如何确定 3 项集列表中是否至少有一个 2 项集？

Question

需要确定一个 3 项集列表是否是至少一个 2 项集的超集。每个具有频繁大小为 2 子集的 3 项集都已经在您的列表中。该列表不包含重复集。

这是我试过的最后一个代码。如果有 subsets/supersets，输出应该是一个小的集合列表。有了这段代码，我似乎得到了一个更大的列表而不是一个更小的列表。已编辑...

itemset2 =[{'', ''}, {'', ''}, {'', ''}, {'', ''}, {'', ''}, {'', ''},
 {'', ''}, {'', ''}, {'', ''}, {'', ''}, {'', ''}]


itemset3 =[{'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, 
{'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''},
 {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}, {'', '', ''}]

stuff = itemset2

final = [set(y) for y in {frozenset(x) for x in stuff}]

final

nostuff=itemset3
ablanklist=[]
ablanklist2=[]

blanklist=set()
for things in nostuff:
    ablanklist.append(list(things))

for stuff in ablanklist:
    for items in final:
        if stuff[0] and stuff[1] in items:
            print(items)

#print(final)

Answer 1

def is_subset(big: set, little: set):
    return little - big == set()

set_a = set([1,2,3])
set_b = set([1,2])
is_subset(set_a, set_b)

编辑：最好使用内置的 set_b.issubset(set_a)

Answer 2

条件是any的简单应用。给定 trio 来自 three-itemset

的元素

if any(pair < trio for pair in two-itemset):

会告诉您是否任何对是给定 3 元素集的子集。

如何确定 3 项集列表中是否至少有一个 2 项集？

How do I find out if at least one 2-itemsets is in a 3-itemsets list?

python

list

set

apriori

pruning