Difference between defining typing.Dict and dict?

I am practicing using type hints in Python 3.5. One of my colleague uses typing.Dict:

    import typing

    def change_bandwidths(new_bandwidths: typing.Dict,
                          user_id: int,
                          user_name: str) -> bool:
        print(new_bandwidths, user_id, user_name)
        return False

    def my_change_bandwidths(new_bandwidths: dict,
                             user_id: int,
                             user_name: str) ->bool:
        print(new_bandwidths, user_id, user_name)
        return True

    def main():
        my_id, my_name = 23, "Tiras"
        simple_dict = {"Hello": "Moon"}
        change_bandwidths(simple_dict, my_id, my_name)
        new_dict = {"new": "energy source"}
        my_change_bandwidths(new_dict, my_id, my_name)

    if __name__ == "__main__":

Both of them work just fine, there doesn't appear to be a difference.

I have read the typing module documentation.

Between typing.Dict or dict which one should I use in the program?

There is no real difference between using a plain typing.Dict and dict, no.

However, typing.Dict is a Generic type that lets you specify the type of the keys and values too , making it more flexible:

    def change_bandwidths(new_bandwidths: typing.Dict[str, str],
                          user_id: int,
                          user_name: str) -> bool:

As such, it could well be that at some point in your project lifetime you want to define the dictionary argument a little more precisely, at which point expanding typing.Dict to typing.Dict[key_type, value_type] is a 'smaller' change than replacing dict.

You can make this even more generic by using Mapping or MutableMapping types here; since your function doesn't need to alter the mapping, I'd stick with Mapping. A dict is one mapping, but you could create other objects that also satisfy the mapping interface, and your function might well still work with those:

    def change_bandwidths(new_bandwidths: typing.Mapping[str, str],
                          user_id: int,
                          user_name: str) -> bool:

Now you are clearly telling other users of this function that your code won't actually alter the new_bandwidths mapping passed in.

Your actual implementation is merely expecting an object that is printable. That may be a test implementation, but as it stands your code would continue to work if you used new_bandwidths: typing.Any, because any object in Python is printable.

From: stackoverflow.com/q/37087457