在 Python 中反转数字的（二进制）数字的巧妙方法？
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Slick way to reverse the (binary) digits of a number in Python?
提问by math4tots
I am looking for a slick function that reverses the digits of the binary representation of a number.
我正在寻找一个巧妙的函数来反转一个数字的二进制表示的数字。
If fwere such a function I would have
如果f是这样的功能，我会
int(reversed(s),2) == f(int(s,2))whenever s is a string of zeros and ones starting with 1.
int(reversed(s),2) == f(int(s,2))每当 s 是一串以 1 开头的 0 和 1 时。
Right now I am using lambda x: int(''.join(reversed(bin(x)[2:])),2)
现在我正在使用 lambda x: int(''.join(reversed(bin(x)[2:])),2)
which is ok as far as conciseness is concerned, but it seems like a pretty roundabout way of doing this.
就简洁性而言，这是可以的，但这样做似乎是一种非常迂回的方式。
I was wondering if there was a nicer (perhaps faster) way with bitwise operators and what not.
我想知道按位运算符是否有更好（也许更快）的方式，而没有。
采纳答案by arshajii
How about
怎么样
int('{0:b}'.format(n)[::-1], 2)
or
或者
int(bin(n)[:1:-1], 2)
The second method seems to be the faster of the two, however both are much faster than your current method:
第二种方法似乎是两者中更快的一种，但是两者都比您当前的方法快得多：
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
回答by NPE
Here is my suggestion:
这是我的建议：
In [83]: int(''.join(bin(x)[:1:-1]), 2)
Out[83]: 9987
Same method, slightly simplified.
相同的方法，稍微简化。
回答by Gareth Latty
I would argue your current method is perfectly fine, but you can lose the list()call, as str.join()will accept any iterable:
我认为您当前的方法非常好，但您可能会失去list()调用，因为str.join()会接受任何迭代：
def binary_reverse(num):
return int(''.join(reversed(bin(num)[2:])), 2)
It would also advise against using lambdafor anything but the simplest of functions, where it will only be used once, and makes surrounding code clearer by being inlined.
它还建议不要使用lambda最简单的函数以外的任何东西，它只会被使用一次，并通过内联使周围的代码更清晰。
The reason I feel this is fine as it describes what you want to do - take the binary representation of a number, reverse it, then get a number again. That makes this code very readable, and that should be a priority.
我觉得这很好，因为它描述了你想要做什么 - 取一个数字的二进制表示，反转它，然后再次得到一个数字。这使得这段代码非常可读，这应该是一个优先事项。
回答by Andrew Mao
There is an entire half chapter of Hacker's Delightdevoted to this issue (Section 7-1: Reversing Bits and Bytes) using binary operations, bit shifts, and other goodies. Seems like these are all possible in Pythonand it should be much quicker than the binary-to-string-and-reverse methods.
Hacker's Delight有整整半章专门讨论这个问题（第 7-1 节：反转位和字节）使用二进制运算、位移和其他优点。似乎这些在 Python 中都是可能的，它应该比二进制到字符串和反向方法快得多。
The book isn't available publicly but I found this blog post that discusses some of it. The method shown in the blog post follows the following quote from the book:
这本书不公开，但我发现这篇博客文章讨论了其中的一些内容。博客文章中显示的方法遵循书中的以下引用：
Bit reversal can be done quite efficiently by interchanging adjacent single bits, then interchanging adjacent 2-bit fields, and so on, as shown below. These five assignment statements can be executed in any order.
通过交换相邻的单个位，然后交换相邻的 2 位字段等，可以非常有效地完成位反转，如下所示。这五个赋值语句可以按任意顺序执行。
http://blog.sacaluta.com/2011/02/hackers-delight-reversing-bits.html
http://blog.sacaluta.com/2011/02/hackers-delight-reversing-bits.html
回答by Kelsey
You could do it with shift operators like this:
你可以用这样的移位运算符来做到这一点：
def revbits(x):
rev = 0
while x:
rev <<= 1
rev += x & 1
x >>= 1
return rev
It doesn't seem any faster than your method, though (in fact, slightly slower for me).
不过，它似乎并不比您的方法快（实际上，对我来说稍微慢一些）。
回答by Eugene Yarmash
>>> def bit_rev(n):
... return int(bin(n)[:1:-1], 2)
...
>>> bit_rev(2)
1
>>>bit_rev(10)
5
