Which is not Valid Namespace in Python?

There are mainly three namespace in python: built-in (outermost), global, and local. A namespace gets created automatically when we execute a module or package.

Python implements the global and local namespaces as Python dictionaries (not the built-in namespace) and provides built-in functions called globals() and locals() to help us access these namespace dictionaries.

We can use the globals() function to access the objects in the global namespace.

Example:

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>>> globals()
{'__name__': '__main__', '__doc__': None, '__package__': None, '__loader__': <class '_frozen_importlib.BuiltinImporter'>, '__spec__': None, '__annotations__': {}, '__builtins__': <module 'builtins' (built-in)>}

As you can see, the interpreter already has several entries in globals(). Now, when you define a variable in the global scope:

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>>> s = 'Mango'
>>> globals()
{'__name__': '__main__', '__doc__': None, '__package__': None, '__loader__': <class '_frozen_importlib.BuiltinImporter'>, '__spec__': None, '__annotations__': {}, '__builtins__': <module 'builtins' (built-in)>, 's': 'Mango'}

A new object (‘Mango’) appears in the global namespace dictionary. The dictionary key is the object’s name, s, and the dictionary value is the object’s value, ‘Mango’.

Similarly, the locals() function helps us access objects in the local namespace.

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>>> def l(p, q):
	s = 'Apple'
	print(locals())

>>> f(15, 0.7)
{'s': 'Apple', 'q': 0.7, 'p': 15}

When called within l() function, locals() returns a dictionary that represents the function’s local namespace. Apart from the local variables, locals() also includes the function parameters p and q since they are local to l().