The hard bit about this appears to be filling in the "gaps" in the histogram. If the histogram was huge, then I suppose that a hash table would be best, filling in the "gaps" on the output. But in this case an array representation appears to be quite good. To find the maximum "peg" value, a numeric sort of the values and taking the last one in that list appears to me to be a good way to go. I think that jwkrahn was very close. Update: Well I suppose it is also possible that I've made the problem more complex than it needed to be!

#!/usr/bin/perl -w
use strict;

my %hash = qw (
        a        2
        b        4
        c        2
        d        1
        e        1
        f        2
        g        4
        h        3
        i        8
        j        10
        k        10
               );
                             
my @values = sort{$a<=>$b}values(%hash);
my $max = (@values)[-1];
my @pegs = (0) x ($max);

for my $value ( @values ) 
{
    $pegs[ $value ]++;
}

for my $peg_num (0..@pegs-1)
{
   print "$peg_num \t=> $pegs[$peg_num]\n";
}
__END__
Prints:
0     => 0
1     => 2
2     => 3
3     => 1
4     => 2
5     => 0
6     => 0
7     => 0
8     => 1
9     => 0
10     => 2

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In reply to Re: count frequency of occurence.... by Marshall
in thread count frequency of occurence.... by cthar

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