Python how to alias module name (rename with preserving backward compatibility)

This forces you to keep a foo directory, but I think it the best way to get this to work.

Directory setup:

bar
├── __init__.py
└── baz.py
foo
└── __init__.py

foo_bar.py

bar/__init__.py is empty.
bar/baz.py: worked = True

foo/__init__.py:

import sys

# make sure bar is in sys.modules
import bar
# link this module to bar
sys.modules[__name__] = sys.modules['bar']

# Or simply
sys.modules[__name__] = __import__('bar')

foo_bar.py:

import foo.baz

assert(hasattr(foo, 'baz') and hasattr(foo.baz, 'worked'))
assert(foo.baz.worked)

import bar
assert(foo is bar)

This answer work with submodules:

import sys
import os
from importlib.abc import MetaPathFinder, Loader
import importlib
from MainModule.SubModule import *

class MyLoader(Loader):
    def module_repr(self, module):
        return repr(module)

    def load_module(self, fullname):
        old_name = fullname
        names = fullname.split(".")
        names[1] = "SubModule"
        fullname = ".".join(names)
        module = importlib.import_module(fullname)
        sys.modules[old_name] = module
        return module


class MyImport(MetaPathFinder):
    def find_module(self, fullname, path=None):
        names = fullname.split(".")
        if len(names) >= 2 and names[0] == "Module" and names[1] == "LegacySubModule":
            return MyLoader()


sys.meta_path.append(MyImport())

Do you mean something like this?

import foo as bar

you can use shortcuts for module imports like:

from numpy import array as arr

in: arr([1,2,3])
out: array([1, 2, 3])

and you can also use more than one alias at a time

from numpy import array as foo
in: foo([1,2,3])
out: array([1, 2, 3])

if your foo is a class you can do:

bar=foo()

and call a subfunction of it by:

bar.conf()

Does this help you?