我要寻找一个灵活的功能反转的一个数字的二进制重新presentation数字。

I am looking for a slick function that reverses the digits of the binary representation of a number.

如果 F 是这样的功能我会

INT（逆转（S），2）== F（INT（S，2））只要s是零和的从1开始的字符串

int(reversed(s),2) == f(int(s,2)) whenever s is a string of zeros and ones starting with 1.

现在，我使用拉姆达X：INT（''联接（逆转（斌（X）[2：]）），2）

这是确定尽可能简明而言，但似乎这样做的pretty的迂回的方式。

which is ok as far as conciseness is concerned, but it seems like a pretty roundabout way of doing this.

我在想，如果有一个更好的（也许更快）的方式与按位运算符，什么不是。

I was wondering if there was a nicer (perhaps faster) way with bitwise operators and what not.

推荐答案

如何

int('{0:b}'.format(n)[::-1], 2)

或

int(bin(n)[:1:-1], 2)

第二种方法似乎是两个的速度更快，但两者都比当前的方法要快得多：

The second method seems to be the faster of the two, however both are much faster than your current method:

import timeit

print timeit.timeit("int('{0:b}'.format(n)[::-1], 2)", 'n = 123456')

print timeit.timeit("int(bin(n)[:1:-1], 2)", 'n = 123456')

print timeit.timeit("int(''.join(reversed(bin(n)[2:])),2)", 'n = 123456')

1.13251614571
0.710681915283
2.23476600647