Regex that matches punctuation at the word boundary including underscore

Question

I am looking for a Python regex for a variable phrase with the following properties: (For the sake of example, let\'s assume the variable phrase here is taking the value

Answer 1

Here is a regex that might solve it:

Regex

(?<=[\W_]+|^)and(?=[\W_]+|$)

Example

# import regex

string = 'this_And'
test = regex.search(r'(?<=[\W_]+|^)and(?=[\W_]+|$)', string.lower())
print(test.group(0))
# prints 'and'

# No match
string = 'Andy'
test = regex.search(r'(?<=[\W_]+|^)and(?=[\W_]+|$)', string.lower())
print(test)
# prints None

strings = [ "this_and", "this.and", "(and)", "[and]", "and^", ";And"]
[regex.search(r'(?<=[\W_]+|^)and(?=[\W_]+|$)', s.lower()).group(0) for s in strings if regex.search(r'(?<=[\W_]+|^)and(?=[\W_]+|$)', s.lower())]
# prints ['and', 'and', 'and', 'and', 'and', 'and']

Explanation

[\W_]+ means we accept before (?<=) or after (?=) and only non-word symbols except the underscore _ (a word symbol that) is accepted. |^ and |$ allow matches to lie at the edge of the string.

Edit

As mentioned in my comment, the module regex does not yield errors with variable lookbehind lengths (as opposed to re).

# This works fine
# import regex
word = 'and'
pattern = r'(?<=[\W_]+|^){}(?=[\W_]+|$)'.format(word.lower())
string = 'this_And'
regex.search(pattern, string.lower())

However, if you insist on using re, then of the top of my head I'd suggest splitting the lookbehind in two (?<=[\W_])and(?=[\W_]+|$)|^and(?=[\W_]+|$) that way cases where the string starts with and are captured as well.

# This also works fine
# import re
word = 'and'
pattern = r'(?<=[\W_]){}(?=[\W_]+|$)|^{}(?=[\W_]+|$)'.format(word.lower(), word.lower())
string = 'this_And'
re.search(pattern, string.lower())