Changing name of the file to parent folder name

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.

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.

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) 

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/"