如何将子字符串括在方括号中 python

Question

我有一个很长的文本，其中的部分用 +++ 括起来，我想用方括号括起来

se1 = "+++TEXT:+++ Moshe Morgenstern is on his way to the main synagogue in the center of Bnei Brak, home to a largely ultra-orthodox - or haredi - community. +++ : Bnei Brak, Tel Aviv + Jerusalem ))+++"

我想将包含在 +++ 中的文本转换为 [[]] 所以，

+++TEXT+++ should become [[TEXT]]

我的代码：

import re


se1 = "+++TEXT:+++ Moshe Morgenstern is on his way to the main synagogue in the center of Bnei Brak, home to a largely ultra-orthodox - or haredi - community. +++ Karte Israel mit: Bnei Brak, Tel Aviv + Jerusalem ))+++"

comments = re.sub(r"\+\+\+.*?\+\+\+", r"[[.*?]]", se1)
print(comments)

但它给出了错误的 output

[[.*?]] Moshe Morgenstern is on his way to the main synagogue in the center of Bnei Brak, home to a largely ultra-orthodox - or haredi - community. [[.*?]]

Answer 1

您需要使用()捕获组，然后使用\1引用该匹配组

这应该可以正常工作：

>>> comments = re.sub(r"\+\+\+(.*?)\+\+\+", r"[[\1]]", se1)
>>> comments
'[[TEXT:]] Moshe Morgenstern is on his way to the main synagogue in the center of Bnei Brak, home to a largely ultra-orthodox - or haredi - community. [[ Karte Israel mit: Bnei Brak, Tel Aviv + Jerusalem ))]]'

考虑到\+\+\+也可以简化为\+{3} 。

Answer 2

你可以使用这个：

re.sub(r'\+\+\+(.*?)\+\+\+',r'[[\1]]',se1)

作为.*? 在第二个字符串中被视为纯字符串而不是替换.*? 在匹配字符串中， (.*?)表示保存这部分以用于替换字符串， \1是保存的数据。