[FIXED] How to initialize a dict from a SimpleNamespace?

Issue

It has been asked: How to initialize a SimpleNamespace from a dict?

My question is for the opposite direction. How to initialize a dict from a SimpleNamespace?

Solution

from types import SimpleNamespace

sn = SimpleNamespace(a=1, b=2, c=3)

vars(sn)
# returns {'a': 1, 'b': 2, 'c': 3}

sn.__dict__
# also returns {'a': 1, 'b': 2, 'c': 3}

Answered By - Николай Гордеев