Import classes from parent folder in Python 3.6

Question

In a certain programme I have an abstract class that should be implemented in two different sub-packages for different purposes. A simplified version of this structure is:

programme/
  __init__.py
  Abstract.py
  pkgA/
    __init__.py
    ClassA.py
  pkgB/
    __init__.py
    ClassB.py

Up to Python 3.2 (I believe), it was possible to import the Abstract class within the a subfolder with a relative reference:

from .. import Abstract

Which with Python 3.6 devolves the error message:

ValueError: attempted relative import beyond top-level package

The classical (and ugly) alternative is to add the parent folder to the path at run time:

import sys
import os

sys.path.append(os.getcwd() + '/..')
from programme import Abstract

But this also fails:

ModuleNotFoundError: No module named 'programme'

How is this now done with Python 3.6? Preferably without modifying the path at run time.

Please re-read the answer you linked: https://stackoverflow.com/a/9331002/2066215 It doesn't mention `parent` anywhere, nor is there a file or folder called `parent` in your directory structure, so there is `No module named 'parent'`, but you can `import Abstract` — ForceBru, Jan 23 '20 at 15:12
Does this answer your question? [Relative imports for the billionth time](https://stackoverflow.com/questions/14132789/relative-imports-for-the-billionth-time) — Chris, Jan 23 '20 at 15:20
@Chris That question is on Python 2.7. The mechanisms for that version no longer apply to more recent versions. — Luís de Sousa, Jan 23 '20 at 16:23
Okay, then [Relative imports in Python 3](https://stackoverflow.com/q/16981921/354577) — Chris, Jan 23 '20 at 16:47

score 2 · Answer 1 · answered Jan 24 '20 at 16:39

You can use a __init__.py at the top level, programme/__init__.py, then import in the sub-module from programme.

Example:

# programme/__init__.py
from .Abstrac import Abstract

# programme/Abstract.py
class Abstract:
    def __init__(self):
        print('Abstract')

# programme/pkgA/ClassA.py
from programme import Abstract

class ClassA(Abstract):
    def __init__(self):
        super().__init__()

if __name__ == '__main__':
    a = ClassA()

Note, if you running as a script you should do:

$ cd programme
$ python -m programme.classA.ClassA
Abstract

This solution does not work. Running as you suggest returns the error: `Error while finding module specification for 'programme.classA.ClassA' (ModuleNotFoundError: No module named 'programme')`. Running `classA.py` from the `pkgA` folder returns: `ModuleNotFoundError: No module named 'programme'`. — Luís de Sousa, Jan 27 '20 at 15:51

score 1 · Answer 2 · answered Jan 23 '20 at 15:56

1

You need to a level higher than programme.

This needs to be added to the code:

import os, sys
current_dir = os.path.dirname(os.path.join(os.getcwd(), __file__))
sys.path.append(os.path.normpath(os.path.join(current_dir, '..', '..')))
from programme import Abstract

This has worked when doing python3.7 ClassA.py on programme/pkgA and it was picked from this response: Ultimate answer to relative python imports

answered Jan 23 '20 at 15:56

Jorge Mendes

309
2
14

Indeed this works. But it would be much cleaner without having to fiddle with the path at run time. – Luís de Sousa Jan 23 '20 at 16:22

Import classes from parent folder in Python 3.6

2 Answers2