Python: remove all punctuation before and after a alphanumeric string, but leave punctuation within-CodePudding

Is there a way with regex to strip all the punctuation of a string while leaving the punctuation and other special symbols within the string unchanged?

for example:

$$//[email protected]<~> becomes [email protected]

While:

       becomes

And:

**A  B** becomes A  B

P.s there are no spaces between the string under analysis

CodePudding user response：

Use re.sub() to remove all non-word characters at the beginning and end.

string = re.sub(r'^\W |\W $', '', string)

^ matches the beginning of the string, and $ matches the end. \W matches a sequence of non-alphanumeric characters.

CodePudding user response：

You could use re.findall here and search for the pattern \w (?:\W \w )*:

inp = ["$$//[email protected]<~>", "**A  B**"]
output = [re.findall(r'\w (?:\W \w )*', x)[0] for x in inp]
print(output)  # ['[email protected]', 'A  B']

The regex pattern used above finds a word, followed by one or more non word characters and another word, zero or more times.

CodePudding user response：

This regex matches everything between the first and last instance of an alphabet including the alphabets. This should work:

[a-zA-Z]([^]*)[a-zA-Z]

Where:

[a-zA-Z] matches upper or lower case alphabet
([^]*) matches 0 or more of any character