Detect missing space after punctuation and space before punctuation

Question

I want to detect a missing space following a punctuation and an extra space before punctuation. I attempted to use a regex to match [A-Za-z0-9][?.,:!][A-Z] and [A-Za-z0-9]\s+[?.,:!], but both of these return None when applied to the string Something is in the air tonight.Or is it ?.

import re

mystring = "Something is in the air tonight.Or is it ?"

missing_space_regex = re.compile('[A-Za-z0-9][?.,:!][A-Z]')
print(missing_space_regex.match(mystring))

extra_space_regex = re.compile('[A-Za-z0-9]\s+[?.,:!]')
print(extra_space_regex.match(mystring))

I realize that the extra_space_regex as above will not detect the case where the text begins with a punctuation, but I can handle that as a special case.

Answer 1

If you can use regex instead of re, you could take advantage of regex Unicode character classes, like \p{P} for a punctuation character:

import regex

mystring = "Something is in the air tonight.Or is it ?"

missing_space_regex = regex.compile(r'.*?\p{P}\S')
print(missing_space_regex.match(mystring))

extra_space_regex = regex.compile(r'.*?\s\p{P}')
print(extra_space_regex.match(mystring))

Outputs:

<regex.Match object; span=(0, 33), match='Something is in the air tonight.O'>
<regex.Match object; span=(0, 42), match='Something is in the air tonight.Or is it ?'>

Or if you do want to use your chosen punctuation characters and re:

punc = "?.,:!"

missing_space_re = re.compile(f".*?[{punc}]\S")
print(missing_space_re.match(mystring))

extra_space_re = re.compile(f'.*?\s[{punc}]')
print(extra_space_re.match(mystring))