monkfan has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks,
I am trying to accumulate the value of HoH (%bucket) below . Such that this code
use strict; use warnings; my %bucket = ( key1 => { '1' => 3}, key2 => { '1' => 3, '2' => 2}, key3 => { '1' => 11, '2' => 36} ); my %spair = (); my $cnt =0; foreach my $kmer ( keys %bucket ) { foreach my $si (keys %{$bucket{$kmer}}) { foreach my $sn (keys %{$bucket{$kmer}}) { next if exists $spair{$sn."-".$si}; if ( $si == $sn ) { $cnt = $bucket{$kmer}{$si}; } else { $cnt = $bucket{$kmer}{$si}*$bucket{$kmer}{$sn}; } $spair{$si."-".$sn} +=$cnt; } } # ----- end foreach ----- } # ----- end foreach ----- use Data::Dumper; print Dumper \%spair;

would give:
$VAR1 = { '1-1' => 17, # 3+3+11 '2-2' => 38, # 2+36 '1-2' => 402 # (3*2)+(11*36), this is correct already };
instead of:
$VAR1 = { '1-1' => 3, #wrong '2-2' => 2, #wrong '1-2' => 402 #correct };

I've been playing around with it, I still can't see where it went wrong. Your advice is very valuable for me. Thanks so much beforehand.

Regards,
Edward

Replies are listed 'Best First'.
Re: Accumulating Count in HoH problem
by pg (Canon) on Oct 06, 2004 at 04:11 UTC

    It is about that 'next' statement. I commented out your next statement, and replaced it with mine:

    use strict; use warnings; my %bucket = ( key1 => {'1' => 3}, key2 => {'1' => 3, '2' => 2}, key3 => {'1' => 11, '2' => 36} ); my %spair = (); my $cnt =0; foreach my $kmer ( keys %bucket ) { foreach my $si (keys %{$bucket{$kmer}}) { foreach my $sn (keys %{$bucket{$kmer}}) { # next if ($spair{$sn."-".$si} != 0); next if ($sn < $si); if ( $si == $sn ) { $cnt = $bucket{$kmer}{$si}; } else { $cnt = $bucket{$kmer}{$si}*$bucket{$kmer}{$sn}; } $spair{$si."-".$sn} +=$cnt; } } # ----- end foreach ----- } # ----- end foreach ----- use Data::Dumper; print Dumper \%spair;

    Here is the result:

    $VAR1 = { '1-1' => 17, '2-2' => 38, '1-2' => 402 };