Python3.x 生成具有偶数个 0 和 1 的给定长度的所有二进制数

如果使用的是Python3.10+，则可使用int.bit_count():

>>> for i in range(16):
...   print(f'{i:04b}, count={i.bit_count()}')
...
0000, count=0
0001, count=1
0010, count=1
0011, count=2
0100, count=1
0101, count=2
0110, count=2
0111, count=3
1000, count=1
1001, count=2
1010, count=2
1011, count=3
1100, count=2
1101, count=3
1110, count=3
1111, count=4

因此，对于4位数字中的偶数个0和1:

>>> for i in range(16):
...   if i.bit_count() == 2:
...     print(f'{i:04b}')
...
0011
0101
0110
1001
1010
1100

或者简单地数一数:

>>> sum(1 for i in range(2**4) if i.bit_count() == 2)
6
>>> sum(1 for i in range(2**24) if i.bit_count() == 12)
2704156

一种更快的计数 Select 是一次生成N位的组合，每次取N/2，这比以上24位数字的.bit_count()求和快约4倍:

>>> from itertools import combinations as C
>>> numbits = 24
>>> target = numbits // 2
>>> sum(1 for c in C(range(numbits), target))
2704156

时序比较:

C:\>py -m timeit -s "from itertools import combinations as C" "sum(1 for c in C(range(24), 12))"
1 loop, best of 5: 383 msec per loop

C:\>py -m timeit "sum(1 for i in range(2**24) if i.bit_count() == 12)"
1 loop, best of 5: 1.53 sec per loop

itertools生成位模式和值的解决方案.例如，下面迭代20次，而不是try 所有64(2⁶)个数字:

from itertools import combinations as C
numbits = 6
target = numbits // 2
for c in C([1<<n for n in range(numbits)], target):
    value = sum(c)
    print(f'{value:0{numbits}b} {value}')

输出:

000111 7
001011 11
010011 19
100011 35
001101 13
010101 21
100101 37
011001 25
101001 41
110001 49
001110 14
010110 22
100110 38
011010 26
101010 42
110010 50
011100 28
101100 44
110100 52
111000 56