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

Posting this as an answer because it would be too long for a comment.

Be careful with this answer. If you try

@kwargs_decorator(a='a', b='b')
def f(x, a, b):
    print(f'x = {x}')
    print(f'a = {a}')
    print(f'b = {b}')

f(1, 2)

it will issue an error:

TypeError: f() got multiple values for argument 'a'

because you are defining a as a positional argument (equal to 2).

I implemented a workaround, even though I'm not sure if this is the best solution:

def default_kwargs(**default):
    from functools import wraps
    def decorator(f):
        @wraps(f)
        def wrapper(*args, **kwargs):
            from inspect import getfullargspec
            f_args = getfullargspec(f)[0]
            used_args = f_args[:len(args)]

            final_kwargs = {
                key: value 
                for key, value in {**default, **kwargs}.items() 
                if key not in used_args
            }

            return f(*args, **final_kwargs)
        return wrapper
    return decorator

In this solution, f_args is a list containing the names of all named positional arguments of f. Then used_args is the list of all parameters that have effectively been passed as positional arguments. Therefore final_kwargs is defined almost exactly like before, except that it checks if the argument (in the case above, a) was already passed as a positional argument.

For instance, this solution works beautifully with functions such as the following.

@default_kwargs(a='a', b='b', d='d')
def f(x, a, b, *args, c='c', d='not d', **kwargs):
    print(f'x = {x}')
    print(f'a = {a}')
    print(f'b = {b}')
    for idx, arg in enumerate(args):
        print(f'arg{idx} = {arg}')
    print(f'c = {c}')
    for key, value in kwargs.items():
        print(f'{key} = {value}')

f(1)
f(1, 2)
f(1, b=3)
f(1, 2, 3, 4)
f(1, 2, 3, 4, 5, c=6, g=7)

Note also that the default values passed in default_kwargs have higher precedence than the ones defined in f. For example, the default value for d in this case is actually 'd' (defined in default_kwargs), and not 'not d' (defined in f).