Python-从字符串列表中删除另一个元素的子字符串中的任何元素 [英] Python - Remove any element from a list of strings that is a substring of another element

问题描述

所以从下面的字符串列表开始

So starting with a list of strings, as below

string_list = [休息"，休息"，看"，看"，它"，吐"]

string_list = ['rest', 'resting', 'look', 'looked', 'it', 'spit']

我想从列表中删除作为另一个元素的子串的任何元素，例如给出结果...

I want to remove any element from the list that is a substring of another element, giving the result for instance...

string_list = [休息"，看上去"，吐"]

string_list = ['resting', 'looked', 'spit']

我有一些实现此目的的代码，但是它很丑陋，而且可能不必要地复杂.有没有一种简单的方法可以在Python中做到这一点?

I have some code that acheives this but it's embarrassingly ugly and probably needlessly complex. Is there a simple way to do this in Python?

推荐答案

第一个构建块:子字符串.

First building block: substring.

您可以使用in进行检查:

>>> 'rest' in 'resting'
True
>>> 'sing' in 'resting'
False

接下来，我们将选择创建新列表的幼稚方法.我们将一项一一添加到新列表中，检查它们是否是子字符串.

Next, we're going to choose the naive method of creating a new list. We'll add items one by one into the new list, checking if they are a substring or not.

def substringSieve(string_list):
    out = []
    for s in string_list:
        if not any([s in r for r in string_list if s != r]):
            out.append(s)
    return out

您可以通过排序以减少比较次数来加快速度(毕竟，较长的字符串永远不能是较短/等长字符串的子字符串):

You can speed it up by sorting to reduce the number of comparisons (after all, a longer string can never be a substring of a shorter/equal length string):

def substringSieve(string_list):
    string_list.sort(key=lambda s: len(s), reverse=True)
    out = []
    for s in string_list:
        if not any([s in o for o in out]):
            out.append(s)
    return out

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