Python中有序集合的交集

Question

我是python的新手，这就是为什么我在思考一个非常基本的问题。 我有两个清单：

a = [0, 1, 2, 3, 4, 5, 6, 7]
b = [1, 2, 5, 6]

在输出中，我需要获取它们之间的所有交集：

c = [[1, 2], [5, 6]]

这有什么算法？

Answer 1

您可以为此使用difflib.SequenceMatcher

#Returns a set of matches from the given list. Its a tuple, containing
#the match location of both the Sequence followed by the size
matches = SequenceMatcher(None, a , b).get_matching_blocks()[:-1]
#Now its straight forward, just extract the info and represent in the manner
#that suits you
[a[e.a: e.a + e.size] for e in matches]
[[1, 2], [5, 6]]

Answer 2

您可以使用支持python中的交集的集

s.intersection(t) s & t new set with elements common to s and t

a = {0, 1, 2, 3, 4, 5, 6, 7}
b = {1, 2, 5, 6}
a.intersection(b)
set([1, 2, 5, 6])

Answer 3

使用集：

In [1]: a = [0, 1, 2, 3, 4, 5, 6, 7]

In [2]: b = [1, 2, 5, 6]

In [4]: set(a) & set(b)

Out[4]: set([1, 2, 5, 6])

Answer 4

您还可以为此使用lambda表达式：

>>> a = [0, 1, 2, 3, 4, 5, 6, 7]
>>> b = [1, 2, 5, 6]
>>> intersect = filter(lambda x: x in a, b)
>>> intersect
[[1, 2, 5, 6]]