Re: du -h, sorted

But then I realized it'd probably be better to use Perl for the sorting, and got this shorter result

Now that sounds like a challenge...

The return is clearly superflous. Off with 7.
[0-9] can be replaced by \d, saving a character.
Hoist the exponentation into the hash values and use a bigger scaling factor (that may be written more concisely).
Replace $n and $u with $1 and $2 thus doing away with their initialisation.
Hang on, moving the exponentiation back into h() means that the hash declaration is shorter still.
If one considers that 'du -h|' adds to the overall count, then we can replace the diamond operator with a backtick of du -h. Overall this saves us the | of the pipe.
We can replace the fat commas with ordinary commas (and fortunately not receive warnings).
In fact, that regexp is way too verbose. /^(...)(.)/ will work just as well, saving a couple of characters.
update: I forgot to add the last step, changing shift to pop (since we only read one arg).

Which gives....

perl -e 'sub h{%h=(K,1,M,4,G,7);pop=~/^(...)(.)/&&$1**$h{$2}}print sort{h($b)<=>h($a)}`du -h`'

ok, ok, I promise I'll leave it alone now, there's just the idea of using an anonymous hash to shave off a couple more characters, and the parens can be dropped from h($a), (but not h($b) because of precedences):

perl -e 'sub h{pop=~/(...)(.)/&&$1**{K,1,M,4,G,7}->{$2}}print sort{h($b)<=>h$a}`du -h`'

... 90 88 characters.

• another intruder with the mooring in the heart of the Perl

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Re^2: du -h, sorted by Roy Johnson (Monsignor) on Feb 25, 2009 at 22:58 UTC
Are you sure yours gets correct results? I'm getting 372K between 4.4M and 3.6M. perl -e 'sub c{pop=~/[GMK]/;ord($&)&7}print sort{c($b)<=>c($a)or$b<=>$ +a}`du -h`' [download] Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re^3: du -h, sorted by grinder (Bishop) on Feb 26, 2009 at 00:17 UTC
You're absolutely correct. Allow me to replace it by the following, clocking in at 87. perl -e'sub h{pop=~/(...)(.)/&&{M,1e3,G,1e6}->{$2}+$1}print sort{h($b) +<=>h$a}`du -h`' [download] update: No wait! There's something in your idea of bitanding... Just have to get rid of the second spaceship comparator... perl -e'sub c{pop=~/[GMK]/;(ord$&&7).1e3+$`}print sort{c($b)<=>c$a}`du + -h`' [download] 76 strokes. • another intruder with the mooring in the heart of the Perl	[reply] [d/l] [select]
Re^4: du -h, sorted by hbm (Hermit) on Feb 26, 2009 at 03:00 UTC
Alas, your last two methods put a 400K file before a 399M file... For the first one, I added a "K" key to your anonymous hash: perl -e 'sub h{pop=~/^([\d.]+)(.)/&&{K,1e1,M,1e3,G,1e6}->{$2}+$1}print + sort{h($b)<=>h$a}`du -h`' [download] For the second one, I think you're in a bind because `ord(k)&7` and `ord(K)&7` are equal... And I offer another method, bumming heavily from yours, longer but possibly faster on large filesystems: perl -e 'print map substr($_,8),reverse sort map sprintf("%8d",(/^([\d +.]+)([kKMG])/)?{K,1e1,M,1e3,G,1e6}->{$2}+$1:$1).$_,`du -h`' [download]	[reply] [d/l] [select]
Re^4: du -h, sorted by bellaire (Hermit) on Feb 26, 2009 at 15:41 UTC
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.	[reply] [d/l] [select]
Re^5: du -h, sorted by hbm (Hermit) on Feb 26, 2009 at 16:28 UTC
Re^6: du -h, sorted by bellaire (Hermit) on Feb 26, 2009 at 19:30 UTC
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Re^2: du -h, sorted by hbm (Hermit) on Feb 25, 2009 at 23:09 UTC
I like your subtle omission of parens from `h$a`, versus `h($b)`. But to sort correctly, I tweaked slightly to use the OP's 10,20,30 values. perl -e 'sub h{pop=~/(...)(.)/&&$12*{K,10,M,20,G,30}->{$2}}print sor +t{h($b)<=>h$a}`du -h`' [download]	[reply] [d/l]