使用以下内容:

> re.sub(r'(.+?)\1+', r'\1', 'xyzzyxxyzzyxxyzzyx')
'xyzzyx'
> re.sub(r'(.+?)\1+', r'\1', 'abcbaccbaabcbaccbaabcbaccba')
'abcbaccba'
> re.sub(r'(.+?)\1+', r'\1', 'iiiiiiiiiiiiiiiiii')
'i'

它基本上匹配重复自身的模式(.+?)\1+，并删除除重复模式之外的所有内容，该模式在第一组\1中被捕获>.还要注意，在这里使用不情愿的限定符，即 +? 会使正则表达式回溯很多.

It basically matches a pattern that repeats itself (.+?)\1+, and removes everything but the repeating pattern, which is captured in the first group \1. Also note that using a reluctant qualifier here, i.e., +? will make the regex backtrack quite a lot.

演示.