Re^4: How do I sort an array by hash value?

Thanks. I really appreciate your help.
Here is how I failed:

@id = sort { $lid->{split(/=/, $a)} cmp $lid->{split(/=/, $b)} } @id;
[download]

Your approach doesn't seem to sort the array unless I change it from

my $a_ = $a
my $b_ = $b
[download]

$a = $a
$b = $b
[download]

But if I do that then the values of the array get cut off at the '=' sign (I understand why).
What does the 'my $a_ = $a' do?

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Re^5: How do I sort an array by hash value? by AnomalousMonk (Archbishop) on Aug 17, 2009 at 22:25 UTC
It should be as ikegami had it, except the `$lid->{$a} cmp $lid->{$b}` statement should be `$lid->{$a_} cmp $lid->{$b_}` (note addition of the trailing underscores). The `my $a_ = $a;` step is so the values in the array aliased by the 'magical' variables `$a` and `$b` are not changed by the substitutions (i.e., the change you noted).	[reply] [d/l] [select]
Re^5: How do I sort an array by hash value? by ikegami (Patriarch) on Aug 18, 2009 at 06:52 UTC
`@id = sort { $lid->{split(/=/, $a)} cmp $lid->{split(/=/, $b)} } @id;` What does split return in scalar context...	[reply] [d/l]
Re^5: How do I sort an array by hash value? by vitoco (Hermit) on Aug 18, 2009 at 03:31 UTC
Here is how I failed: `@id = sort { $lid->{split(/=/, $a)} cmp $lid->{split(/=/, $b)} } @id;` [download] Near to success: `@id = sort { $lid->{(split(/=/, $a))[0]} cmp $lid->{(split(/=/, $b))[0 +]} } @id;` [download] This line has to evaluate $a and $b each time `sort` does a comparison, which isn't efficient. I guess that ikegami's line neither is. Should be used a temporary translation hash? Something like: `my %temp = map { $_ => (split(/=/, $_, 2))[0] } @id; @id = sort { $lid->{$temp{$a}} cmp $lid->{$temp{$b}} } @id;` [download]	[reply] [d/l] [select]