From: Matthew Diephouse <fokke_wulf@...>
Date: 2002-06-04T02:18:23+09:00
Subject: Re: ruby equivalent for perl multi-index sort?

Rick Bradley wrote:
> 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 ;-).

That code seems redundant, because you're doing a seperate sort on the 
fractions when you could instead use just one sort. Consider the 
following functionally equivalent (and a lot more efficient) perl:

@filelist =
   sort { $a <=> $b }
     grep {/^\d+(?:\.\d+)?$/}
        readdir(INDIR);

That would also making translating it to ruby a lot easier.

    md |- m:att d:iephouse