Changing name of the file to parent folder name

Question

I have a bunch of folders in my directory. In each of them there is a file, which you can see below:

Regardless the file extension I would like to have the name of this file to be exactly the same as its parent folder, i.e. when considering folder 2023-10-18 I would like to have the file inside 2023-10-18 instead of occultation....

I tried to rename the multiple files by using this thread:

Renaming multiple files in a directory using Python

and here

https://pynative.com/python-rename-file/#:~:text=Use%20rename()%20method%20of,function%20to%20rename%20a%20file.

but unfortunately after application the code like this:

 import os
 from pathlib import Path
 pth = Path(__file__).parent.absolute()
 files = os.listdir(pth)

 for file in files:
 os.rename(os.pth.join(pth, file), os.pth.join(pth, '' + file + '.kml'))

I have an error:

AttributeError: module 'os' has no attribute 'pth'

described here:

AttributeError: 'module' object has no attribute

which says only a little to me, as I am a novice in Python.

How can I auto change the name of all the files in these directories? I need the same filename as the directory name. Is it possible?

UPDATE:

After hint below, my code looks like this now:

 import os
 from pathlib import Path
 pth = Path(__file__).parent.absolute()
 files = os.listdir(pth)

 for file in files:
  os.rename(os.path.join(pth, file), os.path.join(pth, '' + file + '.kml'))

but instead of changing the filename inside the folder list, all the files in the given directory have been changed to .kml. How can I access to the individual files inside the folderlist?

Answer 1

Reasoning for each line of code are commented! Every answer should use iglob, please read more about it here! The code is also suffix agnostic (.klm as the suffix is not hardcoded), and works in any scenario that requires this utility.

Only standard library functions were used.

The most satisfying method: move out of directory, rename, and delete directory

import os
from shutil import move
from glob import iglob
from pathlib import Path
from concurrent.futures import ThreadPoolExecutor


# The .py file has to be on the same directory as the folders containing the files!
root = Path(__file__).parent

# Using threading in case the operation becomes I/O bound (many files)
with ThreadPoolExecutor() as executor:
    for file in iglob(str(root / "**" / "*")):
        file = Path(file)

        # The new filename is the name of the directory, and the suffix(es) of the original file
        new_filename = f"{file.parent.name}{''.join(file.suffixes)}"

        # Move AND rename simultaneosly
        executor.submit(move, file, root / new_filename)

        # Delete directory because it is empty, and has no utility; ommit this line if not True
        executor.submit(os.rmdir, file.parent)

Less satisfying; OPs request: rename file (keep inside directory)

In case you really want to only rename the files, and keep them in their respective directory:

import os
from shutil import move
from glob import iglob
from pathlib import Path
from concurrent.futures import ThreadPoolExecutor


RENAME_ONLY = True


# The .py file has to be on the same directory as the folders containing the files!
root = Path(__file__).parent

# Using threading in case the operation becomes I/O bound
with ThreadPoolExecutor() as executor:
    for file in iglob(str(root / "**" / "*")):
        file = Path(file)

        # The new filename is the name of the directory, and the suffix(es) of the original file
        new_filename = f"{file.parent.name}{''.join(file.suffixes)}"

        if RENAME_ONLY:
            executor.submit(os.rename, file, file.parent / new_filename)
        else:
            # Move AND rename simultaneosly
            executor.submit(move, file, root / new_filename)

            # Delete directory because it is empty, and has no utility; ommit this line if not True
            executor.submit(os.rmdir, file.parent)

Why ''.join(file.suffixes)?

There are files that have multiple periods; like abc.x.yz. We get .yz with file.suffix, and .x.yz with ''.join(file.suffixes); hence my choice to use the latter.

It is a matter of sensitivity towards sub-suffixes, which are often important. For instance, .tar.gz files file.suffix wouldn't catch .tar, which is detrimental to the file format.

Answer 2

Here is a simple solution using only the os and shutil modules, both already preinstalled. It is cross-platform and works fine and fast for me. It can also handle multiple files being in each of the subfolders.

