Give function defaults arguments from a dictionary in Python

Question

Let\'s imagine I have a dict :

d = {\'a\': 3, \'b\':4}

I want to create a function f that does the exact same thing than this function :

Answer 1

Sure... hope this helps

def funcc(x, **kwargs):
    locals().update(kwargs)
    print(x, a, b, c, d)

kwargs = {'a' : 1, 'b' : 2, 'c':1, 'd': 1}
x = 1

funcc(x, **kwargs)

Answer 2

You can unpack values of dict:

from collections import OrderedDict

def f(x, a, b):
    print(x, a, b)

d = OrderedDict({'a': 3, 'b':4})
f(10, *d.values())

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UPD.

Yes, it's possible to implement this mad idea of modifying local scope by creating decorator which will return class with overriden __call__() and store your defaults in class scope, BUT IT'S MASSIVE OVERKILL.

Your problem is that you're trying to hide problems of your architecture behind those tricks. If you store your default values in dict, then access to them by key. If you want to use keywords - define class.

P.S. I still don't understand why this question collect so much upvotes.