Strip punctuation with regular expression - python

Question

I would like to strip all of the the punctuations (except the dot) from the beginning and end of a string, but not in the middle of it.

For instance for an original string:

@#%%.Hol$a.A.$%

I would like to get the word .Hol$aA removed from the end and beginning but not from the middle of the word.

Another example could be for the string:

@#%%...&Hol$a.A....$%

In this case the returned string should be ..&Hol$aA... because we do not care if the allowed characters are repeated.

The idea is to remove all of the punctuations( except the dot ) just at the beginning and end of the word. A word is defined as \\w and/or a .

A practical example is the string 'Barnes&Nobles' . For text analysis is important to recognize Barnes&Nobles as a single entity, but without the '

How to accomplish the goal using Regex?

Answer 1

Use this simple and easily adaptable regex:

[\w.].*[\w.]

It will match exactly your desired result, nothing more.

• [\\w.] matches any alphanumeric character and the dot
• .* matches any character (except newline normally)
• [\\w.] matches any alphanumeric character and the dot

To change the delimiters, simply change the set of allowed characters inside the [] brackets.

Check this regex out on regex101.com

import re
data = '@#%%.Hol$a.A.$%'
pattern = r'[\w.].*[\w.]'
print(re.search(pattern, data).group(0))
# Output: .Hol$a.A.

Answer 2

Depending on what you mean with striping the punctuation, you can adapt the following code :

import re
res = re.search(r"^[^.]*(.[^.]*.([^.]*.)*?)[^.]*$", "@#%%.Hol$a.A.$%")
mystr = res.group(1)

This will strip everything before and after the dot in the expression. Warning, you will have to check if the result is different of None, if the string doesn't match.