I think you can understand the code from the comments, but feel free to notify me if that's not the case.

import os, shutil
from os.path import * # just to avoid typing "os.path." everywhere

# I am using abspath() here to get the absolute path to the folder.
folder = abspath(input('Enter the main folder: '))

# index through all elements in the main folder that are directories
for subfolder in os.listdir(folder):
    abs_subfolder = join(folder, subfolder) # get the folder's absolute path
    if not isdir(abs_subfolder):
        continue # go to the next element because this one wasn't a folder

    # index through all the files in this subfolder
    for file in os.listdir(abs_subfolder):
        # get the extension of the file to move
        extension = splitext(file)[1]

        new_file_path = abs_subfolder + '.' + extension

        # move the file to its parent directory, and change its name
        shutil.move(join(abs_subfolder, file), new_file_path)

    # delete the directory the files were in
    # you can comment the next line if you don't want that to happen
    os.rmdir(abs_subfolder)

Basically, what this code does is to index through every directory inside the folder that contains all of these subfolders with files inside them.

Then, it searches for every file in each of these subfolders, then it moves these files to the main folder while changing their name to the subfolder's that they were in.

Finally, once every file of that subfolder has been moved and renamed, it removes the empty directory. You can just comment the last line if you don't want that to happen.

I hope that helps.

---

Also, I don't know where you got your code from, but the reason why you are getting .kml everywhere is because all that the code does is to rename all of your folders into their name + .kml. It doesn't even touch the files in the subfolders. I don't think I can make your code work as you want without changing pretty much everything in it.

---

If you want to learn more about the os module, check out this page as well as this one for os.path. I would say shutil is just a "complement" to the os module, and it shares some similarities with is, but you can see the full documentation here.

If you would like to learn Python in general, I think w3schools is the best place to go.

Answer 3

Here a sample code using pathlib module. Be sure to modify the base_folder.

Solution 1

"""
rename_filename.py

Rename filename inside the folders.

https://stackoverflow.com/questions/71408697/changing-name-of-the-file-to-parent-folder-name


Example:

base_folder
F:/Tmp/s13/

sub_folders
F:/Tmp/s13/2022-05-01
F:/Tmp/s13/2022-08-01

files under subfolder
F:/Tmp/s13/2022-05-01/aa.txt
F:/Tmp/s13/2022-08-01/bb.txt


Usage:

Be sure to modify first the "base_folder" value in the main()

command lines:
python rename_filename.py or
python3 rename_filename.py
"""

from pathlib import Path  # python version >= 3.4


def rename_file(base_folder):
    """
    Rename the filename of the file under the sub-folders of the base_folder.
    """
    base_path = Path(base_folder).glob('*/*')

    # Get the file path in every sub-folder.
    for file in base_path:
        # print(file)
        sub_folder_abs_path = file.parent  # sub-folder path, F:/Tmp/s13/2022-05-01
        sub_folder_name = file.parent.name  # sub-folder name, 2022-05-01

        # Rename the file to sub-folder name.
        new_file = Path(sub_folder_abs_path, sub_folder_name)
        file.rename(new_file)


def main():
    # Change the base folder according to your case.
    base_folder = 'F:/Tmp/s13/'
    rename_file(base_folder)


if __name__ == '__main__':
    main()

Solution 2
Uses pathlib and argparse modules. It offers --base-folder option to locate the base folder no need to modify the source. See usage.

