[FIXED] Python regex match space-separated words that contain two or fewer o characters

Issue

I am new to python and trying to solve some problems (in the way to learn).
I want to match space-separated words that contain two or fewer o characters.
That is what I actually did:

import re

pattern = r'\b(?:[^a\s]*o){1}[^a\s]*\b'

text = "hop hoop hooop hoooop hooooop"
print(re.findall(pattern, text))

When I run my code it does match all the words in the string..
Any suggestion?

Solution

You can use

import re
pattern = r'(?<!\S)(?:[^\so]*o){0,2}[^o\s]*(?!\S)'
text = "hop hoop hooop hoooop hooooop"
print(re.findall(pattern, text))

# Non  regx solution:
print([x for x in text.split() if x.count("o") < 3])

See the Python demo. Both yield ['hop', 'hoop'].

The (?<!\S)(?:[^\so]*o){0,2}[^o\s]*(?!\S) regex matches

• (?<!\S) - a left-hand whitespace boundary
• (?:[^\so]*o){0,2} - zero, one or two occurrences of any zero or more chars other than whitespace and o char, and then an o char
• [^o\s]* - zero or more chars other than o and whitespace
• (?!\S) - a right-hand whitespace boundary

Answered By - Wiktor Stribiżew