检查两个无序列表是否相等 [英] Check if two unordered lists are equal

问题描述

我正在寻找一种简单(快速)的方法来确定两个无序列表是否包含相同的元素:

I'm looking for an easy (and quick) way to determine if two unordered lists contain the same elements:

例如:

['one', 'two', 'three'] == ['one', 'two', 'three'] :  true
['one', 'two', 'three'] == ['one', 'three', 'two'] :  true
['one', 'two', 'three'] == ['one', 'two', 'three', 'three'] :  false
['one', 'two', 'three'] == ['one', 'two', 'three', 'four'] :  false
['one', 'two', 'three'] == ['one', 'two', 'four'] :  false
['one', 'two', 'three'] == ['one'] :  false

我希望不使用地图就可以做到这一点.

I'm hoping to do this without using a map.

推荐答案

Python有一个内置数据类型，用于存储(可哈希)事物的无序集合，称为set.如果将两个列表都转换为集合，则比较将是无序的.

Python has a built-in datatype for an unordered collection of (hashable) things, called a set. If you convert both lists to sets, the comparison will be unordered.

set(x) == set(y)

set

@mdwhatcott指出您要检查重复项. set会忽略这些内容，因此您需要一个类似的数据结构，该数据结构还应跟踪每个列表中的项目数.这称为多重集；标准库中最好的近似值是 collections.Counter :

@mdwhatcott points out that you want to check for duplicates. set ignores these, so you need a similar data structure that also keeps track of the number of items in each list. This is called a multiset; the best approximation in the standard library is a collections.Counter:

>>> import collections
>>> compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
>>> 
>>> compare([1,2,3], [1,2,3,3])
False
>>> compare([1,2,3], [1,2,3])
True
>>> compare([1,2,3,3], [1,2,2,3])
False
>>> 

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