在 Python 中使用 *args 和 **kwargs 验证具有默认值的参数

Question

I have been looking for the answer to this for sometime but I always end up looking at tutorials covering the basics. My problem is -

If I have a function:

def foo(a, b, *args, **kwargs):

    if args[0]:
        if args[0] in range(1, 4):
            x=args[0]
        else:
            raise ValueError("Eww that is one ugly x!")
    else:
        x = kwargs.get('x', 3)
    if args[1]:
        if args[1] in ['some','list','of','strings']:
             y = args[1]
        else:
           raise ValueError("Invalid y")
    else:        
        y = kwargs.get('y', "some")

    if x == 1:
       print("Good")
    elif x == 2:
       print("Bad")
    elif x == 3:
       print("Clint")
    else:
       raise ValueError("Eww that is one ugly x!")

    if y == 'some':
       print(y + str(x))
    elif y == 'list':
       print("happy")
    elif y == 'of':
       print("healthy")
    elif y == 'strings':
       print(y + 'me')
    else:
       raise ValueError("Invalid y")

I am looking for a simpler way of treating args[0] and kwargs.get('x') as equivalent - insofar as I would like to perform the same type and value validation checks on which ever is assigned. In short how do I map the value of both args[i] and kwargs.get(k) to the same object.

Answer 1

Assuming you really want your function to depend on args and kwargs, you could do a try/except statement (telling you if args[0] exist) and use the fact that your ValueError is the same for both cases.

def foo(*args, **kwargs):
    try:
        x = args[0] if args[0] in range(1,4) else None
    except:
        x = kwargs.get('x', 3)

    if x == 1:
       print("Good")
    elif x == 2:
       print("Bad")
    elif x == 3:
       print("Clint")
    else:
       raise ValueError("Eww that is one ugly x!")

Answer 2

I'm curious what your real question is?

If I were simply trying to improve upon your example, I think I would choose the following approach.

from typing import Tuple, Dict, Iterable, Any


def value_from_args_or_kwargs(args: Tuple, index: int, cond: Iterable, error_msg: str,
                              kwargs: Dict, key: str, val: Any):
    if len(args) - 1 >= index and args[index]:
        return args[index] if args[index] in cond else ValueError(error_msg)
    else:
        return kwargs.get(key, val)


def bar(a, b, *args, **kwargs):
    x = value_from_args_or_kwargs(args, 0, range(1, 4), 'Eww that is one ugly x!',
                                  kwargs, 'x', 3)
    y = value_from_args_or_kwargs(args, 1, ['some', 'list', 'of', 'strings'], 'Invalid y',
                                  kwargs, 'y', 'some')
    print('Good') if x == 1 else \
        print('Bad') if x == 2 else \
        print('Client') if x == 3 else ValueError('Eww that is one ugly x!')

    print(f'{y}{x}') if y == 'some' else \
        print('happy') if y == 'list' else \
        print('healthy') if y == 'of' else \
        print(f'{y}me') if y == 'strings' else ValueError('Invalid y')

Answer 3

You should just define your function like this:

def foo(x):
    if x == 1:
        print("Good")
    elif x == 2:
        print("Bad")
    elif x == 3:
        print("Clint")
    else:
        raise ValueError("Eww that is one ugly x!")

Now you can call this function like:

foo(3)

or as:

foo(x=3)

There's no need to use *args or **kwargs if you know exactly what arguments you are expecting.

If you need more arguments, then just name them:

def foo(x, y, z, a, b, c):
    ...

There's no need for *args or **kwargs if you know what arguments you are expecting.