Better way / regex to extract values form a dictionary

[Ganesh Pal]

I have one of the dictionary values in the below format

'/usr/local/ABCD/EDF/ASASAS/GTH/HELLO/MELLO/test04_Failures.log'
'/usr/local/ABCD/EDF/GTH/HEL/OOLO/MELLO/test02_Failures.log'
'/usr/local/ABCD/EDF/GTH/BEL/LO/MELLO/test03_Failures.log'

I need to extract the file name in the path example, say test04_Failure.log
and testcase no i.e test04


Here is my solutions:

gpal-cwerzvd-1# vi filename.py
import re

Common_dict = {}
Common_dict['filename'] =
'/usr/local/ABCD/EDF/GTH/HELLO/MELLO/test04_Failures.log'

def return_filename_test_case(filepath):
    if filepath:
       filename = re.findall(r'(test\d{1,4}_\S+)', filepath)
    if filename:
       testcase = re.findall(r'(test\d{1,4})', ''.join(filename))

    return filename, testcase


if Common_dict['filename']:
   path = Common_dict['filename']
   fname, testcase = return_filename_test_case(path)
   print fname, testcase


op:
qerzvd-1# python filename.py
['test04_Failures.log']
['test04']


Please suggest how can this code can be optimized  further looks messy ,
what would be your one liner or a simple solution to return both test-case
no and filename

I am on Python 2.7 and Linux


Regards,
Ganesh