Find String Not Preceded By Other String

I want to get only ['bar'] here: >>> re.findall(r'(?

Solution 1:

The current regex matches oo in foo because oo( is not preceded with "def ".

To stop the pattern from matching inside a word, you may use a a word boundary, \b and the fix might look like r"\b(?<!\bdef )([a-zA-Z0-9.]+?)\(".

Note that identifiers can be matched with [a-zA-Z_][a-zA-Z0-9_], so your pattern can be enhanced like

re.findall(r'\b(?<!\bdef\s)([a-zA-Z_]\w*(?:\.[a-zA-Z_]\w*)*)\(', s, re.A)

Note that re.A or re.ASCII will make \w match ASCII only letters, digits and _.

See the regex demo.

Details

• \b - a word boundary
• (?<!\bdef\s) - no def + space allowed immediately to the left of the current location
• ([a-zA-Z_]\w*(?:\.[a-zA-Z_]\w*)*) - Capturing group 1 (its value will be the result of re.findall call):
  • [a-zA-Z_] - an ASCII letter or _
  • \w* - 1+ word chars
  • (?: - start of a non-capturing group matching a sequence of...
    • \. - a dot
    • [a-zA-Z_] - an ASCII letter or _
    • \w* - 1+ word chars
• )* - ... zero or more times
• \( - a ( char.

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