我知道有很多帖子有类似的问题，并且已经看完了所有这些帖子。但是，我无法执行所需的操作。
我有列表说 l1=[0,1,2,3,4] 我想将其分成一对元组，如下所示:
[(0, 1), (2, 3), 4],
[(0, 1), (2, 4), 3],
[(0, 1), (3, 4), 2],
[(0, 2), (1, 3), 4],
[(0, 2), (1, 4), 5],
[(0, 2), (3, 4), 1],
[(0, 3), (1, 2), 4],
[(0, 3), (2, 4), 1],
[(0, 3), (1, 4), 2],
[(0, 4), (1, 2), 3],
[(0, 4), (1, 3), 2],
[(0, 4), (2, 3), 1]
我尝试了帖子 how-to-split-a-list-into-pairs-in-all-possible-ways 中的解决方案.
def all_pairs(lst):
if len(lst) < 2:
yield lst
return
a = lst[0]
for i in range(1,len(lst)):
pair = (a,lst[i])
for rest in all_pairs(lst[1:i]+lst[i+1:]):
yield [pair] + rest
我得到以下输出:
[(0, 1), (2, 3), 4]
[(0, 1), (2, 4), 3]
[(0, 2), (1, 3), 4]
[(0, 2), (1, 4), 3]
[(0, 3), (1, 2), 4]
[(0, 3), (1, 4), 2]
[(0, 4), (1, 2), 3]
[(0, 4), (1, 3), 2]
我发现我想要的列表中缺少一些组合。
我会很感激任何建议？
请您参考如下方法:
您可以使用 itertools.permutations 并使用 frozenset 过滤掉重复项:
In [173]: d = {frozenset([frozenset(x[:2]), frozenset(x[2:4]), x[-1]]) for x in itertools.permutations(l1,
...: len(l1))}
In [174]: d2 = [sorted(x, key=lambda x: (not isinstance(x, frozenset), x)) for x in d]
In [175]: sorted([[tuple(x[0]), tuple(x[1]), x[-1]] for x in d2])
Out[175]:
[[(0, 4), (2, 3), 1],
[(1, 2), (0, 3), 4],
[(1, 2), (0, 4), 3],
[(1, 2), (3, 4), 0],
[(1, 3), (0, 2), 4],
[(1, 3), (0, 4), 2],
[(1, 3), (2, 4), 0],
[(1, 4), (0, 2), 3],
[(1, 4), (0, 3), 2],
[(2, 3), (0, 1), 4],
[(2, 3), (1, 4), 0],
[(2, 4), (0, 1), 3],
[(2, 4), (0, 3), 1],
[(3, 4), (0, 1), 2],
[(3, 4), (0, 2), 1]]
