我会做类似的事情......:
>>> import itertools
>>> x = [[1, 1, 2], [1, 1, 1, 2], [1, 1, 2], 0, 0, 0]
>>> numzeros = x.count(0)
>>> listlen = len(x)
>>> where0s = itertools.combinations(range(listlen), numzeros)
>>> nonzeros = [y for y in x if y != 0]
>>> for w in where0s:
... result = [0] * listlen
... picker = iter(nonzeros)
... for i in range(listlen):
... if i not in w:
... result[i] = next(picker)
... print result
...
[0, 0, 0, [1, 1, 2], [1, 1, 1, 2], [1, 1, 2]]
[0, 0, [1, 1, 2], 0, [1, 1, 1, 2], [1, 1, 2]]
[0, 0, [1, 1, 2], [1, 1, 1, 2], 0, [1, 1, 2]]
[0, 0, [1, 1, 2], [1, 1, 1, 2], [1, 1, 2], 0]
[0, [1, 1, 2], 0, 0, [1, 1, 1, 2], [1, 1, 2]]
[0, [1, 1, 2], 0, [1, 1, 1, 2], 0, [1, 1, 2]]
[0, [1, 1, 2], 0, [1, 1, 1, 2], [1, 1, 2], 0]
[0, [1, 1, 2], [1, 1, 1, 2], 0, 0, [1, 1, 2]]
[0, [1, 1, 2], [1, 1, 1, 2], 0, [1, 1, 2], 0]
[0, [1, 1, 2], [1, 1, 1, 2], [1, 1, 2], 0, 0]
[[1, 1, 2], 0, 0, 0, [1, 1, 1, 2], [1, 1, 2]]
[[1, 1, 2], 0, 0, [1, 1, 1, 2], 0, [1, 1, 2]]
[[1, 1, 2], 0, 0, [1, 1, 1, 2], [1, 1, 2], 0]
[[1, 1, 2], 0, [1, 1, 1, 2], 0, 0, [1, 1, 2]]
[[1, 1, 2], 0, [1, 1, 1, 2], 0, [1, 1, 2], 0]
[[1, 1, 2], 0, [1, 1, 1, 2], [1, 1, 2], 0, 0]
[[1, 1, 2], [1, 1, 1, 2], 0, 0, 0, [1, 1, 2]]
[[1, 1, 2], [1, 1, 1, 2], 0, 0, [1, 1, 2], 0]
[[1, 1, 2], [1, 1, 1, 2], 0, [1, 1, 2], 0, 0]
[[1, 1, 2], [1, 1, 1, 2], [1, 1, 2], 0, 0, 0]
>>>
当然,可以通过多种方式进行微优化,但我希望总体思路清晰:识别所有可能为零的索引集,并将原始列表的非零项按顺序放置在其他位置。