Python3 super() and generic class

Question

I believe a test case is worth a thousand words:

#!/usr/bin/env python3

def generate_a(key):
    class A(object):
        def method(self):
            return {'key': key,}
    return A

BaseForB = generate_a(1337)

class B(BaseForB):
    def method(self):
        dict = super(BaseForB, self).method()
        dict.update({'other_key': 0,})
        return dict

EXPECTED = {'other_key': 0, 'key': 1337,}
RESULT = B().method()

if EXPECTED == RESULT:
    print("Ok")
else:
    print("EXPECTED: ", EXPECTED)
    print("RESULT: ", RESULT)

This raises:

AttributeError: 'super' object has no attribute 'method'

The question is - how to run A.method() in B.method() (the thing I tried to do with super())

edit

Here's more appropriate test case:

#!/usr/bin/env python3

def generate_a(key):
    class A(object):
        def method(self):
            return {'key': key,}
    return A

class B(object):
    def method(self):
        return {'key': 'thisiswrong',}

BaseForC = generate_a(1337)

class C(B, BaseForC):
    def method(self):
        dict = super(C, self).method()
        dict.update({'other_key': 0,})
        return dict

EXPECTED = {'other_key': 0, 'key': 1337,}
RESULT = C().method()

if EXPECTED == RESULT:
    print("Ok")
else:
    print("EXPECTED: ", EXPECTED)
    print("RESULT: ", RESULT)

The question is - how do I choose which parent class I'm interested in?

Solution

Your super() call is wrong. it should be

super(B, self).method()

or in Python 3.x also just

super().method()

Furthermore, don't use dict as a variable name -- this will shadow the built-in class.