in reply to List::Ranking

I usually just code this in-line. A module seems overkill: 
my %data = ( 
      'slashdot.org' => 180, 
      'cpan.org' => 150, 
      'perl.com' => 150, 
      'apache.org' => 120, 
  );
my @ranks; my @keys = keys %data;
@ranks[ sort { $data{$keys[$b]} cmp $data{$keys[$a]} } 0..$#keys ] = 1
+..@keys;
print "key $keys[$_] has rank $ranks[$_]\n" for 0..$#ranks;

[download]
</code>

-- Randal L. Schwartz, Perl hacker

Replies are listed 'Best First'.
Re: Re: List::Ranking
by miyagawa (Chaplain) on Dec 10, 2001 at 20:31 UTC
    Usually do I also. But this code doesn't care the fact that "cpan.org" and "perl.com" have the same value (150). What this module does is that.

    --
    Tatsuhiko Miyagawa
    miyagawa@cpan.org

[reply]

      Right. So instead, try this:

      my %data = (
      'slashdot.org' => 180,
      'cpan.org'     => 150,
      'perl.com'     => 150,
      'apache.org'   => 120,
  );
my @keys = keys %data;

# invert the %data hash, creating arrayrefs each with list
# of elements with the same ranking
my %ranks;
push @{$ranks{$data{$_}}}, $_ for @keys;

# count/sort the rankings
my @ranks = reverse sort {$a<=>$b} keys %ranks;

for my $rank ( 0 .. $#ranks ) {  # print them
  my @tied = @{$ranks{ $ranks[$rank] }};
  if (@tied > 1) {  # more than one with this ranking
    local $"=', ';
    my $last = pop @tied;
    print "@tied and $last all have rank ". 1+$ranks[$rank] ."\n";
  } else {
    print "@tied has rank ". 1+$ranks[$rank] ."\n";
  }
}

      [download]

      Update: Rankings should be 1-based.

      dmm

      You can give a man a fish and feed him for a day ...
      Or, you can       teach him to fish and feed him for a lifetime
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In Section Meditations