Python subpackage in arbitrary folder. How to write the init.py?

Question

If I have this hierarchy

A
|- __init__.py
|-B
  |- __init__.py
  |- funcitons.py 
     (which contains def my_function(): pass)

and I have package A installed, I can do the following

from A.B import functions
functions.my_function()

or

from A.B.functions import my_function
my_function()

How to accomplish the same result if B is encapsulated in several subfolders (which are not packages)?

touching `sys.path` is always wrong. Which python version is this? Python 2? Python 3? Modify `__path__` in the `A/__init__.py` — Antti Haapala -- Слава Україні, Jul 25 '17 at 13:18
One is https://stackoverflow.com/questions/2699287/what-is-path-useful-for - the other is to use https://www.python.org/dev/peps/pep-0420/ — Antti Haapala -- Слава Україні, Jul 25 '17 at 13:20
then the former. Naturally compatibility would be useful when writing in a language that has already been given a death sentence. — Antti Haapala -- Слава Україні, Jul 25 '17 at 13:25

score 1 · Accepted Answer · answered Jul 26 '17 at 08:18

1

As suggested by Antti Haapala, in Python 2 this is what __path__ is for. The following A/__init__.py did the job.

import os
__path__.append( os.path.join(os.path.dirname(__file__), 'res/path'))

Python subpackage in arbitrary folder. How to write the __init__.py?