我知道如何得到两个平面列表的交集:

b1 = [1,2,3,4,5,9,11,15]
b2 = [4,5,6,7,8]
b3 = [val for val in b1 if val in b2]

or

def intersect(a, b):
    return list(set(a) & set(b))
 
print intersect(b1, b2)

但当我必须为嵌套列表找到交集时,我的问题就开始了:

c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]

最后我希望收到:

c3 = [[13,32],[7,13,28],[1,6]]

你们能帮我一下吗?

相关的

在python中扁平一个浅列表


当前回答

对于只想找到两个列表交集的人,Asker提供了两个方法:

B1 = [1,2,3,4,5,9,11,15] B2 = [4,5,6,7,8] B3 = [val for val in b1 if val in b2] 而且 Def相交(a, b): 返回列表(set(a) & set(b)) 打印相交(b1, b2)

但是有一种混合方法更有效,因为你只需要在list/set之间做一次转换,而不是三次:

b1 = [1,2,3,4,5]
b2 = [3,4,5,6]
s2 = set(b2)
b3 = [val for val in b1 if val in s2]

这将在O(n)中运行,而他最初的包含列表理解的方法将在O(n²)中运行

其他回答

c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
c3 = [list(set(i) & set(c1)) for i in c2]
c3
[[32, 13], [28, 13, 7], [1, 6]]

对我来说,这是非常优雅和快速的方法。

你认为[1,2]与[1,[2]]相交吗?也就是说,你只关心数字,还是也关心列表结构?

如果只有数字,研究如何“扁平化”列表,然后使用set()方法。

你不需要定义交集。它已经是集合的一流部分了。

>>> b1 = [1,2,3,4,5,9,11,15]
>>> b2 = [4,5,6,7,8]
>>> set(b1).intersection(b2)
set([4, 5])

考虑到:

> c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]

> c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]

我发现下面的代码工作得很好,如果使用set操作可能会更简洁:

> c3 = [list(set(f)&set(c1)) for f in c2] 

它有:

> [[32, 13], [28, 13, 7], [1, 6]]

如需订购:

> c3 = [sorted(list(set(f)&set(c1))) for f in c2] 

我们有:

> [[13, 32], [7, 13, 28], [1, 6]]

顺便说一下,对于更python的风格,这个也很好:

> c3 = [ [i for i in set(f) if i in c1] for f in c2]

python获取两个列表的交集的方法是:

[x for x in list1 if x in list2]