更改文件夹中的文件名

Question

I have a folder that contains a lot of files that has a lot of copies which make them unreadable.

Example:

cow.txt
cow.txt(1)
cow.txt(2)
cow.txt(3)
dog.txt
dog.txt(1)

I would like to to have all the files structured in away that makes them able to be opened. Example

cow.txt
cow(1).txt
cow(2).txt
cow(3).txt
dog.txt
dog(1).txt

Any help you can provided would be greatly appreciated. I am just looking to make sure there name is changed, and am not looking to read each individual file. In addition if possible I would like to break up the files into 20k blocks. Thank you in advance.

I have tried using os.rename to simply rename the file but I am confused on how to do the efficiently as the numbers come after the.txt I then decided to read all the files and convert them to a pandas data frame and fix it that way. However I am confused on how to pull the files and make them with that name.

list_of_files = os.listdir()
df = pd.DataFrame(list_of_files, columns  = ['File_Name'])
df['.txt_removed'] = df.replace(to_replace = '.txt', value = '', regex = True)
df['txt_add'] = df['.txt_removed'] + '.txt'

To pull the files I would do something like this

for filewant_in df['txt_add']:
     if filewant in os.listdir():
         sutil.copy(os.path.join(filewant), 'new location')

I do not think this option will work even though it gives me my intended result. As I would like to change the overall file names.

Answer 1

You can use python's standard library, the os module has the os.rename function.

Like this:

It works like this:

os.rename('cow.txt(1)', 'cow(1).txt')

Answer 2

Create a.py file and paste the code below then run it. Change /mydir path with the path to the directory having the files. The code will loop through the directory finding all the containing have.txt as part of the file extension and renaming them to a.txt file. I hope it works.

import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt*"):
    file_name = os.path.basename(file)
    part_name = file_name.split(".", 1)
    new_name = part_name[0]+'.txt'
    os.rename(file,new_name)