Python列表列表的所有组合

Question

所以我有一个字符串列表列表

[['a','b'],['c','d'],['e','f']]

我想得到所有可能的组合，这样的结果是

[['a','b'],['c','d'],['e','f'],
 ['a','b','c','d'],['a','b','e','f'],['c','d','e','f'],
 ['a','b','c','d','e','f']]

到目前为止，我已经想出了这个代码片段

input = [['a','b'],['c','d'],['e','f']]
combs = []
for i in xrange(1, len(input)+1):
    els = [x for x in itertools.combinations(input, i)]
    combs.extend(els)
print combs

主要遵循 this post 中的答案.

但这会导致

[(['a','b'],),(['c','d'],),(['e','f'],),
 (['a','b'],['c','d']),(['a','b'],['e','f']),(['c','d'],['e','f']),
 (['a','b'],['c', 'd'],['e', 'f'])]

我现在很困惑，试图找到一种优雅的 pythonic 方式来解压这些元组。

Answer 1

您可以使用 itertools.chain.from_iterable将列表的元组展平为列表。示例 -

import itertools
input = [['a','b'],['c','d'],['e','f']]
combs = []
for i in xrange(1, len(input)+1):
    els = [list(itertools.chain.from_iterable(x)) for x in itertools.combinations(input, i)]
    combs.extend(els)

---

演示 -

>>> import itertools
>>> input = [['a','b'],['c','d'],['e','f']]
>>> combs = []
>>> for i in range(1, len(input)+1):
...     els = [list(itertools.chain.from_iterable(x)) for x in itertools.combinations(input, i)]
...     combs.extend(els)
...
>>> import pprint
>>> pprint.pprint(combs)
[['a', 'b'],
 ['c', 'd'],
 ['e', 'f'],
 ['a', 'b', 'c', 'd'],
 ['a', 'b', 'e', 'f'],
 ['c', 'd', 'e', 'f'],
 ['a', 'b', 'c', 'd', 'e', 'f']]