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
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]
其他回答
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]
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]
要定义正确考虑元素基数的交集，请使用Counter:
from collections import Counter
>>> c1 = [1, 2, 2, 3, 4, 4, 4]
>>> c2 = [1, 2, 4, 4, 4, 4, 5]
>>> list((Counter(c1) & Counter(c2)).elements())
[1, 2, 4, 4, 4]
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。
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
