How to make word boundary \b not match on dashes

Question

I simplified my code to the specific problem I am having.

import re
pattern = re.compile(r'\bword\b')
result = pattern.sub(lambda x: "match", "-word- word")

I am getting

'-match- match'

but I want

'-word- match'

edit:

Or for the string "word -word-"

I want

"match -word-"

Answer 1

What you need is a negative lookbehind.

pattern = re.compile(r'(?<!-)\bword\b')
result = pattern.sub(lambda x: "match", "-word- word")

To cite the documentation:

(?<!...) Matches if the current position in the string is not preceded by a match for ....

So this will only match, if the word-break \b is not preceded with a minus sign -.

If you need this for the end of the string you'll have to use a negative lookahead which will look like this: (?!-). The complete regular expression will then result in: (?<!-)\bword(?!-)\b

Answer 2

\b basically denotes a word boundary on characters other than [a-zA-Z0-9_] which includes spaces as well. Surround word with negative lookarounds to ensure there is no non-space character after and before it:

re.compile(r'(?<!\S)word(?!\S)')