python – How to exactly match a word with one regexp?

Using Python 3.9.13.

How can I match an exact word without preceding or trailing character with a simple one regexp expression?

There are answers using the boundary word \b. However, according to the Python doc, it can match non word characters.

\b: Matches the empty string, but only at the beginning or end of a word. A word is defined as a sequence of word characters. Note that formally, \b is defined as the boundary between a \w and a \W character (or vice versa), or between \w and the beginning/end of the string.

For instance, I want to get a match only with the word cool, but no match with #cool or cool..

re.findall(
    pattern=r'\bcool\b', 
    string=" #cool", 
    flags=re.IGNORECASE|re.ASCII
)
---
['cool']

Alternative can be listing multiple patterns but it is not clean.

re.findall(
    pattern=r'\Acool\Z|\s+cool\s+|^cool\s+|\s+cool\Z', 
    string=" #cool", 
    flags=re.IGNORECASE|re.ASCII
)
---

Please advise if there is better and simple one regexp pattern to achieve this.

As answered by The fourth bird

re.findall(
    pattern=r'(?<!\S)cool(?!\S)', 
    string=" #cool cool& cool cool\n (cool)", 
    flags=re.IGNORECASE|re.ASCII
)
---
['cool', 'cool']

re.findall(
    pattern=r'(?<=\s|^)cool(?=\s|$)', 
    string=" #cool cool& cool cool\n (cool)", 
    flags=re.IGNORECASE|re.ASCII
)
---
['cool', 'cool']