如何删除除26个字母之外的所有字母，以及。 ，（）'“？！来自Python中的字符串？

Question

I have :

string = 'Here it is, your gif! am a bot. [^(Report an issue)] ❤ that bot,I ❤ ur mom **YEET** 😎 ,GOTTEM!"'

and I try :

string = re.sub(r'\W+', ' ', string)

and that gives me :

'Here it is your gif am a bot Report an issue that bot I ur mom YEET GOTTEM'  

But I would like this :

'Here it is, your gif! am a bot. (Report an issue) that bot,I ur mom YEET ,GOTTEM!"'

Just the 26 letters, no numbers and only the most used symbols in this group: .,()'"?!

Answer 1

对要接受的事物进行字符分类（使用[] ）并将其反转（使用前导^使其变为[^stuff] ）：

string = re.sub(r'[^a-zA-Z.,()\'"?! ]+', '', string)

Answer 2

Use this for your regex instead : [^a-zA-Z?!.,()\\'" ]+

The brakets define a collection of elements you wish to select, the caret at the front defines the negation of what is inside.

Thus leaving you with

pattern = r'[^a-zA-Z?!.,()\'" ]+'
string = re.sub(pattern, ' ', string)