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
运算符&取两个集合的交集。
{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}
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]
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)))
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
你应该使用这段代码(来自http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks)，这段代码未经测试，但我很确定它是有效的:
def flatten(x):
"""flatten(sequence) -> list
Returns a single, flat list which contains all elements retrieved
from the sequence and all recursively contained sub-sequences
(iterables).
Examples:
>>> [1, 2, [3,4], (5,6)]
[1, 2, [3, 4], (5, 6)]
>>> flatten([[[1,2,3], (42,None)], [4,5], [6], 7, MyVector(8,9,10)])
[1, 2, 3, 42, None, 4, 5, 6, 7, 8, 9, 10]"""
result = []
for el in x:
#if isinstance(el, (list, tuple)):
if hasattr(el, "__iter__") and not isinstance(el, basestring):
result.extend(flatten(el))
else:
result.append(el)
return result
在你平摊了列表之后，你用通常的方式执行交叉:
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]]
def intersect(a, b):
return list(set(a) & set(b))
print intersect(flatten(c1), flatten(c2))
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)
