我想写一个脚本来获取类别列表,并返回将类别拆分为两组的独特方法。现在我用元组形式(list\u a,list\u b),其中list\u a和list\u b的联合表示类别的完整列表。
下面我以类别['A','B','C','D]为例,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'])相同的拆分。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'])]
我需要以下内容:
[(['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'])]