How to make an "always relative to current module" file path?

Solution 1:

The solution is to use __file__ and it's pretty clean:

import os

TEST_FILENAME = os.path.join(os.path.dirname(__file__), 'test.txt')

Solution 2:

For normal modules loaded from .py files, the __file__ should be present and usable. To join the information from __file__ onto your relative path, there's a newer option than os.path interfaces available since 2014:

from pathlib import Path

here = Path(__file__).parent
fname = here / "test.txt"
with fname.open() as f:
    ...

pathlib was added to Python in 3.4 - see PEP428. For users still on Python 2.7 wanting to use the same APIs, a backport is available.

Note that when you're working with a Python package, there are better approaches available for reading resources - you could consider moving to importlib-resources. This requires Python 3.7+, for older versions you can use pkgutil. One advantage of packaging the resources correctly, rather than joining data files relative to the source tree, is that the code will still work in cases where it's not extracted on a filesystem (e.g. a package in a zipfile). See How to read a (static) file from inside a Python package? for more details about reading/writing data files in a package.