From: Rick Bradley <rick@...>
Date: 2002-06-03T23:42:51+09:00
Subject: ruby equivalent for perl multi-index sort?

Coming from a perl background I have a lot of programs which do
operations on arrays, including sorts.  For example today I came across
one of mine where I wrote:

@filelist =   
  map { $_->[0] . ($_->[1] ? '.'.$_->[1] : '') }
    sort {$a->[0] <=> $b->[0] || $a->[1] <=> $b->[1] }
      map { /(\d+)(?:\.(\d+))?/; [ $1, $2 || 0 ] } 
        grep {/^\d+(?:\.\d+)?$/} 
            readdir(INDIR);

It takes a list of files, extracts the numeric ones and orders them
numerically.  If they have fractional parts it numbers them by treating
the fractional parts as integers (don't ask why ;-).

The bulk of that is easy to rewrite the Ruby way, except for
the multi-index sort:

  sort { $a->[0] <=> $b->[0] 
                  ||
         $a->[1] <=> $b->[1] }

The spaceship returns 0 when the first pair of elements are 0 and this
perl idiom takes advantage of the fact that 0 is false and therefore
'||' will evaluate its second operand.  It's akin to saying "if the
first elements are the same then order on the second elements".

In Ruby 0 isn't false so the comparable Ruby expression won't trigger
the comparison on later indices.  I've been wondering if there's a way
to efficiently implement this idiom in Ruby.

Ideas?

Rick
-- 
 http://www.rickbradley.com    MUPRN: 957    (82F/90F)
                       |  like being trapped in
   random email haiku  |  a video game. i woke
                       |  up thrashing around.