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 :

def f(x, a=d['a'], b=d['b']):
  print(x, a, b)

(Not necessarily print, but do some stuff with the variable and calling directly from their name).

But I would like to create this function directly from the dict, that is to say, I would like to have something that look likes

def f(x, **d=d):
  print(x, a, b)

and that behaves like the previously defined function. The idea is that I have a large dictionary that contains defaults values for arguments of my function, and I would like not to have to do

def f(a= d['a'], b = d['b'] ...)

I don't know if it's possible at all in python. Any insight is appreciated !

Edit : The idea is to be able to call f(5, a=3). Edit2 : The question is not about passing arguments stored in a dict to a function but to define a function whose arguments names and defaults values are stored in a dict.

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.

Answer 2

try this:

# Store the default values in a dictionary
>>> defaults = {
...     'a': 1,
...     'b': 2,
... }
>>> def f(x, **kwa):
        # Each time the function is called, merge the default values and the provided arguments
        # For python >= 3.5:
        args = {**defaults, **kwa}
        # For python < 3.5:
        # Each time the function is called, copy the default values
        args = defaults.copy()
        # Merge the provided arguments into the copied default values
        args.update(kwa)
...     print(args)
... 
>>> f(1, f=2)
{'a': 1, 'b': 2, 'f': 2}
>>> f(1, f=2, b=8)
{'a': 1, 'b': 8, 'f': 2}
>>> f(5, a=3)
{'a': 3, 'b': 2}

Thanks Olvin Roght for pointing out how to nicely merge dictionaries in python >= 3.5

Answer 3

Python is designed such that the local variables of any function can be determined unambiguously by looking at the source code of the function. So your proposed syntax

def f(x, **d=d):
  print(x, a, b)

is a nonstarter because there's nothing that indicates whether a and b are local to f or not; it depends on the runtime value of the dictionary, whose value could change across runs.

If you can resign yourself to explicitly listing the names of all of your parameters, you can automatically set their default values at runtime; this has already been well covered in other answers. Listing the parameter names is probably good documentation anyway.

If you really want to synthesize the whole parameter list at run time from the contents of d, you would have to build a string representation of the function definition and pass it to exec. This is how collections.namedtuple works, for example.

Variables in module and class scopes are looked up dynamically, so this is technically valid:

def f(x, **kwargs):
    class C:
        vars().update(kwargs)  # don't do this, please
        print(x, a, b)

But please don't do it except in an IOPCC entry.

Answer 4

How about the **kwargs trick?

def function(arg0, **kwargs):
    print("arg is", arg0, "a is", kwargs["a"], "b is", kwargs["b"])

d = {"a":1, "b":2}
function(0., **d)

outcome:

arg is 0.0 a is 1 b is 2

Answer 5

This question is very interesting, and it seemed different people have their different own guess about what the question really want.

I have my own too. Here is my code, which can express myself:

# python3 only
from collections import defaultdict

# only set once when function definition is executed
def kwdefault_decorator(default_dict):
    def wrapper(f):
        f.__kwdefaults__ = {}
        f_code = f.__code__
        po_arg_count = f_code.co_argcount
        keys = f_code.co_varnames[po_arg_count : po_arg_count + f_code.co_kwonlyargcount]
        for k in keys:
            f.__kwdefaults__[k] = default_dict[k]
        return f

    return wrapper

default_dict = defaultdict(lambda: "default_value")
default_dict["a"] = "a"
default_dict["m"] = "m"

@kwdefault_decorator(default_dict)
def foo(x, *, a, b):
    foo_local = "foo"
    print(x, a, b, foo_local)

@kwdefault_decorator(default_dict)
def bar(x, *, m, n):
    bar_local = "bar"
    print(x, m, n, bar_local)

foo(1)
bar(1)
# only kw_arg permitted
foo(1, a=100, b=100)
bar(1, m=100, n=100)

output:

1 a default_value
1 m default_value
1 100 100
1 100 100

Answer 6

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).

Answer 7

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())

---

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.

Answer 8

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)