Proper way of getting path of main script from library in Python

Question

I'm writing a Python library which which you can load an object from a file and do something with it. For convenience, I'd like to make it so that people can provide three kinds of paths:

1. A path starting with a "/", which will be interpreted as an absolute path

2. A path starting with a "~/", which will be interpreted as relative to the user's home directory (for which I plan to use os.path.expanduser)

3. A path starting with neither, which would be interpreted as relative to the directory of the top-level script that is importing my library.

My question is: what is the best way of approaching the third case? The two ways I have come across on Stack Overflow are both kind of klugey:

1)

import __main__
if hasattr(__main__, "__file__"):
    script_dir = os.path.dirname(os.path.abspath(__main__.__file__))

import inspect
frame_info = inspect.stack()[-1]
mod = inspect.getmodule(frame_info[0])
script_dir = os.path.dirname(mod.__file__)

Obviously both of these will fail in the case that we're running from an interactive terminal or something, and in that case I would want to fall back to just treating it as an absolute path.

Anyway, I get the sense that using inspect like this in a library is frowned upon, but the other way seems klugey as well. Is there a better way to do this that I'm unaware of?

Answer 1

Use the built-in package pathlib's getcwd method. It defaults to the root folder of the top package.

Will not work if you change the working directory. Although, an unprefixed path is common practice to be relative to the working directory.

import pathlib
print(pathlib.Path.cwd())

>>> A:\Programming\Python\generalfile