Python:有效的字节数组增量

问题描述

如何在Python中迭代bytearray of length = n的所有可能值?在最坏的情况下n <= 40bytes

How to iterate all possible values of bytearray of length = n in Python ?in worst case n <= 40bytes

例如，迭代n = 4:

00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000001
00000000 00000000 00000000 00000010

.
.
.
.

11111111 11111111 11111111 11111110
11111111 11111111 11111111 11111111

推荐答案

灵感来自 https://stackoverflow.com/a/15538456 /1219006

n = 2
[[[i>>k&1 for k in range(j, j-8, -1)] for j in range(8*n-1, 0, -8)]
 for i in range(2**(8*n))]

对于大型n，您需要在Python 3上运行它，因为xrange不支持大型整数.

You'll need to run this on Python 3 for large n cause xrange doesn't support big ints.

作为生成器:

def byte_array(n):
    for i in range(2**(8*n)):
        yield [[i>>k&1 for k in range(j, j-8, -1)] for j in range(8*n-1, 0, -8)]

---

>>> i = byte_array(4)
>>> next(i)
[[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0]]
>>> next(i)
[[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1]]

或者，如果您不希望它们分组，则更简单:

Or if you don't want them grouped it's simpler:

[[i>>j&1 for j in range(8*n-1, -1, -1)] for i in range(2**(8*n))]

等效生成器:

def byte_array(n):
    for i in range(2**(8*n)):
        yield [i>>j&1 for j in range(8*n-1, -1, -1)]

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