Using \b in a regex, trying not to match words that start with $

Question

I'm having trouble getting the desired output using negative lookahead.

import re
text = "$FOO FOO $BAR BAR"

# Expected. Return words without 'F'.
re.findall(r"\b(?!F)\w+", text)
> ['BAR', 'BAR']

# Expected. Return words without 'B'.
re.findall(r"\b(?!B)\w+", text)
> ['FOO', 'FOO']

# Unexpected. Return words without '$'.
re.findall(r"\b(?!\$)\w+", text)
> ['FOO', 'FOO', 'BAR', 'BAR']

The first two work as expected. I expect the last one to return the list ['FOO', 'BAR'] matching words without the "$" character. Because it's a special character, I've tried various ways to escape it but haven't found the right solution.

Answer 1

You actually need to fix the pattern in the following way:

\b(?<!\$)\w+

See the Python demo .

The reason is that \b(?!\$)\w+ is equal to \b\w+ since $ cannot be matched with \w , so no need to restrict the first char matched with \w with the (?!\$) negative lookahead. You need to restrict the char that comes immediately before the first char matched wit \w , and that is done with a negative lookbehind , here, (?<!\$) .

import re
text = "$FOO FOO $BAR BAR"
print(re.findall(r"\b(?<!\$)\w+", text))
# > ['FOO', 'BAR']

Now, as you say (?<=^)(??\$)\w+|(?<=\s)(?!\$)\w+ works for you, you can now see that you may safely remove the lookaheads from the regex as they do not do anything meaningful, and the regex becomes (?<=^)\w+|(?<=\s)\w+ . This expression can be shrunk further into a slim (?<!\S)\w+ pattern that matches any one or more word chars that are immediately preceded with start of string or a whitespace.

Answer 2

Thanks to Charles for putting me on the right track. I had an incorrect understanding of how boundary characters function.

import re
text = "FOO $FOO FOO $BAR BAR"


re.findall('(?<=^)(?!\$)\w+|(?<=\s)(?!\$)\w+', text)
> ['FOO', 'FOO', 'BAR']

Replacing \b with a negative look-behind that matched a space or the beginning of a string gives the desired output.