将文件夹的文件夹中的文件重命名为其父文件夹？

Question

I have a batch of folders that have a name based on the date. Each folder has a folder where they have file names which are all the same.

Is there a way rename the files so they become unique based on the directory structure (which appears is the parent folder (the first folder) which is based on the date) that they are held within.

 \user\date\1_2_2019\ABC\0001.csv -> abc_1_2_2019.csv

 \user\date\1_3_2019\JKL\0001.csv -> JKL_1_3_2019.csv

 \user\date\1_4_2019\XYZ\0001.csv -> XYZ_1_4_2019.csv

 \user\date\1_5_2019\123\0001.csv -> 123_1_5_2019.csv

 \user\date\1_6_2019\456\0001.csv -> 456_1_6_2019.csv

I know the basic python code to get to all the files is this

  import os
   for dirname, _, filenames in os.walk('\user\date'):
     for filename in filenames:
       print(os.path.join(dirname, filename))

But is there a python code to change all the names of the files to add at the very least have the date of the parent file in the beginning name.

thanks in advance!

Answer 1

Here is one-way using pathlib from python 3.4+ and f-strings from python 3.6+

first you need to set your path at the top-level directory, so we can recursively find all the csv files and rename with a simple for loop.

from pathlib import Path

files = Path(r'C:\Users\datanovice\Documents\Excels').rglob('*.csv')
# remove 'r' string if you're on macos.

for file in files:
    parent_1 = file.parent.name
    parent_2 = file.parent.parent.name
    file.rename(Path(file.parent,f"{parent_1}_{parent_2}{file.suffix}"))
    print(f"{file.name} --> {parent_1}_{parent_2}{file.suffix}")
#1.csv --> ABC_1_2_2019.csv
#1.csv --> ABC_2_2_2019.csv

---

result

for f in files:
    print(f)
C:\Users\datanovice\Documents\Excels\1_2_2019\ABC\1.csv
C:\Users\datanovice\Documents\Excels\2_2_2019\ABC\1.csv
#after
for f in files:
    print(f)
C:\Users\datanovice\Documents\Excels\1_2_2019\ABC\ABC_1_2_2019.csv
C:\Users\datanovice\Documents\Excels\2_2_2019\ABC\ABC_2_2_2019.csv