I found the following snippet of code to sort alphanumeric keys with a default key breaking the alphabetic order
#!/usr/bin/perl
#
use strict;
use Data::Dumper;

my $test = { k4 => 1, k2 => 1, k3 => 1, k1 => 1};
my $first = 'k4';
my @list = map  { [$_->[0], ($_->[0] eq $first ? '' : $_->[0])] } 
 sort { $a->[1] cmp $b->[1] } 
 map  { [$_, ($_ eq $first ? '' : $_)] } 
 keys %$test;
print "Erg: " . Dumper(@list);

[download]

As I'm fairly new to perl, it took some of my time to see whats going on. And then I thought: Wait! the leftmost map is unnessecary. I tested this with
#
# above code repeated
#
my @list2 = sort { $a->[1] cmp $b->[1] } 
 map  { [$_, ($_ eq $first ? '' : $_)] } 
 keys %$test;
print "Erg2: " . Dumper(@list2);

[download]

and it yielded as expected the same result.

My Question is: Why used the (way more seasoned) perl-hacker this second (the leftmost) map-statement? I'm quite sure I'm missing a crucial point here but can't get it.

Thanks for your time,
regards,
Tomte

In reply to Mapping, Sorting, Mapping on hash-keys by Tomte

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