使用 python 根据包含日期的文件名创建文件夹和子文件夹

create folder and subfolder based on filename that contain the date using python

我有多个包含日期的文件。例如文件名是:

one08172021patient.txt
two09182021patient.txt
three09182021patient.txt

我想在 python 中编写一个脚本来为年、月、日创建文件夹,并根据文件名中已有的日期移动文件。 我在 google 中四处寻找这个逻辑,但我没有找到。本案例的结果。

在 运行 代码之后,结果将类似于:

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

这可以通过 re 模块使用正则表达式简单地完成,例如:

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

这将给出以下输出

08 17 2021
09 18 2021
09 18 2021

您可以使用这个脚本:

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}')