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
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。
其他回答
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]
运算符&取两个集合的交集。
{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}
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]
# Problem: Given c1 and c2:
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]]
# how do you get c3 to be [[13, 32], [7, 13, 28], [1, 6]] ?
这里有一种不涉及集合的方法来设置c3:
c3 = []
for sublist in c2:
c3.append([val for val in c1 if val in sublist])
但如果你喜欢只用一行，你可以这样做:
c3 = [[val for val in c1 if val in sublist] for sublist in c2]
它是列表推导式中的列表推导式，这有点不寻常，但我认为你应该不会有太多的问题。
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])
