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

Hi Monks,
I have hash structure like this :
%devLst = ("dev1" => 2001,"dev2" => 1791, "dev3" => 32000,"dev4" => 1791, "dev5" => 32000,"dev6" => 1791, "dev7" => 32000 );
where dev1 = device name and 2001 = size of device
Want to make one array where all device name (dev1...devn). Should be in ascending order by size.
 My final array should look like this:
ex: my @resArr = (dev2,dev4,dev6,dev1,dev3,dev5,dev7); Any help will be appriciated.
Thanks.

Replies are listed 'Best First'.
Re: Sorting of hash by value
by ikegami (Patriarch) on Jan 12, 2009 at 06:25 UTC 
    my @resArr = sort { $devLst{$a} <=> $devLst{$b} } keys %devLst;

    [download]
[reply]
[d/l]
Re: Sorting of hash by value
by moritz (Cardinal) on Jan 12, 2009 at 06:23 UTC
    See perlfaq4, "How do I sort a hash (optionally by value instead of key)?"
[reply]
Re: Sorting of hash by value
by BrowserUk (Patriarch) on Jan 12, 2009 at 06:26 UTC 
    my @resArr =  sort{ 
    $devLst{ $a } <=> $devLst{ $b } 
} sort keys %devLst;;

    [download]

    Update: That said, if your keys are really Dev1 ... DevN, then you'd same memory and avoid one level of sort by storing your data in an array to start with: 

    @devLst = (undef, 2001, 1791, 32000, 1791, 32000, 1791, 32000 );;

print "dev$_" for sort{ $devLst[ $a ] <=> $devLst[ $b ] } 1 .. $#devLs
+t;
dev2
dev4
dev6
dev1
dev3
dev5
dev7

    [download]
    Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
    "Science is about questioning the status quo. Questioning authority".
    In the absence of evidence, opinion is indistinguishable from prejudice.
[reply]
[d/l]
[select]
Re: Sorting of hash by value
by wfsp (Abbot) on Jan 12, 2009 at 06:48 UTC 
    my @name_by_size = sort {
  $devLst{$a} <=> $devLst{$b}
    ||
  $a cmp $b
} keys %devLst;

    [download]
[reply]
[d/l]
      Thanks all for prompt reply.
[reply]

Back to Seekers of Perl Wisdom