Create folder and subfolder based on filename that contain the date

Question

I have multiple files that contain the date. for example the file names are :

one08172021patient.txt
two09182021patient.txt
three09182021patient.txt

I would to write a script in Python to create folder for year ,month,day and move the file based on the date that’s already in the file name.
I have looked around for this logic in google but i didn’t get any. result for this case.

After running the code the result will be similar to this :

+---2021 
|   ---08 
|       ----17   one08172021patient.txt
|
---2021
    +---09
        ------18
                  two09182021patient.txt

Asked By: pirate

||

Source

Answer 1

This can be simply done using regexes through re module , an example of that would be :

import re


data= ["one08172021patient.txt","two09182021patient.txt","three09182021patient.txt"]


for file in data:
    date = re.match(r"^D+(d{2})(d{2})(d{4})patient.txt",file)
    month = date[1]
    day = date[2]
    year = date[3];
    print(month,day,year)
    # Your code to save file

if __name__ == "__main__":
    pass

which would give the following ouput

08 17 2021
09 18 2021
09 18 2021

Answered By: majdsalloum

Answer 2

You can use this script:

import re
from os import walk, rename
from pathlib import Path

src_path = './src'
des_path = './dst'
filenames = next(walk(src_path), (None, None, []))[2]

for file in filenames:
    match = re.search(r'^D+(d{2})(d{2})(d{4})', file)
    if match:
        month = match.group(1)
        day = match.group(2)
        year = match.group(3)
        path = f'{des_path}/{year}/{month}/{day}'
        Path(path).mkdir(parents=True, exist_ok=True)
        rename(f'{src_path}/{file}', f'{path}/{file}')

Answered By: Alireza MH

Create folder and subfolder based on filename that contain the date

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