集合的元组列表到集合列表

Question

这是数据集：

data=[(frozenset({'I1'}), frozenset({'I2'})), (frozenset({'I1'}), frozenset({'I3'})), (frozenset({'I1'}), frozenset({'I4'})), (frozenset({'I2'}), frozenset({'I3'})), (frozenset({'I2'}), frozenset({'I4'})), (frozenset({'I3'}), frozenset({'I4'}))]

我希望它将其转换为如下列表：

[ frozenset({'I1','I2'}), frozenset({'I1','I3'}),...]

尝试转换

data=[(frozenset({'I1'}), frozenset({'I2'})), (frozenset({'I1'}), frozenset({'I3'})), (frozenset({'I1'}), frozenset({'I4'})), (frozenset({'I2'}), frozenset({'I3'})), (frozenset({'I2'}), frozenset({'I4'})), (frozenset({'I3'}), frozenset({'I4'}))]
for x in data:
    for y in x:
        #tests

这就是我想要做的

[ frozenset({'I1','I2'}), frozenset({'I1','I3'}),...]

Answer 1

这个怎么样？

sets_lst = [(frozenset({'I1'}), frozenset({'I2'})), (frozenset({'I1'}), frozenset({'I3'})),
                   (frozenset({'I1'}), frozenset({'I4'})), (frozenset({'I2'}), frozenset({'I3'})),
                   (frozenset({'I2'}), frozenset({'I4'})), (frozenset({'I3'}), frozenset({'I4'}))]


result_lst = [frozenset().union(*curr_set_group) for curr_set_group in sets_lst]

如果有任何不清楚或有任何疑问，请告诉我！

Answer 2

您想将每个元组中的frozenset chain在一起，然后将每个链转换为frozenset

from itertools import chain
result = list(map(frozenset, map(chain.from_iterable, data)))
# [frozenset({'I1', 'I2'}), frozenset({'I1', 'I3'}), frozenset({'I1', 'I4'}), frozenset({'I3', 'I2'}), frozenset({'I2', 'I4'}), frozenset({'I3', 'I4'})]