Python regex words boundary with unexpected result

Question

import re
sstring = "ON Any ON Any"
regex1 = re.compile(r''' \bON\bANY\b''', re.VERBOSE)
regex2 = re.compile(r'''\b(ON)?\b(Any)?''', re.VERBOSE)
regex3 = re.compile(r'''\b(?:ON)?\b(?:Any)?''', re.VERBOSE)
for a in regex1.findall(sstring): print(a)
print("----------")
for a in regex2.findall(sstring): print(a)
print("----------")
for a in regex3.findall(sstring): print(a)
print("----------")

---

('ON', '') ('', '') ('', 'Any') ('', '') ('ON', '') ('', '') ('', 'Any')

('', '')

ON

Any

ON

Any

---

Having read many articles on the internet and S.O. I think I still don't understand the regex word boundary: \b

The first regex doesn't give me the expected result I think it's must give me "ON Any On Any" but it still not give me that.

The second regex gives me tuples and I don't know why or understand the meaning of: ('', '')

The third regex gives prints the results on separated lines and empty lines in betweens

Could you please help me to understand that.

Answer 1

Note that to match ON ANY you need to add an escaped (since you are using re.VERBOSE flag) space between ON and ANY as \b word boundary being a zero-width assertion does not consume any text, just asserts a position between specific characters. That is the reason for your first re.compile(r''' \bON\bANY\b''', re.VERBOSE) approach failure.

Use

rx = re.compile(r''' \bON\ ANY\b ''', re.VERBOSE|re.IGNORECASE)

See the Python demo

The re.compile(r'''\b(ON)?\b(Any)?''', re.VERBOSE) returns tuples since you defined (...) capturing groups in the pattern.

The re.compile(r'''\b(?:ON)?\b(?:Any)?''', re.VERBOSE) matches optional sequences, either ON or Any, so you get those words as values. You get empty values as well because this regex can match just a word boundary (all other subpatterns are optional).

More details about word boundaries:

• Word boundaries at Regular-Expressions.info
• Java Regex Word Boundaries (this is still a word boundary in a regex, also applicable here)