python return matching and non-matching patterns of string

Question

I would like to split a string into parts that match a regexp pattern and parts that do not match into a list.

For example

import re
string = 'my_file_10'
pattern = r'\d+$'
#  I know the matching pattern can be obtained with :
m = re.search(pattern, string).group()
print m
'10'
#  The final result should be as following
['my_file_', '10']

Answer 1

Put parenthesis around the pattern to make it a capturing group, then use re.split() to produce a list of matching and non-matching elements:

pattern = r'(\d+$)'
re.split(pattern, string)

Demo:

>>> import re
>>> string = 'my_file_10'
>>> pattern = r'(\d+$)'
>>> re.split(pattern, string)
['my_file_', '10', '']

Because you are splitting on digits at the end of the string, an empty string is included.

If you only ever expect one match, at the end of the string (which the $ in your pattern forces here), then just use the m.start() method to obtain an index to slice the input string:

pattern = r'\d+$'
match = re.search(pattern, string)
not_matched, matched = string[:match.start()], match.group()

This returns:

>>> pattern = r'\d+$'
>>> match = re.search(pattern, string)
>>> string[:match.start()], match.group()
('my_file_', '10')