in reply to Re^3: du -h, sorted
in thread du -h, sorted

Oh, I see. Could I then also change the regex to save two more characters? 
perl -e'sub c{pop=~/.\s/;(ord$&&7).1e3+$`}print sort{c($b)<=>c$a}`du -
+h`'

[download]
74 characters.

And, in fact, since we are only using .1e3 to make sure the sizes sort properly, why not save one more: 
perl -e'sub c{pop=~/.\s/;(ord$&&7)**7+$`}print sort{c($b)<=>c$a}`du -h
+`'

[download]
73 characters.

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Re^5: du -h, sorted
by hbm (Hermit) on Feb 26, 2009 at 16:28 UTC

    Slick! But how about 70, using 'x' instead of '**', and 'die' instead of 'print': 

    perl -e'sub c{pop=~/.\s/;(ord$&&7)x7+$`}die sort{c($b)<=>c$a}`du -h`'

    [download]
[reply]
[d/l]
      Very impressive. Made me waste a lot of time trying to beat it!

      I think I've got it just a little shorter. I don't know if the time I spent on this has brought me any closer to enlightenment, but it did make me fast (as in skip a meal)... 
      perl -e'%h=map{/.\s/;7**(ord$&&30)-$`,$_}`du -h`;die @h{sort %h}'

      [download]
      66 characters.

      The key is the 7 to the power of ord(char) bitwise 30 part. I had to write a little perl script that would find the magic values 7 and 30 which would work for a lexicographic (default) sort. That is, I take 7 to the power of those values, and subtract the value of $` ... to produce hash keys which sort properly.

      Then I take a hash slice on the sorted result. The sort actually returns the entire flattened hash sorted, keys and values, but since the values in that hash don't map to anything, they are silent.

      I think I need to go get something to eat.
[reply]
[d/l]

        Beautiful!! And I hear you about the food thing. (Where are you? We could see how small a lunch we could get.)

        But I'm afraid you threw me a softball, easily shorted to 64: 

        perl -e'%h=map{/.\s/;7**(ord$&&30)-$`,$_}`du -h`;die@h{sort%h}'

        [download]
[reply]
[d/l]

In Section Cool Uses for Perl