如何删除仅包含字符串而不是特殊字符的方括号

Question

这是我的文字

(he needs to buy him an) [/-] (.) this rabbit <I> [/] is (..) woke  up (...) (be)cause he needs to go home with his father .

所需的输出是

he needs to buy him an [/-] (.) this rabbit <I> [/] is (..) woke    up (...) because he needs to go home with his father .

Answer 1

尝试一下：

\((?=[A-z])|(?<=[A-z])\)

我同时使用了正向超前(?= )和正向超前(?<= )来匹配其中带有字符串的括号。

• 正向超前(?= )

• 正向(?<= )后面

• [A-z]仅匹配字符串

演示：https://regex101.com/r/pc3JDu/1/

Answer 2

这是一种方法，它对re.sub进行了两次调用：

input = "(he needs to buy him an) [/-] (.) this rabbit <I> [/] is (..) woke [BLAH] up (...) (be)cause he needs to go home with his father ."
output = re.sub(r'\[([A-Za-z0-9]+?)\]', '\\1', re.sub(r'\(([A-Za-z0-9]+?)\)', '\\1', input))

Demo

这与模式\[([A-Za-z0-9]+)\]或\(([A-Za-z0-9]+)\)匹配，然后仅替换括号或方括号内的内容。

Answer 3

re.sub('\(\w\)', '', s)

是-是您的字符串

'\ w'-匹配单词的正则表达式