Greedy match with negative lookahead in a regular expression

Question

I have a regular expression in which I'm trying to extract every group of letters that is not immediately followed by a "(" symbol. For example, the following regular expression operates on a mathematical formula that includes variable names (x, y, and z) and function names (movav and movsum), both of which are composed entirely of letters but where only the function names are followed by an "(".

re.findall("[a-zA-Z]+(?!\()", "movav(x/2, 2)*movsum(y, 3)*z")

I would like the expression to return the array

['x', 'y', 'z']

but it instead returns the array

['mova', 'x', 'movsu', 'y', 'z']

I can see in theory why the regular expression would be returning the second result, but is there a way I can modify it to return just the array ['x', 'y', 'z']?

Answer 1

An alternate approach: find strings of letters followed by either end-of-string or by a non-letter, non-bracket character; then capture the letter portion.

re.findall("([a-zA-Z]+)(?:[^a-zA-Z(]|$)", "movav(x/2, 2)*movsum(y, 3)*z")

Answer 2

Another solution which doesn't rely on word boundaries:

Check that the letters aren't followed by either a ( or by another letter.

>>> re.findall(r'[a-zA-Z]+(?![a-zA-Z(])', "movav(x/2, 2)*movsum(y, 3)*z")
['x', 'y', 'z']

Answer 3

You need to limit matches to whole words. So use \b to match the beginning or end of a word:

re.findall(r"\b[a-zA-Z]+\b(?!\()", "movav(x/2, 2)*movsum(y, 3)*z")

Answer 4

Add a word-boundary matcher \b:

>>> re.findall(r'[a-zA-Z]+\b(?!\()', "movav(x/2, 2)*movsum(y, 3)*z")
['x', 'y', 'z']

\b matches the empty string in between two words, so now you're looking for letters followed by a word boundary that isn't immediately followed by (. For more details, see the re docs.