python非零傳回怎麼解決_在python計數非零位的快速方法

I need a fast way to count the number of bits in an integer in python.

My current solutions is

bin(n).count("1")

but I am wondering if there is any faster way of doing this?

PS: (i am representing a big 2D binary array as a singe list of numbers and doing bitwise operations, and that brings the time down from hours to minutes. and now i would like to get rid of those extra minutes.

Edit:

1. it has to be in python 2.7 or 2.6

and optimizing for small numbers does not matter that much since that would not be a clear bottle neck, but I do have numbers with 10 000 + bits at some places

for example this is a 2000 bit case:

12448057941136394342297748548545082997815840357634948550739612798732309975923280685245876950055614362283769710705811182976142803324242407017104841062064840113262840137625582646683068904149296501029754654149991842951570880471230098259905004533869130509989042199261339990315125973721454059973605358766253998615919997174542922163484086066438120268185904663422979603026066685824578356173882166747093246377302371176167843247359636030248569148734824287739046916641832890744168385253915508446422276378715722482359321205673933317512861336054835392844676749610712462818600179225635467147870208L

解決方案

For arbitrary-length integers, bin(n).count("1") is the fastest I could find in pure Python.

I tried adapting Óscar's and Adam's solutions to process the integer in 64-bit and 32-bit chunks, respectively. Both were at least ten times slower than bin(n).count("1") (the 32-bit version took about half again as much time).

On the other hand, gmpy popcount() took about 1/20th of the time of bin(n).count("1"). So if you can install gmpy, use that.