简体   繁体   English

python重新引用重复元素

[英]python re backreference repeated elements

Let's say I have a string like this... 假设我有一个像这样的字符串......

myStr = 'START1(stuff); II(morestuff); 8(lessstuff)'

...and I want to extract the string immediately before the parentheses, as well as the string within the parentheses: 1 , stuff , II , morestuff , 8 , lessstuff . ... ...我想在括号前面提取字符串,以及括号内的字符串: 1stuffIImorestuff8lessstuff I can achieve this using split(';') , etc., but I want to see if I can do it in one fell swoop with re.search() . 我可以使用split(';')等来实现这一点,但我想看看我是否可以用re.search()一举完成它。 I have tried... 我努力了...

test = re.search( r'START(?:([I0-9]+)\(([^)]+?)\)(?:; )?)*', myStr ).groups()

...or in a more readable format... ...或者以更易读的格式......

test = re.search( r'''
                  START         # This part begins each string
                  (?:           # non-capturing group
                    ([I0-9]+)   # capture label before parentheses
                    \(
                      ([^)]+?)  # any characters between the parentheses
                    \)
                    (?:; )?     # semicolon + space delimiter
                  )*
                  ''', myStr, re.VERBOSE ).groups()

...but I only get the last hit: ('8', 'lessstuff') . ......但我只获得最后一击:( ('8', 'lessstuff') Is there a way to backreference multiple hits of the same part of the expression? 有没有办法反向引用表达式相同部分的多个命中?

You can use this regex in findall to capture your text: 您可以在findall使用此正则表达式来捕获您的文本:

>>> myStr = 'START1(stuff); II(morestuff); 8(lessstuff)'
>>> print re.findall(r'(?:START)?(\w+)\(([^)]*)\)', myStr)
[('1', 'stuff'), ('II', 'morestuff'), ('8', 'lessstuff')]

RegEx Demo RegEx演示

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM