fake! you dont need to sort the array to times... because u know where the lowest hand the higest is. Less of the half of 11 wallclocks would be what i expect.
> -----Ursprüngliche Nachricht----- > Von: Sudarshan Raghavan [mailto:[EMAIL PROTECTED]] > Gesendet am: Samstag, 31. August 2002 04:11 > An: Perl beginners > Betreff: Re: find the lowest number? > > On 30 Aug 2002, Felix Geerinckx wrote: > > > There is no need to sort the full array (as others suggested) to just > > find the lowest and highest number when an O(n) operation will > > suffice: > > Precisely the comment that I had a few months back for a similar question. > Read through this thread > http://archive.develooper.com/beginners%40perl.org/msg22716.html > > > > > my ($min, $max); > > $min = $max = $array[0]; > > for (1..@array-1) { > > $min = $array[$_] if $array[$_] < $min; > > $max = $array[$_] if $array[$_] > $max; > > } > > print "min = $min, max = $max\n"; > > For further proof here is a benchmarking result > #!/usr/local/bin/perl -w > use strict; > use Benchmark; > > my @arr = qw(34 12 23 45 11 91 32); > sub using_sort { > my $min = (sort {$a <=> $b} @arr)[0]; > my $max = (reverse sort {$a <=> $b} @arr)[0]; > } > > sub using_for { > my ($min, $max); > $min = $max = $arr[0]; > for (1..@arr-1) { > $min = $arr[$_] if $arr[$_] < $min; > $max = $arr[$_] if $arr[$_] > $max; > } > } > > timethese (1000000, { using_sort => \&using_sort, > using_for => \&using_for }); > > =cut > Benchmark: timing 1000000 iterations of using_for, using_sort... > using_for: 11 wallclock secs (11.60 usr + 0.04 sys = 11.64 CPU) @ > 85910.65/s (n=1000000) > using_sort: 8 wallclock secs ( 7.63 usr + 0.06 sys = 7.69 CPU) @ > 130039.01/s (n=1000000) > > Note that even with two sorts and a reverse the using_sort subroutine > is faster in terms of speed. > > > -- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > > -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
