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

You cannot achieve this at function definition because Python determines the scope of a function statically. Although, it is possible to write a decorator to add in default keyword arguments.

from functools import wraps

def kwargs_decorator(dict_kwargs):
    def wrapper(f):
        @wraps(f)
        def inner_wrapper(*args, **kwargs):
            new_kwargs = {**dict_kwargs, **kwargs}
            return f(*args, **new_kwargs)
        return inner_wrapper
    return wrapper

Usage

@kwargs_decorator({'bar': 1})
def foo(**kwargs):
    print(kwargs['bar'])

foo() # prints 1

Or alternatively if you know the variable names but not their default values...

@kwargs_decorator({'bar': 1})
def foo(bar):
    print(bar)

foo() # prints 1

Caveat

The above can be used, by example, to dynamically generate multiple functions with different default arguments. Although, if the parameters you want to pass are the same for every function, it would be simpler and more idiomatic to simply pass in a dict of parameters.