问题描述
假设我有一个字符串，例如
Suppose I have a string such as
'I hate *some* kinds of duplicate. This string has a duplicate phrase, duplicate phrase.'
我想删除第二次出现的重复短语而不删除其组成部分的其他出现，例如duplicate的其他用法.
I want to remove the second occurrence of duplicate phrase without removing other occurrences of its constituent parts, such as the other use of duplicate.
此外，我需要删除所有潜在重复短语，而不仅仅是我事先知道的某些特定短语的重复.
Moreover, I need to remove all potential duplicate phrases, not just the duplicates of some specific phrase that I know in advance.
我发现了几篇关于类似问题的帖子，但都没有帮助我解决我的特定问题:
I have found several posts on similar problems, but none that have helped me solve my particular issue:
我曾希望从那里的最后一个链接中调整方法 (re.sub(r'\b(.+)(\s+\1\b)+', r'\1', s)) 用于我的目的，但无法弄清楚如何这样做.
I had hoped to adapt the approach from the last link there (re.sub(r'\b(.+)(\s+\1\b)+', r'\1', s)) for my purposes, but could not figure out how to do so.
如何从 Python 中的字符串中删除两个或多个单词的所有任意重复短语?
推荐答案
感谢大家的尝试和评论.我终于找到了解决方案:
Thanks everyone for your attempts and comments. I have finally found a solution:
s = 'I hate *some* kinds of duplicate. This string has a duplicate phrase, duplicate phrase.'
re.sub(r'((\b\w+\b.{1,2}\w+\b)+).+\1', r'\1', s, flags = re.I)
# 'I hate *some* kinds of duplicate. This string has a duplicate phrase.'
说明
正则表达式
r'((\b\w+\b.{1,2}\w+\b)+).+\1'
查找由一个或两个 [任何字符] 分隔的多个字母数字字符的每次出现(涵盖单词不仅由空格分隔，还可能是句点或逗号和空格分隔的情况)，然后重复跟随一些不确定长度的[任何字符].然后
finds every occurrence of multiple runs of alphanumeric characters separated by one or two [any character] (to cover the case where words are separated not just by a space, but perhaps a period or comma and a space), and then repeated following some run of [any character] of indeterminate length. Then
re.sub(r'((\b\w+\b.{1,2}\w+\b)+).+\1', r'\1', s, flags = re.I)
用由一个或两个 [任何字符] 分隔的第一个多组字母数字字符替换此类事件，确保忽略大小写(因为重复的短语有时会出现在句子的开头).
replaces such occurrences with the first multiple run of alphanumeric characters separated by one or two [any character], being sure to ignore case (since the duplicate phrase could sometimes occur at the beginning of a sentence).
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