在可选括号之前和括号内部提取单词

Question

I have a bunch of strings that look like the following two sentences:

A couple of words (abbreviation)
A couple of words

I am trying to get python to extract the 'a couple of words' part and the 'abbreviation' part with a single regex, while also allowing strings where no abbreviation is given.

I've come up with this:

re_both = re.compile(r"^(.*)(?:\((.*)\))$")

It works for the first case, but not for the second case:

[in]   re_both.findall('a couple of words (abbreviation)')
[out]  [('a couple of words ', 'abbreviation')]

[in]   re_both.findall('a couple of words')
[out]  []

I would like the second case to yield:

[out] [('a couple of words','')]

Can this be done somehow?

Answer 1

Your regex is fine except you have to make the second part optional and make the first part non greedy.:

re_both = re.compile(r"^(.*?)(?:\((.*)\))?$")
#                   here __^      here __^

Answer 2

You need to make the second part as optional by adding a quantifier ? , and also you need to add the quantifer ? inside the first capturing group just after to .* so that it would do a non-greedy match.

^(.*?)(?:\((.*)\))?$
    ^             ^

DEMO

If you don't want to capture the space which was just before to the ( by the first capturing group then you could try the below regex,

^(.*?)(?: \((.*)\))?$

DEMO

>>> import re
>>> s = """A couple of words (abbreviation)
... A couple of words"""
>>> m = re.findall(r'^(.*?)(?: \((.*)\))?$', s, re.M)
>>> m
[('A couple of words', 'abbreviation'), ('A couple of words', '')]
>>> m = re.findall(r'^(.*?)(?:\((.*)\))?$', s, re.M)
>>> m
[('A couple of words ', 'abbreviation'), ('A couple of words', '')]