Python中的正则表达式

Python中的正则表达式 - 使用单个“re.sub”调用的子串(Regex in Python - Substring with single “re.sub” call) python

我正在研究Python中的Regex函数。 作为其中的一部分，我试图从字符串中提取子字符串。

例如，假设我有字符串：

<place of birth="Stockholm">

有没有办法通过一个正则表达式调用来提取斯德哥尔摩？

到目前为止，我有：

location_info = "<place of birth="Stockholm">" #Remove before location_name1 = re.sub(r"<place of birth=\"", r"", location_info) #location_name1 --> Stockholm"> #Remove after location_name2 = re.sub(r"\">", r"", location_name1) #location_name2 --> Stockholm

关于如何在不使用两个“re.sub”调用的情况下提取字符串斯德哥尔摩的任何建议都非常受欢迎。

I am looking into the Regex function in Python. As part of this, I am trying to extract a substring from a string.

For instance, assume I have the string:

<place of birth="Stockholm">

Is there a way to extract Stockholm with a single regex call?

So far, I have:

location_info = "<place of birth="Stockholm">" #Remove before location_name1 = re.sub(r"<place of birth=\"", r"", location_info) #location_name1 --> Stockholm"> #Remove after location_name2 = re.sub(r"\">", r"", location_name1) #location_name2 --> Stockholm

Any advice on how to extract the string Stockholm, without using two "re.sub" calls is highly appreciated.

最满意答案

当然，您可以将开头与双引号相匹配，然后匹配并捕获除双引号之外的所有字符：

import re p = re.compile(r'<place of birth="([^"]*)') location_info = "<place of birth=\"Stockholm\">" match = p.search(location_info) if match: print(match.group(1))

请参阅IDEONE演示

<place of birth="匹配为文字， ([^"]*)是捕获组1，匹配0或更多字符而不是" 。该值使用.group(1)访问。

这是一个REGEX演示 。

Sure, you can match the beginning up to the double quotes, and match and capture all the characters other than double quotes after that:

import re p = re.compile(r'<place of birth="([^"]*)') location_info = "<place of birth=\"Stockholm\">" match = p.search(location_info) if match: print(match.group(1))

See IDEONE demo

The <place of birth=" is matched as a literal, and ([^"]*) is a capture group 1 matching 0 or more characters other than ". The value is accessed with .group(1).

Here is a REGEX demo.