"""
rename_file.py

Rename filename inside the folders.

https://stackoverflow.com/questions/71408697/changing-name-of-the-file-to-parent-folder-name


Example:

base_folder
F:/Tmp/s13/

sub_folders
F:/Tmp/s13/2022-05-01
F:/Tmp/s13/2022-08-01

files under subfolder
F:/Tmp/s13/2022-05-01/aa.txt
F:/Tmp/s13/2022-08-01/bb.txt


Usage:

command line:

python rename_file.py --base-folder "F:/Tmp/s13/"
"""

from pathlib import Path  # python version >= 3.4
import argparse


def rename_file(base_folder):
    """
    Rename the filename of the file under the sub-folders of the base_folder.
    """
    base_path = Path(base_folder).glob('*/*')

    # Get the file path in every sub-folder.
    for file in base_path:
        # print(file)
        sub_folder_abs_path = file.parent  # sub-folder path, F:/Tmp/s13/2022-05-01
        sub_folder_name = file.parent.name  # sub-folder name, 2022-05-01

        # Rename the file to sub-folder name.
        new_file = Path(sub_folder_abs_path, sub_folder_name)
        file.rename(new_file)


def main():
    parser = argparse.ArgumentParser(description='Rename file to sub-folder name.')
    parser.add_argument('--base-folder', required=True,
                        help='the base folder, example: --base-folder "f:/tmp/so13/"')

    args = parser.parse_args()
    rename_file(args.base_folder)


if __name__ == '__main__':
    main()

Usage:
Open command prompt and CD to the location of rename_file.py.

python rename_file.py --base-folder "f:/tmp/s13/"

Answer 4

So, based on what I understood, you have a single file in each folder. You would like to rename the file with the same folder name and preserve the extension.

import os

# Passing the path to your parent folders
path = r'D:\bat4'

# Getting a list of folders with date names
folders = os.listdir(path)

for folder in folders:
    files = os.listdir(r'{}\{}'.format(path, folder))

    # Accessing files inside each folder
    for file in files:

        # Getting the file extension
        extension_pos = file.rfind(".")
        extension = file[extension_pos:]

        # Renaming your file
        os.rename(r'{}\{}\{}'.format(path, folder, file),
                  r'{}\{}\{}{}'.format(path, folder, folder, extension))

I have tried it on my own files as follows:

This is an example for the output:

I hope I got your point. :)

Answer 5

You need to save it to a file (for example, rename.py) and call it using the python interpreter with an additional parameter - the name of the parent directory for which you want to rename the files in its subdirectories. For example: "python rename.py parent_dir". The directory name can be either absolute or relative. As an additional parameter, you can also specify the key for saving the extension when renaming a file (0 - do not save, 1 - save). Extensions are not saved by default. Here's an example with saving an extension: "python rename.py parent_dir 1".

Script in rename.py:

import os
import sys

def rename_first_file_in_dir(dir_path, new_file_name, keep_extension = False):
  for current_file_name in os.listdir(dir_path):
    current_file_path = os.path.join(dir_path, current_file_name) # full or relative path to the file in dir
    if not os.path.isfile(current_file_path):
      break
    # rename only base name of file to the name of directory
    if keep_extension:
      file_extension = os.path.splitext(current_file_name)[1]
      if len(file_extension) > 0:
        new_file_name = new_file_name + file_extension 
        
    new_file_path = os.path.join(dir_path, new_file_name)
    print("File " + current_file_name + " renamed to " + new_file_name + " in " + os.path.basename(dir_path) + " directory");
    os.rename(current_file_path, new_file_path)
    # exit after first processed file
    break

if len(sys.argv) < 2:
  print("Usage: python " + os.path.basename(__file__) + " <directory> [keep_files_extensions]") # help for usage
  exit(0)
scan_dir = sys.argv[1]
keep_extension = False if len(sys.argv) < 3 else not (int(sys.argv[2]) == 0) # optional parameter 0 - False, 1 - True, by default - False
if not os.path.exists(scan_dir):
  print("Error: directory " + scan_dir + " does not exists")
  exit(-1)
if not os.path.isdir(scan_dir):
  print("Error: file " + scan_dir + " is not a directory")
  exit(-1)
print("Scanning directory " + scan_dir)
for file_name in os.listdir(scan_dir): # walk through directory
  file_path = os.path.join(scan_dir, file_name)
  if os.path.isdir(file_path):
    rename_first_file_in_dir(file_path, file_name, keep_extension)