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
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]]
其他回答
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)))
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]]
查找迭代对象之间的差异和交集的简单方法
如果重复很重要，就使用这种方法
from collections import Counter
def intersection(a, b):
"""
Find the intersection of two iterables
>>> intersection((1,2,3), (2,3,4))
(2, 3)
>>> intersection((1,2,3,3), (2,3,3,4))
(2, 3, 3)
>>> intersection((1,2,3,3), (2,3,4,4))
(2, 3)
>>> intersection((1,2,3,3), (2,3,4,4))
(2, 3)
"""
return tuple(n for n, count in (Counter(a) & Counter(b)).items() for _ in range(count))
def difference(a, b):
"""
Find the symmetric difference of two iterables
>>> difference((1,2,3), (2,3,4))
(1, 4)
>>> difference((1,2,3,3), (2,3,4))
(1, 3, 4)
>>> difference((1,2,3,3), (2,3,4,4))
(1, 3, 4, 4)
"""
diff = lambda x, y: tuple(n for n, count in (Counter(x) - Counter(y)).items() for _ in range(count))
return diff(a, b) + diff(b, a)
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。
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
