根据参数数量选择我的函数

Question

这就是我今天所做的；

def some_function_with_one_argument(x): ...

def almost_the_same_function_but_with_two_arguments(x, y): ...

def same_again_but_now_with_three_arguments(x, y, z): ...

def some_function(*args):
    match len(args):
        case 1:
            return some_function_with_one_argument(*args)
        case 2:
            return almost_the_same_function_but_with_two_arguments(*args)
        case 3:
            same_again_but_now_with_three_arguments(*args)
        case _:
            raise TypeError(f"some_function() takes between one and three positional arguments but {len(args)} were given")

但在我看来，这是非常尴尬和丑陋的。该怎么办？

我可以创建某种“重载”装饰器等。但这不是我的代码的重点，我不想首先包含仅用于构造我的代码的函数。

Answer 1

我认为一个不太难看的方法是使用一个字典（下面代码中的

function_mapper

）将参数数量映射到函数引用，然后你可以调用一次

function_mapper[len(args)](*args)

。这样代码会更短，并且您不必从匹配案例中显式调用每个函数

def some_function_with_one_argument(x): ...

def almost_the_same_function_but_with_two_arguments(x, y): ...

def same_again_but_now_with_three_arguments(x, y, z): ...

def some_function(*args):
    function_mapper = {
        1: some_function_with_one_argument,
        2: almost_the_same_function_but_with_two_arguments,
        3: same_again_but_now_with_three_arguments
    }

    if len(args) not in function_mapper:
        raise TypeError(f"some_function() takes between one and three positional arguments but {len(args)} were given")
    else:
        function_mapper[len(args)](*args)