how to import a submodule in Python (without `exec`)

Question

I would like to import a submodule without knowing its name beforehand,

>>> __import__("os.path")
<module 'os' from '/usr/lib/python3.3/os.py'>

Doesn't work as you might expect, returning os, not os.path.

I came up with this solution.

def import_submodule(mod, submod):
    ns = {}
    exec_str = "from %s import %s as submod" % (mod, submod)
    exec(exec_str, ns, ns)
    return ns["submod"]

This gives the result:

>>> import_submodule("os", "path")
<module 'posixpath' from '/usr/lib/python3.3/posixpath.py'>

However I would rather not use exec() because its rather bad practice and seems unnecessary when Pythons import mechanisms are available already through __import__, imp and importlib modules.

Is there a way in Python3.x to do this kind of import though a function call, rather then using exec() ?

Answer 1

Use importlib.import_module:

>>> import importlib
>>> importlib.import_module('os.path')
<module 'posixpath' from '/usr/lib/python2.7/posixpath.pyc'>

This should work in python2.7+ and 3.1+.

Answer 2

Note that if you want do: from A import B as C as a function call, importlib.import_module won't always work, since B may not be a module.

Heres a function which uses importlib and getattr.

def my_import_from(mod_name, var_name):
    import importlib
    mod = importlib.import_module(mod_name)
    var = getattr(mod, var_name)
    return var

So this:

from os.path import dirname as var

Can be replaced with this:

var = my_import_from("os.path", "dirname")

Which avoids exec and allows both submodules and any variables defined in the module.

Since my question explicitly says importing a submodule, the answer from @Bakuriu is correct, however including this for completeness and it may help others who run into the same problem.