有没有办法在 Python 中获取元组或列表的差异和交集?-开发者之家 | 您所在的位置:网站首页 › python的差集 › 有没有办法在 Python 中获取元组或列表的差异和交集?-开发者之家 |
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}) |
CopyRight 2018-2019 实验室设备网 版权所有 |