Python Regex \pL matching issues

Question

I'm trying to match a list of keywords I have, taking care to include all Latin characters (e.g accented).

Here's an example

import regex as re
p = r'((?!\pL)|^)blah((?!\pL)|$)'
print(re.search(p, "blah u"))
print(re.search(p, "blahé u"))
print(re.search(p, "éblah u"))
print(re.search(p, "blahaha"))

gives:


None
None
None

Which looks correct. However:

print(re.search(p, "u blah"))

gives:

None

This is wrong, as I expect a match for "u blah".

I've tried to also use Pythons built in re module, but I cannot get it to work with \pL or \p{Latin} due to "bad-escape" errors. I've also tried to use unicode strings (without the "r") but despite adding slashes to \\pL, I just can't get this to work right.

Note: I'm using Python 3.8

user_3832970 · Answer

The problem with your ((?!\pL)|^)blah((?!\pL)|$) regex is that the ((?!\pL)|^) group contains two alternatives where the first one always fails the regex (why? Because (?!\pL) is a negative lookahead that fails the match if the next char is a letter, and the next char to match is b in blah) and only ^ works all the time, i.e. your regex is equal to ^blah((?!\pL)|$) and only matches at the start of string.

Note (?!\pL) already matches a position at the end of string, so ((?!\pL)|$) = (?!\pL).

You should use

(?


See the regex demo (switched to PCRE for the demo purposes).
Note that the re-compatible version of the regex is
(?

See the regex demo.

Python Regex \pL matching issues

1 Answers