正则表达式：用条件替换子字符串
[英]Regex: replace substring with condition
问题描述
I have text, it's example 
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]
Desire output 需求输出
<address> 680031, Khabarovsk Territory, Khabarovsk, ul. District, 6, building b, office 3.<\address> and [04.1_OGRN660050463454]
I need if re.findall(r'\\[[\\d\\.]+_(?:Filialy|MN)[^]]+] in str, replace first r'\\[[\\d\\.]+_(?:Filialy|MN)' and r']' to <address> and <\\address> . 我需要re.findall(r'\\[[\\d\\.]+_(?:Filialy|MN)[^]]+] ，替换第一个r'\\[[\\d\\.]+_(?:Filialy|MN)'和r']'到<address>和<\\address> 。
How can I do that? 我怎样才能做到这一点？
1 个解决方案
解决方案1
1 已采纳 2018-02-28 16:14:47
Place [^]]+ in capturing group () and use re.sub() . 将[^]]+放入捕获组()然后使用re.sub() 。
Regex : \\[[\\d\\.]+_(?:Filialy|MN)([^]]+)\\] Substitution: <address>\\1<\\\\address> 正则表达式 ： \\[[\\d\\.]+_(?:Filialy|MN)([^]]+)\\]替代： <address>\\1<\\\\address>
Python code : Python代码 ：
re.sub(r'\[[\d\.]+_(?:Filialy|MN)([^]]+)\]', r'<address>\1<\\address>', str)
Output: 输出：
<address> 680031, Khabarovsk Territory, Khabarovsk, ul. District, 6, building b, office 3.<\address> and [04.1_OGRN660050463454]
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