Python test given path to file in given path of folder

Question

User input two strings, one is filename and the other one is folder. The filename variable is a path to some file on disk. And the folder variable is a path to some folder on disk. I want to know if the file is located in the folder (either directly or in its sub folders).

For example:

isContains("C:\\a.txt", "C:\\") # True
isContains("C:\\a.txt", "C:\\a") # False
isContains("C:\\a.txt", "D:\\") # False
isContains("C:\\a\\b\\c\\d.txt", "C:\\") # True
isContains("C:\\a\\b\\c\\d.txt", "C:\\a\\b") # True

What I have done yet:

import os
isContains = lambda filename, folder: os.path.abspath(filename).startswith(os.path.join(os.path.abspath(folder), ''))

But I believe there must be some more elegant ways I didn't find out. As these code looks too complex. How should I implement this function?

My program is running on Windows. But I want the code be platform independent.

It's already implemented in pathlib: Path.is_relative_to(). You can just do Path(filename).is_relative_to(folder). Alternatively you can use os.path.commonprefix() and check for non-empty return: commonprefix((filename, folder)) — Olvin Roght
– Olvin Roght, Commented Jan 16, 2022 at 14:30
@OlvinRoght But os.path.commonprefix(('c:\\a\\b\\c\\d.txt', 'c:\\a\\d')) is not empty (should be false) — tsh
– tsh, Commented Jan 16, 2022 at 14:38
Exactly, I've not thought about all possible cases writing first comment. Option with commonprefix() requires additional return validation to meet all conditions, so if you don't want additional (unnecessary) headache - use pathlib. — Olvin Roght
– Olvin Roght, Commented Jan 16, 2022 at 14:38

Olvin Roght · Accepted Answer · 2022-01-16 14:49:13Z

2

If you have no strict requirement to use os.path, I'd recommend for all path-related work use pathlib, it will save you lot of time.

There's special method of Path(PurePath) class which does exactly what you're trying to implement - Path.is_relative_to(). Basically, you just need to initialize Path from your filename and call this method with folder:

Path(filename).is_relative_to(folder)

answered Jan 16, 2022 at 14:49

Olvin Roght

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