我有文字，这是一个例子
[04.1_Filialy 680031, Khabarovsk Territory, Khabarovsk, ul. District, 6, building b, office 3.] and [04.1_OGRN660050463454]
欲望输出
<address> 680031, Khabarovsk Territory, Khabarovsk, ul. District, 6, building b, office 3.<\address> and [04.1_OGRN660050463454]
我需要在str中使用re.findall(r'\[[\d\.]+_(?:Filialy|MN)[^]]+]，将第一个r'\[[\d\.]+_(?:Filialy|MN)'和r']'替换为<address>和<\address>。
我怎样才能做到这一点？
将[^]]+放入捕获组()并使用re.sub()。
正则表达式：\[[\d\.]+_(?:Filialy|MN)([^]]+)\]换人：<address>\1<\\address>
Python代码：
re.sub(r'\[[\d\.]+_(?:Filialy|MN)([^]]+)\]', r'<address>\1<\\address>', str)
输出：
<address> 680031, Khabarovsk Territory, Khabarovsk, ul. District, 6, building b, office 3.<\address> and [04.1_OGRN660050463454]