Python/Regex - How to extract date from filename using regular expression?

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滥情空心 2020-11-27 08:05

I need to use python to extract the date from filenames. The date is in the following format:

month-day-year.somefileextension

Examples:

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  •  野性不改
    2020-11-27 08:47

    I think you can extract the date using re.split as follows

    $ ipython
    
    In [1]: import re
    
    In [2]: input_file = '10-12-2011.zip'
    
    In [3]: file_split = re.split('(\d{2}-\d{2}-\d{4})', input_file, 1)
    
    In [4]: file_split
    Out[4]: ['', '10-12-2011', '.zip']
    
    In [5]: file_split[1]
    Out[5]: '10-12-2011'
    
    In [6]: input_file = 'somedatabase-10-04-2011.sql.tar.gz'
    
    In [7]: file_split = re.split('(\d{2}-\d{2}-\d{4})', input_file, 1)
    
    In [8]: file_split
    Out[8]: ['somedatabase-', '10-04-2011', '.sql.tar.gz']
    
    In [9]: file_split[1]
    Out[9]: '10-04-2011'
    

    I ran the tests with Python 3.6.6, IPython 5.3.0

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