以所有可能的方式將列表分成所有對

Question

我知道有很多類似問題的帖子，並且已經遍歷了所有問題。 但是，我無能為力。

我有一個列表說l1=[0,1,2,3,4]我想將其划分為一對元組，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]

我嘗試了如何在所有可能的方式中將列表拆分成對的方法 。

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

我得到以下輸出：

[(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]

我發現我想要的列表中缺少一些組合。

我將不勝感激任何建議嗎？

Answer 1

您可以使用itertools.permutations並使用frozenset過濾出重復frozenset ：

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]]

Answer 2

您可以使用itertools.permutations ，然后使用列表推導從每個置換的前4個項目中創建對：

l1=[0,1,2,3,4]
from itertools import permutations
l2 = permutations(l1)
l3 = [[(x[0], x[1]), (x[2], x[3]), x[4]] for x in l2]

Answer 3

[[(0, i), tuple(item for item in l if item not in {0, i ,j}), j] for i in range(1, 5) for j in [item for item in l if item not in {0, i}]]

[[(0, 1), (3, 4), 2],
 [(0, 1), (2, 4), 3],
 [(0, 1), (2, 3), 4],
 [(0, 2), (3, 4), 1],
 [(0, 2), (1, 4), 3],
 [(0, 2), (1, 3), 4],
 [(0, 3), (2, 4), 1],
 [(0, 3), (1, 4), 2],
 [(0, 3), (1, 2), 4],
 [(0, 4), (2, 3), 1],
 [(0, 4), (1, 3), 2],
 [(0, 4), (1, 2), 3]]