問題描述

給定一個字符串列表，我想刪除重復的單詞和原始單詞.

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']

輸出應刪除重復項，所以像這樣 ['a', 'b', 'd']

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

我不需要保留訂單.

推薦答案

使用 collections.Counter() object，然后只保留那些計數為 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]

這是一個 O(N) 算法；您只需要遍歷 N 個項目的列表一次，然后再對更少的項目 (< N) 進行第二次循環，以提取那些只出現一次的值.

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.

如果順序很重要并且您正在使用 Python <3.6、分離步驟:

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']

兩種方法的順序相同是巧合；對于 Python 3.6 之前的 Python 版本，其他輸入可能會導致不同的順序.

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，字典的實現發生了變化，現在保留了輸入順序.

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

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