这是数据集：
data=[(frozenset({'I1'}),frozenset({'I2'})),(frozenset({'I1'}),frozenset({'I3'})),frozenset({'I4'})),(frozenset({'I2'}),(frozenset({'I3'}),frozenset({'I4'}))]
，我希望它将其转换为如下所示的列表：
[ frozenset({'I1','I2'}),frozenset({'I1','I3'}),...]
尝试转换
data=[(frozenset({'I1'}),frozenset({'I4'}))]
for x in data:
for y in x:
#tests
这就是我要制作的
[ frozenset({'I1',...]
怎么样？
sets_lst = [(frozenset({'I1'}),frozenset({'I2'})),(frozenset({'I1'}),frozenset({'I3'})),frozenset({'I4'})),(frozenset({'I2'}),(frozenset({'I3'}),frozenset({'I4'}))]
result_lst = [frozenset().union(*curr_set_group) for curr_set_group in sets_lst]
让我知道不清楚的地方还是有任何疑问！
,您想将每个元组中的frozenset chain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'}),'I4'}),frozenset({'I3',frozenset({'I2','I4'})]