Python 从列表中删除列表的重复元组

我想写一个脚本来获取类别列表，并返回将类别分成两组的独特方法。现在我用元组形式（list_a，list_b）表示它，其中list_a和list_b的并集表示类别的完整列表

下面我用一个类别为['A'、'B'、'C'、'D']的例子来展示，我可以得到所有的组。但是，有些是重复的（['A']，['B'，'C'，'D']）表示与（['B'，'C'，'D']，['A']）相同的分割。如何仅保留唯一拆分？还有，这篇文章的更好标题是什么

import itertools
def getCompliment(smallList, fullList):
    compliment = list()
    for item in fullList:
        if item not in smallList:
            compliment.append(item)
    return compliment

optionList = ['A','B','C','D']
combos = list()
for r in range(1,len(optionList)):
    tuples = list(itertools.combinations(optionList, r))
    for t in tuples:
        combos.append((list(t),getCompliment(list(t), optionList)))

print(combos)

[(['A'], ['B', 'C', 'D']),
 (['B'], ['A', 'C', 'D']), 
 (['C'], ['A', 'B', 'D']),
 (['D'], ['A', 'B', 'C']),
 (['A', 'B'], ['C', 'D']),
 (['A', 'C'], ['B', 'D']),
 (['A', 'D'], ['B', 'C']),
 (['B', 'C'], ['A', 'D']),
 (['B', 'D'], ['A', 'C']),
 (['C', 'D'], ['A', 'B']),
 (['A', 'B', 'C'], ['D']),
 (['A', 'B', 'D'], ['C']),
 (['A', 'C', 'D'], ['B']),
 (['B', 'C', 'D'], ['A'])]

我需要以下资料：

[(['A'], ['B', 'C', 'D']),
 (['B'], ['A', 'C', 'D']), 
 (['C'], ['A', 'B', 'D']),
 (['D'], ['A', 'B', 'C']),
 (['A', 'B'], ['C', 'D']),
 (['A', 'C'], ['B', 'D']),
 (['A', 'D'], ['B', 'C'])]

---

你很接近。您需要的是一组结果

由于

set

元素必须是可散列的，并且

list

对象不可散列，因此可以改用

tuple

。这可以通过对代码进行一些微不足道的更改来实现

import itertools

def getCompliment(smallList, fullList):
    compliment = list()
    for item in fullList:
        if item not in smallList:
            compliment.append(item)
    return tuple(compliment)

optionList = ('A','B','C','D')
combos = set()
for r in range(1,len(optionList)):
    tuples = list(itertools.combinations(optionList, r))
    for t in tuples:
        combos.add(frozenset((tuple(t), getCompliment(tuple(t), optionList))))

print(combos)

{frozenset({('A',), ('B', 'C', 'D')}),
 frozenset({('A', 'C', 'D'), ('B',)}),
 frozenset({('A', 'B', 'D'), ('C',)}),
 frozenset({('A', 'B'), ('C', 'D')}),
 frozenset({('A', 'C'), ('B', 'D')}),
 frozenset({('A', 'D'), ('B', 'C')}),
 frozenset({('A', 'B', 'C'), ('D',)})}

如果需要将结果转换回列表列表，可以通过列表理解：

res = [list(map(list, i)) for i in combos]

[[['A'], ['B', 'C', 'D']],
 [['B'], ['A', 'C', 'D']],
 [['A', 'B', 'D'], ['C']],
 [['A', 'B'], ['C', 'D']],
 [['B', 'D'], ['A', 'C']],
 [['B', 'C'], ['A', 'D']],
 [['A', 'B', 'C'], ['D']]]

---

非常感谢。这太棒了！