How to run code based on file name in Python?

Question

I have a folder at the moment containing various files that I need to run some code on and convert into a csv. I'm trying to see if there is a way to check the name of the file in the folder and then execute the respective code. I have tried:

for xlsb_file in pathlib.Path('Directory Path').glob('Filename*.xlsb'):
RUN CODE
convert dataset into a csv

For some reason this code seems to execute without any error but no csv file is outputted. Ideally what I would like is when I run the code, for the code to detect the filename, run the respective code and then spit out a csv.

Is there anyway to do this? All help is appreciated!

Could a solution be to supply the directory name via command line argument? — Reedinationer
– Reedinationer, Commented Mar 31, 2019 at 23:29
Of course you can analyze the file names and take appropriate action. What’s your question about doing so? — Davis Herring
– Davis Herring, Commented Apr 1, 2019 at 2:15

edinho · Accepted Answer · 2019-04-01 03:15:35Z

1

You can try:

import os

def convert_to_csv(filename):
    """this function converts and save a csv version of the file"""

def convert_this_file(filename):
    """this functions checks if a file should be converted"""
    return filename.endswith('.xlsb')

for root, folders, files in os.walk('Directory Path'):
    for file in files:
        filename = os.path.join(root, file)
        if convert_this_file(filename):
            convert_to_csv(filename)
    break

Remove the 'break' if you want the code scan sub directories recursively

answered Apr 1, 2019 at 3:15

edinho

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