Re: ruby equivalent for perl multi-index sort?

> 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);

Except for two things:

(1) the results are different:

  $ perl -e '@a=(1,2,"2.1","2.10", "2.2", 3); @a = sort { $a <=> $b } @a; print join(", ", @a)."\n";'
  1, 2, 2.1, 2.10, 2.2, 3

  whereas the other sort will produce:

    1, 2, 2.1, 2.2, 2.10, 3

  on the same input.  It's not a numeric sort even though the strings
  look like fractions (think of them as document section numbers, which
  is quite similar to what they represented in the problem domain).

(2) I asked the question primarily to find out how to convert

      sort { foo($a) <=> foo($b) || bar($a) <=> bar($b) || ... }

    into a Ruby idiom.  Were that example reducible to a simple sort
    there would still be plenty of similar cases which won't be.


Thanks to Pit's suggestion I converted the original perl expression to
the Ruby equivalent:

indir.entries.grep(/^\d+(?:\.\d+)?$/).collect{ |x| 
       x += '.0' unless x =~ /\./
       x 
    }.sort{ |a,b| 
        (a.to_i <=> b.to_i).nonzero? ||
        (a.sub(/^.*\./, '').to_i <=> b.sub(/^.*\./,'').to_i) 
    }.collect{|x| 
        x.sub(/\.0$/, '')
        x
    }

Rick
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