python regex to get all text until a (, and get text inside brackets

Question

I need help with 2 regex.

1. get all text until a open bracket.

e.g. this is so cool (234) => 'this is so cool'

1. get the text inside the brackets, so the # '234'

Answer 1

Up until the paren: regex = re.compile("(.*?)\s*\(")

Inside the first set of parens: regex = re.compile(".*?\((.*?)\)")

Edit: Single regex version: regex = re.compile("(.*?)\s*\((.*?)\)")

Example output:

>>> import re
>>> r1 = re.compile("(.*?)\s*\(")
>>> r2 = re.compile(".*?\((.*?)\)")
>>> text = "this is so cool (234)"
>>> m1 = r1.match(text)
>>> m1.group(1)
'this is so cool'
>>> m2 = r2.match(text)
>>> m2.group(1)
'234'
>>> r3 = re.compile("(.*?)\s*\((.*?)\)")
>>> m3 = r3.match(text)
>>> m3.group(1)
'this is so cool'
>>> m3.group(2)
'234'
>>> 

Note of course that this won't work right with multiple sets of parens, as it's only expecting one parenthesized block of text (as per your example). The language of matching opening/closing parens of arbitrary recurrence is not regular.

Answer 2

No need for regular expression.

>>> s="this is so cool (234)"
>>> s.split("(")[0]
'this is so cool '
>>> s="this is so cool (234) test (123)"
>>> for i in s.split(")"):
...  if "(" in i:
...     print i.split("(")[-1]
...
234
123

Answer 3

Here is my own library function version without regex.

def between(left,right,s):
    before,_,a = s.partition(left)
    a,_,after = a.partition(right)
    return before,a,after

s="this is so cool (234)"
print('\n'.join(between('(',')',s)))

Answer 4

Sounds to me like you could just do this:

re.findall('[^()]+', mystring)

Splitting would work, too:

re.split('[()]', mystring)

Either way, the text before the first parenthesis will be the first item in the resulting array, and the text inside the first set of parens will be the second item.