python - 如何在单引号内获取字符串但忽略“'s”和“'t”？

Question

我想检索单引号内的字符串，例如在句子中：

Play 'My lady's on fire' by Ty Segall

我想检索：

My lady's on fire

我想忽略带有“不要”和“女士”之类的's词：'t

我试过这个：

re.findall(r"\s\'.*?\'", user_input)

但是，我得到：

[ 'My lady']

我想得到：

[My lady's on fire]

Answer 1

\B'(?:[^']*(?:'\b)?)+'

\B assert position where \b does not match
' matches the character ' literally (case sensitive)
Non-capturing group (?:[^']*(?:'\b)?)+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character not present in the list below [^']*
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
' matches the character ' literally (case sensitive)
Non-capturing group (?:'\b)?
? Quantifier — Matches between zero and one times, as many times as possible, giving back as needed (greedy)
' matches the character ' literally (case sensitive)
\b assert position at a word boundary: (^\w|\w$|\W\w|\w\W)
' matches the character ' literally (case sensitive)

Answer 2

您可以使用此正则表达式 --- >\b\s'(.*?)(?=\'\s|\'$) 您可以在https://pythex.org/上对其进行测试

蟒蛇代码：

import re user_input = "Play 'My lady's on fire' by Nipsey Hussle Play 'My lady's on fire'" print(re.findall(r"\b\s'(.*?)(?=\'\s|\'$)",user_input))

Answer 3

根据您的要求，一种替代方案（可能不是性能最高的）是这个：

\'(?:(?!\'[^st]).)*\'

从本质上讲，您正在获取字符，直到您找到一个以s或t'开头但不跟在其后的序列。

如果您需要考虑单词边界的更一般情况，请查看其他答案。