python popcnt

popcnt - count bits set ("1") in a integer, of course using python

I've implemented only four ways to do this:

• using look up table 0..255 (no iterations, oen sum and a few shifting by 8 bits)
• naive (number of iterations ==  number of bits)
• Brian Kernighan way (number of iterations == number of bits set)
• pythonic by ugly (count "1"s after calling bin()...)

All function were run on 1000 samples made by 100 random numbers:



And now testing results:

• naive (Zzzzz...)

[Wed, 05 Jun 22:07 E0][cibak@localhost:~/popcnt]> time python testNaive.py
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175.885u 0.055s 2:56.35 99.7%    0+0k 0+0io 0pf+0w

• BK way (much better!)

[Wed, 05 Jun 22:04 E0][cibak@localhost:~/popcnt]> time python testBK.py
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75.853u 0.028s 1:16.03 99.7%    0+0k 0+0io 0pf+0w 

• look up table (niiiice!)

[Wed, 05 Jun 22:06 E0][cibak@localhost:~/popcnt]> time python testLUT.py
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42.828u 0.030s 0:42.93 99.8%    0+0k 0+0io 0pf+0w 

• ugly but pythonic (wut?)

[Wed, 05 Jun 22:04 E1][cibak@localhost:~/popcnt]> time python testUgly.py
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40.880u 0.040s 0:41.06 99.6%    0+0k 0+0io 0pf+0w


And guess what? Ugly (but pythonic) won! Coincidence? I don't think so... ;)

EDIT:
And of course I can completely remove lines with lambda and all this stuff within popcntLUT  and it could be coded just like that:



But it doesn't help to much.

Also one can check say 16 bits look up table, say:



 This should run at least 2 times faster that original popcntLUT.