,\@a2)," of a1\n";
print "a3 is ",issubset(\@a1,\@a3)," of a1\n";

__END__
Output is:
a2 is a subset of a1
a3 is not a subset of a1

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    HTH

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[d/l]

      Array "lookups" are not faster than hash key lookups. That's the point. It's always O(1) {always takes the same amount of time, regardless of set size} to detect the presence of a key in a hash, while it's O(N) {grows linearly slower as you have more elements} to detect the presence of an element in an array.

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