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

from random import *

a = sample(range(0, 1000), 100)
b = sample(range(0, 1000), 100)
print(a)
print(b)
print(set(a).intersection(set(b)))

其他回答

运算符&取两个集合的交集。

{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}

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

我们可以使用set方法:

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]]

   result = [] 
   for li in c2:
       res = set(li) & set(c1)
       result.append(list(res))

   print result

我也在寻找一种方法来做到这一点,最终结果是这样的:

def compareLists(a,b):
    removed = [x for x in a if x not in b]
    added = [x for x in b if x not in a]
    overlap = [x for x in a if x in b]
    return [removed,added,overlap]

python获取两个列表的交集的方法是:

[x for x in list1 if x in list2]