Python regex assistance

Question

I have a string that looks like this

24 (prem)-42-48 (6 ext)

and what I want to get out of it is

['24 prem', '42', '48', '6 ext']

I can get the numbers like this:

import re
MyString = r'24 (prem)-42-48 (6 ext)'
Splits = re.findall( r'(\d+)', MyString) # ['24','42','48','6']

but I lose the succeeding text.

I can also do this:

import re
MyString = r'24 (prem)-42-48 (6 ext)'
Splits = re.findall( r'[\\s:\\-]', MyString) # ['24 (prem)','42', '48 (6 ext)']

but that misses the (6 ext) item.

EDIT after seeing responses:

I think perhaps the simplest way for me to handle this would be to split on numbers and then just use str.replace to get rid of the "(" and " " characters.

So, is there a simple regex statement to split the string before the first character of a number?

The result from performing it on

'24 (prem)-42-48 (6 ext)'

would be

['24 (prem)-','42-',48 (', '6 ext)]

Answer 1

to get that result, you do not need regexps, all you need to do is remove the unwanted chars by replacing them with spaces and split the string on spaces:

>>> s ="24 (prem)-42-48 (6 ext)"
>>> l = s.replace('(',' ').replace('-',' ').replace('(',' ').replace(')',' ').split()
>>> l
['24', 'prem', '42', '48', '6', 'ext']

here's a version using translate for python3:

>>> s.translate(s.maketrans("()-", "   ")).split()
['24', 'prem', '42', '48', '6', 'ext']

here's a version using regexps:

>>> list(filter(lambda x: x is not '', re.findall('[^-() ]*', s)))
['24', 'prem', '42', '48', '6', 'ext']

though, I'm considering that the '24 prem' and '6 ext' in the result list is a typo you made, otherwise there's no generic way to do what you want, though you can achieve this by doing:

>>> [" ".join(l[:2])] + l[2:-2] + [" ".join(l[-2:])]
['24 prem', '42', '48', '6 ext']