Update: Yes, there's something going on here -- my example below has to do with a one-level hash, whereas the original is about a two-level hash. Nothing to see here. This is not the Perl code you're looking for. You can go about your business.
There must be something else going on here. Here's a simpler example that shows that I end up with fewer keys than I started with -- using just a one-dimensional hash. Code:
#!/usr/bin/perl
use warnings;
use strict;
use Data::Dumper;
{
my %hash = ( 1 => 5, 3 => 12, 7 => 19, 9 => 44 );
print 'At '
. __LINE__
. ' there are '
. ( scalar keys %hash )
. " keys: ";
print Dumper(\%hash);
my %hash2 = ( 1 => 5, 9 => 22 );
print "This should delete one of the keys, using this hash ..\
+n";
print Dumper ( \%hash2 );
foreach my $key ( keys %hash2 ) {
if ( exists $hash{ $key } ) {
$hash{ $key } -= $hash2{ $key };
if ( $hash{ $key } <= 0 ) {
delete $hash{ $key };
}
}
}
print 'At '
. __LINE__
. ' there are '
. ( scalar keys %hash )
. " keys: ";
print Dumper(\%hash);
}
.. and the result:
tab@music4:~/Pianoforte/Development/Perlmonks/11113721 $ perl hashkeys
+.pl
At 11 there are 4 keys: $VAR1 = {
'9' => 44,
'1' => 5,
'3' => 12,
'7' => 19
};
This should delete one of the keys, using this hash ..
$VAR1 = {
'1' => 5,
'9' => 22
};
At 33 there are 3 keys: $VAR1 = {
'9' => 22,
'3' => 12,
'7' => 19
};
tab@music4:~/Pianoforte/Development/Perlmonks/11113721 $
I set the example up so that one of the keys would end up with zero (and get deleted), and another would be reduced, but not to zero. I started up with four keys, and finished with three.
Check your code -- I'm guessing you're missing something.
Alex / talexb / Toronto
Thanks PJ. We owe you so much. Groklaw -- RIP -- 2003 to 2013.