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

Let's say you have a module which contains

myfile = open('test.txt', 'r')

And the 'test.txt' file is in the same folder. If you'll run the module, the file will be opened successfully.

Now, let's say you import that module from another one which is in another folder. The file won't be searched in the same folder as the module where that code is.

So how to make the module search files with relative paths in the same folder first?

There are various solutions by using "__file__" or "os.getcwd()", but I'm hoping there's a cleaner way, like same special character in the string you pass to open() or file().

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

import os

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

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.