我知道如何获得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]
要么
def intersect(a, b): return list(set(a) & set(b)) print intersect(b1, b2)
但是当我必须找到嵌套列表的交集时,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]]
最后我想收到:
c3 = [[13,32],[7,13,28],[1,6]]
你们能帮我解决这个问题吗?
如果你想:
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 = [[13, 32], [7, 13, 28], [1,6]]
那么这里是 Python 2 的解决方案:
c3 = [filter(lambda x: x in c1, sublist) for sublist in c2]
在 Python 3filter中返回一个可迭代对象而不是list,因此您需要使用 包装filter调用list():
filter
list
list()
c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2]
解释:
过滤器部分获取每个子列表的项目并检查它是否在源列表 c1 中。对 c2 中的每个子列表执行列表推导。