use Benchmark;
@Array1 = ("A", "B", "C", "D");
@Array2 = ("J", "C", "T", "A");
%seen = ();

timethese(1_000_000, {
  'stephen'=> sub {
    %seen=();foreach (@Array1) { $seen{$_}++ };
    my $count = 0;
    foreach (@Array2) { exists $seen{$_} and $count++; }
  },

  'Russ'=> sub {
    @Hash1{@Array1} = undef;
    $matches=scalar grep {exists $Hash1{$_}} @Array2;
  },

  'BUU'=> sub {
    $matches=grep{$x=$_;grep{$_ eq $x} @Array1} @Array2;
  },

  'Perlfaq'=> sub       {
    @union = @intersection = @difference = ();
    %count = ();
    foreach $element (@Array1, @Array2) {$count{$element}++ 
    foreach $element (keys %count) {
      push @union, $element;
      push @{ $count{$element} > 1 ? \@intersection : \@difference }, 
+$element;
    }
  }
});

__DATA__

Benchmark: timing 1000000 iterations of BUU, Perlfaq, Russ, stephen...
       BUU: 14 wallclock secs (15.17 usr +  0.00 sys = 5.17 CPU) @ 659
+19.58/s (n=1000000)
   Perlfaq: 58 wallclock secs (57.13 usr +  0.00 sys = 7.13 CPU) @ 175
+03.94/s (n=1000000)
      Russ:  6 wallclock secs ( 5.12 usr +  0.00 sys = 5.12 CPU) @ 195
+312.50/s (n=1000000)
   stephen: 17 wallclock secs (15.99 usr +  0.01 sys = 6.00 CPU) @ 625
+00.00/s (n=1000000)

[download]
Editor's Note: Corrected minor code error and re-ran results. Also, note that Perlfaq computes more than the simple intersection, so its results may be misleading at first glance. Using a hash is still the clear winner, Russ' comment notwithstanding. :-)

In reply to Re: Count of words that match in an array by BUU
in thread Count of words that match in an array by Al Shiferaw

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