Callable modules

Question

Python doesn't allow modules to override or add any magic method, because keeping module objects simple, regular and lightweight is just too advantageous considering how rarely strong use cases appear where you could use magic methods there.

When such use cases do appear, the solution is to make a class instance masquerade as a module. Specifically, code your mod_call.py as follows:

import sys

class mod_call:
    def __call__(self):
        return 42

sys.modules[__name__] = mod_call()

Now your code importing and calling mod_call works fine.

Special methods are only guaranteed to be called implicitly when they are defined on the type, not on the instance. (__call__ is an attribute of the module instance mod_call, not of <type 'module'>.) You can't add methods to built-in types.

https://docs.python.org/reference/datamodel.html#special-lookup

As Miles says, you need to define the call on class level. So an alternative to Alex post is to change the class of sys.modules[__name__] to a subclass of the type of sys.modules[__name__] (It should be types.ModuleType).

This has the advantage that the module is callable while keeping all other properties of the module (like accessing functions, variables, ...).

import sys

class MyModule(sys.modules[__name__].__class__):
    def __call__(self):  # module callable
        return 42

sys.modules[__name__].__class__ = MyModule

Note: Tested with python3.6.

Christoph Böddeker's answer seems to be the best way to create a callable module, but as a comment says, it only works in Python 3.5 and up.

The benefit is that you can write your module like normal, and just add the class reassignment at the very end, i.e.

# coolmodule.py
import stuff

var = 33
class MyClass:
   ...
def function(x, y):
   ...

class CoolModule(types.ModuleType):
    def __call__(self):
        return 42
sys.modules[__name__].__class__ = CoolModule

and everything works, including all expected module attributes like __file__ being defined. (This is because you're actually not changing the module object resulting from the import at all, just "casting" it to a subclass with a __call__ method, which is exactly what we want.)

To get this to work similarly in Python versions below 3.5, you can adapt Alex Martelli's answer to make your new class a subclass of ModuleType, and copy all the module's attributes into your new module instance:

#(all your module stuff here)

class CoolModule(types.ModuleType):
    def __init__(self):
        types.ModuleType.__init__(self, __name__)
        # or super().__init__(__name__) for Python 3
        self.__dict__.update(sys.modules[__name__].__dict__)
    def __call__(self):
        return 42

sys.modules[__name__] = CoolModule()

Now __file__, __name__ and other module attributes are defined (which aren't present if just following Alex's answer), and your imported module object still "is a" module.

All answers work only for import mod_call. To get it working simultaneously for from mod_call import *, the solution of @Alex Martelli can be enhanced as follow

import sys

class mod_call:
    def __call__(self):
        return 42
    mod_call = __call__
    __all__ = list(set(vars().keys()) - {'__qualname__'})   # for python 2 and 3

sys.modules[__name__] = mod_call()

This solution was derived with the discussion of an answer of a similar problem.

To turn the solution into a convenient reusable function:

def set_module(cls, __name__):
    import sys

    class cls2(sys.modules[__name__].__class__, cls):
        pass

    sys.modules[__name__].__class__ = cls2

save it to, say, util.py. Then in your module,

import util

class MyModule:
    def __call__(self):  # module callable
        return 42

util.set_module(MyModule, __name__)

Hurray!

I wrote this because I need to enhance a lot of modules with this trick.

---

P.S. Few days after I wrote this answer, I removed this trick from my code, since it is so tricky for tools like Pylint to understand.