in reply to Re: minimum, maximum and average of a list of numbers at the same time
in thread minimum, maximum and average of a list of numbers at the same time

I had similar results even with longer arrays:
#!/usr/bin/perl use strict; use warnings; use Benchmark qw( cmpthese ); my @array = map { int rand 100 } 1 .. 1_000_000; cmpthese -10, { op => sub { my ($min, $max, $avg) = min_max_avg(\@array); }, simple => sub { my ($min, $max, $avg) = min_max_avg_simple(\@array); }, }; sub min_max_avg_simple { my $ref = shift; my $n_elements = scalar @$ref; return unless $n_elements > 0; my ( $min, $max, $sum ) = ($ref->[0]) x 3; my $i = 1; # Start from next element while ($i < $n_elements) { my $value = $ref->[$i]; if ($value < $min) { $min = $value } elsif ($value > $max) { $max = $value } $sum += $value; ++$i; } return ( $min, $max, $sum / $n_elements ); } sub min_max_avg { my $ref = shift; my ( $min, $max, $agv, $i ); my ( $current_min, $current_max ); if ( @{$ref} % 2 == 0 ) { ( $min, $max ) = $ref->[0] < $ref->[1] ? ( $ref->[0], $ref->[1] ) : ( $ref->[1], $ref->[0] ); $agv = $ref->[0] + $ref->[1]; $i = 2; } else { $min = $max = $agv = $ref->[0]; $i = 1; } while ( $i < @{$ref} ) { ( $current_min, $current_max ) = $ref->[$i] < $ref->[ $i + 1 ] ? ( $ref->[$i], $ref->[ $i + 1 ] ) : ( $ref->[ $i + 1 ], $ref->[$i] ); $min = $current_min if ( $current_min < $min ); $max = $current_max if ( $current_max > $max ); $agv += $ref->[$i] + $ref->[ $i + 1 ]; $i += 2; } return ( $min, $max, $agv / @{$ref} ); } __END__ s/iter op simple op 1.13 -- -37% simple 0.707 60% --
Next time I'll wait a bit more before coding :)

Flavio
perl -ple'$_=reverse' <<<ti.xittelop@oivalf

Don't fool yourself.
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