Better way / regex to extract values form a dictionary

Paul Moore p.f.moore at gmail.com
Sat Jul 21 08:19:03 EDT 2018


def return_filename_test_case(filepath):
    filename = os.path.basename(filepath)
    testcase = filename.partition('_')[0]
    return filename, testcase

On 21 July 2018 at 12:37, Ganesh Pal <ganesh1pal at gmail.com> wrote:
> 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
> --
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