Re: double sort HoH by value/key

You have been close. Johngg did already post one solution (he was faster than me ;-)

Another variant would be:

use strict;
use warnings;

my %HoH = ( a => { value1 => "foo", value2 => 1 },
            e => { value1 => "bar", value2 => 2 },
            b => { value1 => "foo", value2 => 2 },
            f => { value1 => "bar", value2 => 2 },
            d => { value1 => "foo", value2 => 3 },
            c => { value1 => "bar", value2 => 5 },
            h => { value1 => "foo", value2 => 4 },
            g => { value1 => "foo", value2 => 4 } );  

my @sorted = sort HoHsort keys %HoH;

sub HoHsort { $HoH{$b}{value2} <=> $HoH{$a}{value2}  ||  $a cmp $b }

print map "$_ =>=> { $HoH{$_}{value2} } \n", @sorted;
[download]

Regards

mwa

Comment on Re: double sort HoH by value/key Download Code

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Re^2: double sort HoH by value/key by rb1980 (Initiate) on Nov 21, 2007 at 22:31 UTC
I think my trouble is in understanding what $a and $b are referring to. I find it hard to comprehend how sort can "see outside" of the foreach loop. I think I get it now, though. I am going to try to read up more on sort.	[reply]
Re^3: double sort HoH by value/key by shmem (Chancellor) on Nov 21, 2007 at 23:40 UTC
I think my trouble is in understanding what $a and $b are referring to. They are referring to the elements you pass into the sort routine. Which element of that list gets associated when to $a or to $b is sort's business. doc;//sort assigns elements of that list to $a and $b as aliases, as the underlying sort algorithm demands. You were passing `keys %HoH` to the sort function, hence the keys of the outer hash, the chars (a .. g) in the order perl's hashing implementation sees fit. $a and $b, inside the sort function block, will never be anything else than an alias to an element of the list you passed in. --shmem _($_=" "x(1<<5)."?\n".q·/)Oo. G°\ / /\_¯/(q / ---------------------------- \__(m.====·.(_("always off the crowd"))."· ");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}	[reply] [d/l]
Re^3: double sort HoH by value/key by ikegami (Patriarch) on Nov 21, 2007 at 23:55 UTC
I find it hard to comprehend how sort can "see outside" of the foreach loop. That's common for all blocks. `my $var = "Hi!"; { print("$var\n"); }` [download] `my $var = "Hi!"; for (1..2) { print("$var\n"); }` [download] `my $var = "Hi!"; print(map { "$var\n" } 1..2);` [download] `my $var = "Hi!"; sub func { print("$var\n"); } func();` [download] Yes, the following works even though `$var` goes out of scope before `func` is called. `{ my $var = "Hi!"; sub func { print("$var\n"); } } func();` [download]	[reply] [d/l] [select]