Printing file path of lift elements

[dyerlee91]

Say I generated a list of all the files in a specific directory, but instead of just the file names, I wanted to print the complete file path for each element into a .txt file. Is is possible to associate a file path with something that's now included in a list? I'm very much a novice, so apologies if this doesn't make sense.

[snippsat]

I'm very much a novice, so apologies if this doesn't make sense.

Yes it make sense,but you should give try with some code then you show some effort and easier to help.
Here a couple of examples using pathlib

import pathlib

lst = []
mydir = r'G:\div_code\answer\weather'
for file in pathlib.Path(mydir).iterdir():
    if file.is_file():
        lst.append(file.name)

>>> lst
['file_1_0_.txt', 'file_33.txt', 'key.ini', 'open_weather.py', 'to_zip.txt']
>>> 
>>> for file in lst:
...     print(f'{mydir}{file}')
...     
G:\div_code\answer\weatherfile_1_0_.txt
G:\div_code\answer\weatherfile_33.txt
G:\div_code\answer\weatherkey.ini
G:\div_code\answer\weatheropen_weather.py
G:\div_code\answer\weatherto_zip.txt

So over generate absolute path for files in the list.

If remove .name then get absolute path with one go.

import pathlib

lst = []
mydir = r'G:\div_code\answer\weather'
for file in pathlib.Path(mydir).iterdir():
    if file.is_file():
        lst.append(file)

>>> lst
[WindowsPath('G:/div_code/answer/weather/file_1_0_.txt'),
 WindowsPath('G:/div_code/answer/weather/file_33.txt'),
 WindowsPath('G:/div_code/answer/weather/key.ini'),
 WindowsPath('G:/div_code/answer/weather/open_weather.py'),
 WindowsPath('G:/div_code/answer/weather/to_zip.txt')]