以正整数计算非零位的快速方法

Fast way of counting non-zero bits in positive integer(以正整数计算非零位的快速方法)

问题描述

我需要一种快速的方法来计算 python 中整数的位数.我目前的解决方案是

bin(n).count("1")
但我想知道是否有更快的方法来做到这一点?
PS:(我将一个大的二维二进制数组表示为一个数字列表并进行按位运算，这将时间从几小时缩短到几分钟.现在我想摆脱那些额外的分钟.
1. 它必须在 python 2.7 或 2.6 中
优化小数字并不重要，因为这不是一个明显的瓶颈，但我确实在某些地方有 10 000 + 位的数字
例如这是一个 2000 位的情况:
<预> <代码> 12448057941136394342297748548545082997815840357634948550739612798732309975923280685245876950055614362283769710705811182976142803324242407017104841062064840113262840137625582646683068904149296501029754654149991842951570880471230098259905004533869130509989042199261339990315125973721454059973605358766253998615919997174542922163484086066438120268185904663422979603026066685824578356173882166747093246377302371176167843247359636030248569148734824287739046916641832890744168385253915508446422276378715722482359321205673933317512861336054835392844676749610712462818600179225635467147870208L

对于任意长度的整数，bin(n).count("1") 是我在纯 Python 中能找到的最快的.

我尝试采用 Óscar 和 Adam 的解决方案分别处理 64 位和 32 位块中的整数.两者都比 bin(n).count("1") 慢了至少十倍(32 位版本又花了大约一半的时间).

另一方面，gmpy popcount()大约是 bin(n).count("1") 时间的 1/20.因此，如果您可以安装 gmpy，请使用它.

要回答评论中的问题，对于字节，我会使用查找表.您可以在运行时生成它:

counts = bytes(bin(x).count("1") for x in range(256)) # py2: 使用 bytearray

或者直接定义它:

计数 = (b'x00x01x01x02x01x02x02x03x01x02x02x03x02x03x03x04'b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'b'x04x05x05x06x05x06x06x07x05x06x06x07x06x07x07x08')

然后是 counts[x] 得到 x 中 1 的位数，其中 0 ≤ x ≤ 255.

I need a fast way to count the number of bits in an integer in python. My current solution 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 single 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 bottleneck, but I do have numbers with 10 000 + bits at some places

for example this is a 2000 bit case:

12448057941136394342297748548545082997815840357634948550739612798732309975923280685245876950055614362283769710705811182976142803324242407017104841062064840113262840137625582646683068904149296501029754654149991842951570880471230098259905004533869130509989042199261339990315125973721454059973605358766253998615919997174542922163484086066438120268185904663422979603026066685824578356173882166747093246377302371176167843247359636030248569148734824287739046916641832890744168385253915508446422276378715722482359321205673933317512861336054835392844676749610712462818600179225635467147870208L

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.

To answer a question in the comments, for bytes I'd use a lookup table. You can generate it at runtime:

counts = bytes(bin(x).count("1") for x in range(256))  # py2: use bytearray

Or just define it literally:

counts = (b'x00x01x01x02x01x02x02x03x01x02x02x03x02x03x03x04'
          b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'
          b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'
          b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'
          b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'
          b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'
          b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'
          b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'
          b'x01x02x02x03x02x03x03x04x02x03x03x04x03x04x04x05'
          b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'
          b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'
          b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'
          b'x02x03x03x04x03x04x04x05x03x04x04x05x04x05x05x06'
          b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'
          b'x03x04x04x05x04x05x05x06x04x05x05x06x05x06x06x07'
          b'x04x05x05x06x05x06x06x07x05x06x06x07x06x07x07x08')

Then it's counts[x] to get the number of 1 bits in x where 0 ≤ x ≤ 255.

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