I have the following bizarre set up. Consider 3 scripts in different directories:
- root1/folderA/scriptA.py
- root2/folderB/scriptB.py
- root2/folderC/scriptC.py
The first file and it’s location are fully modifiable. The second and third are completely fixed.
scriptA.py contains a parent class:
class A: def get_path(self): # ... # code to determine "my_path" # ... return my_path
scriptB.py contains a child class:
from root1.folderA.scriptA import A class B(A): pass
scriptC.py contains the code to execute:
from root2.folderB.scriptB import B if __name__ == "__main__": b = B() print(b.get_path()) # want print root/folderB/scriptB.py
In scriptC.py what I want is the behaviour to get the path of the child class’s declaration file (without any hard-coding). Is there any way to program the A.get_path()
to have this behavoir?
I’ve tried looking into the inspect
, os.path
and pathlib
modules but I haven’t had any luck.
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Answer
It looks like the trick is to use __init_subclass__
which runs whenever a class is sub-classed, in conjunction with a class’s __module__
attribute to retrieve its containing module, and __file__
to retrieve the absolute path to the python script or module.
For example, in script_a.py
:
import sys from pathlib import Path class PathInfo: __slots__ = () path: Path def __init_subclass__(cls, **kwargs): mod = sys.modules[cls.__module__] # `__file__` is a string, so we want to convert to a `Path` so it's easier to work with cls.path = Path(mod.__file__) class A(PathInfo): __slots__ = ()
script_b.py
:
from script_a import A class B(A): pass # prints: /Users/<name>/PycharmProjects/my-project/script_a.py print(A().path) # prints: /Users/<name>/PycharmProjects/my-project/script_b.py print(B.path)