Change output filename accordingly to previous file versions in a folder

Question

In a Python script, I create an output file. I want its name to change accordingly to the its old versions in the output folder. For example:

output_file_name = 'output.txt'
if output.txt already exists --> output_file_name = output_(2).txt
if output_(2).txt already exists --> output_file_name = output_(3).txt
...

My trial:

import os.path
def check_versions(last_version_filename):
    # if an output_file_name_(n).txt is present then
    # return output_file_name(n+1).txt

    #extract the version of the file inside the output folder
    try:
        old_version = last_version_filename.split('(')[1].split(')')[0]

    # case in which no output_(n).txt version are present
    except IndexError:
        old_version = 1

    if old_version == 1:
        to_be_replaced = '.txt'
    else:
        to_be_replaced = '_({}).txt'.format(old_version)

    new_version = int(old_version) + 1
    last_version_filename = last_version_filename.replace(to_be_replaced,'_({}).txt'.format(new_version))


OUTPUT_FILE_NAME = 'output.txt'

# if output.txt file already exists, then we need to modify its name
if os.path.isfile('./' + OUTPUT_FILE_NAME):

    # last_file_version is the name of the last version file in the folder 
    OUTPUT_FILE_NAME = check_versions(last_version_filename)

So how can I check the last version of the file in a folder? Consider that its format will always be output_(n).txt. So the greatest n determines last file version.

In addition, the main fault of this code is that it works fine if

output.txt
output_(2).txt
.
.
.
output_(n).txt

are ALL present. But if output.txt is missing, the code will create output.txt as latest version output file. So you will have output.txt which is newer than output_(2).txt. And this could result tricky.

Again, it is clear I need to know which is the latest version and then naming the output file accordingly.

How would you solve this?

Answer 1

You could sort the filenames present in the directory that conform to your pattern, then add one to the number in the last one:

import os
import re

valid_filenames = [filename for filename in os.listdir(os.curdir) if filename.startswith('output_')]

n_valid = len(valid_filenames)

if n_valid == 0:
    if os.path.isfile('output.txt'):
        next_filename = 'output_(2).txt'

    else:
        next_filename = 'output.txt'

else:
    regex = re.compile(r'\((\d+)\)')
    tagged_filenames = sorted((int(regex.search(filename).group(1)), filename) for filename in valid_filenames)
    next_filename = f'output_({int(tagged_filenames[-1][0]) + 1}).txt'