Question

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

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

Answer 1

您可以使用itertools.product：

In [11]: from itertools import product

In [15]: for x in product('01',repeat=4): #for your n=4 change repeat to 32 
    print "".join(x)
   ....:     
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

Answer 2

受https://stackoverflow.com/a/15538456/1219006启发

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))]

您需要在Python 3上针对较大的n运行此操作，因为xrange不支持大整数。

作为发电机：

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：

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

等效发电机：

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

Python：高效的bytearray增量

2 个答案: