问题描述
如果我有清单:
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
c = a * b
应该给我:
c = [4, 5]
和
c = a - b
应该给我:
c = [1, 2, 3]
这适用于 Python 还是我必须自己编写?
Is this available for Python or do I have to write it myself?
元组也可以这样吗?我可能会使用列表，因为我将添加它们，但只是想知道.
Would the same work for tuples? I will likely use lists as I will be adding them, but just wondering.
推荐答案
如果顺序无所谓，可以使用set 为此.它实现了交集和差异.
If the order doesn't matter, you can use set for this. It has intersection and difference implemented.
>>> a = set([1, 2, 3, 4, 5])
>>> b = set([4, 5, 6, 7, 8])
>>> a.intersection(b)
set([4, 5])
>>> a.difference(b)
set([1, 2, 3])
以下是这些操作的时间复杂度信息:https://wiki.python.org/moin/TimeComplexity#set.请注意，减数的顺序会改变运算复杂度.
Here is the info of time complexities of these operations: https://wiki.python.org/moin/TimeComplexity#set. Notice, that the order of subtrahends changes operation complexity.
如果元素可以出现多次(正式名称为 multiset)，您可以使用 Counter:
If element can occur several times (formally it is called multiset), you can use Counter:
>>> from collections import Counter
>>> a = Counter([1, 2, 3, 4, 4, 5, 5])
>>> b = Counter([4, 4, 5, 6, 7, 8])
>>> a - b
Counter({1: 1, 2: 1, 3: 1, 5: 1})
>>> a & b
Counter({4: 2, 5: 1})
这篇关于有没有办法在 Python 中获取元组或列表的差异和交集?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！