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
纯列表理解版本
>>> 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]]
>>> c1set = frozenset(c1)
平变体:
>>> [n for lst in c2 for n in lst if n in c1set]
[13, 32, 7, 13, 28, 1, 6]
嵌套的变体:
>>> [[n for n in lst if n in c1set] for lst in c2]
[[13, 32], [7, 13, 28], [1, 6]]
其他回答
# 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。
函数方法:
input_list = [[1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7]]
result = reduce(set.intersection, map(set, input_list))
它可以应用于更一般的1+列表
既然定义了intersect，一个基本的列表推导式就足够了:
>>> c3 = [intersect(c1, i) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]
多亏了S. Lott的评论和TM。的相关评论:
>>> c3 = [list(set(c1).intersection(i)) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]
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(c2[i]).intersection(set(c1))) for i in xrange(len(c2))]
c3
->[[32, 13], [28, 13, 7], [1, 6]]
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])
