问题描述

给定一个字符串列表，我想删除重复的单词和原始单词.

Given a list of strings I want to remove the duplicates and original word.

例如:

lst = ['a', 'b', 'c', 'c', 'c', 'd', 'e', 'e']

输出应删除重复项，所以像这样 ['a', 'b', 'd']

The output should have the duplicates removed, so something like this ['a', 'b', 'd']

我不需要保留订单.

推荐答案

使用 collections.Counter() object，然后只保留那些计数为 1 的值:

Use a collections.Counter() object, then keep only those values with a count of 1:

from collections import counter

[k for k, v in Counter(lst).items() if v == 1]

这是一个 O(N) 算法；您只需要遍历 N 个项目的列表一次，然后再对更少的项目 (< N) 进行第二次循环，以提取那些只出现一次的值.

This is a O(N) algorithm; you just need to loop through the list of N items once, then a second loop over fewer items (< N) to extract those values that appear just once.

如果顺序很重要并且您正在使用 Python <3.6、分离步骤:

If order is important and you are using Python < 3.6, separate the steps:

counts = Counter(lst)
[k for k in lst if counts[k] == 1]

演示:

>>> from collections import Counter
>>> lst = ['a', 'b', 'c', 'c', 'c', 'd', 'e', 'e']
>>> [k for k, v in Counter(lst).items() if v == 1]
['a', 'b', 'd']
>>> counts = Counter(lst)
>>> [k for k in lst if counts[k] == 1]
['a', 'b', 'd']

两种方法的顺序相同是巧合；对于 Python 3.6 之前的 Python 版本，其他输入可能会导致不同的顺序.

That the order is the same for both approaches is a coincidence; for Python versions before Python 3.6, other inputs may result in a different order.

在 Python 3.6 中，字典的实现发生了变化，现在保留了输入顺序.

In Python 3.6 the implementation for dictionaries changed and input order is now retained.

这篇关于从列表中删除重复项和原始项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！