python - Python - 比较两个集合列表

Question

我有两个清单：

list1 = [
    set(['3105']),
    set(['3106', '3107']),
    set(['3115']),
    set(['3122']),
    set(['3123', '3126', '286'])
]

和

list2 = [
    set(['400']),
    set(['3115']),
    set(['3100']),
    set(['3107']),
    set(['3123', '3126'])
]

如何比较这些列表的交集，例如，如果 3126 在两个列表的任何一组中的某个位置，它将附加另一个列表与 3126。我的最终目标是附加一个单独的列表，然后取长度列表，所以我知道列表之间有多少匹配项。

Answer 1

您必须合并所有集合；取两个列表中集合的并集，然后取这两个并集的交集：

sets_intersection = reduce(set.union, list1) & reduce(set.union, list2)

if 3126 in sets_intersection:
    # ....

Answer 2

>>> common_items = set().union(*list1) & set().union(*list2)
>>> common_items
set(['3123', '3115', '3107', '3126'])
>>> '3126' in common_items
True

时间比较：

>>> %timeit reduce(set.union, list1) & reduce(set.union, list2)
100000 loops, best of 3: 11.7 us per loop
>>> %timeit set().union(*list1) & set().union(*list2)      #winner
100000 loops, best of 3: 4.63 us per loop
>>> %timeit set(s for x in list1 for s in x) & set(s for x in list2 for s in x)
10000 loops, best of 3: 11.6 us per loop
>>> %timeit import itertools;set(itertools.chain.from_iterable(list1)) & set(itertools.chain.from_iterable(list2))
100000 loops, best of 3: 9.91 us per loop

Answer 3

您可以将两个集合列表展平为集合：

l1 = set(s for x in list1 for s in x)
l2 = set(s for x in list2 for s in x)

然后你可以计算交集：

common = l1.intersection(l2)  # common will give common elements
print len(common) # this will give you the number of elements in common.

结果：

>>> print common
set(['3123', '3115', '3107', '3126'])
>>> len(common)
4