以编程方式创建函数规范

Question

为了娱乐，我想知道如何实现以下目标：

functionA = make_fun(['paramA', 'paramB'])
functionB = make_fun(['arg1', 'arg2', 'arg3'])

相当于

^{pr2}$

这意味着需要以下行为：

functionA(3, paramB=1)       # Works
functionA(3, 2, 1)           # Fails
functionB(0)                 # Fails

问题的焦点是变量argspec-I是否可以轻松地使用常用的装饰器技术创建函数体。在

对于那些感兴趣的人，我试着用编程方式创建如下类。同样，困难在于生成带有编程参数的__init__方法——类的其余部分使用修饰符或元类看起来很简单。在

class MyClass:
    def __init__(self, paramA=None, paramB=None):
        self._attr = ['paramA', 'paramB']
        for a in self._attr:
            self.__setattr__(a, None)

    def __str__(self):
        return str({k:v for (k,v) in self.__dict__.items() if k in self._attributes})

Answer 1

可以使用^{}从包含Python代码的字符串构造函数对象：

def make_fun(parameters):
    exec("def f_make_fun({}): pass".format(', '.join(parameters)))
    return locals()['f_make_fun']

示例：

^{pr2}$

如果您想要更多的功能（例如，默认的参数值），那就需要调整包含代码的字符串并让它表示所需的函数签名。在

免责声明：正如下面所指出的，重要的是要验证parameters的内容，并且生成的Python代码字符串可以安全地传递给exec。您应该自己构造parameters，或者设置适当的限制，以防止用户为parameters构造恶意值。在

Answer 2

下面是另一种方法，使用functools.wrap，它保留了签名和docstring，至少在Python3中是这样。诀窍是在从未被调用的伪函数中创建签名和文档。这里有几个例子。在

基本示例

import functools

def wrapper(f):
    @functools.wraps(f)
    def template(common_exposed_arg, *other_args, common_exposed_kwarg=None, **other_kwargs):
        print("\ninside template.")
        print("common_exposed_arg: ", common_exposed_arg, ", common_exposed_kwarg: ", common_exposed_kwarg)
        print("other_args: ", other_args, ",  other_kwargs: ", other_kwargs)
    return template

@wrapper
def exposed_func_1(common_exposed_arg, other_exposed_arg, common_exposed_kwarg=None):
    """exposed_func_1 docstring: this dummy function exposes the right signature"""
    print("this won't get printed")

@wrapper
def exposed_func_2(common_exposed_arg, common_exposed_kwarg=None, other_exposed_kwarg=None):
    """exposed_func_2 docstring"""
    pass

exposed_func_1(10, -1, common_exposed_kwarg='one')
exposed_func_2(20, common_exposed_kwarg='two', other_exposed_kwarg='done')
print("\n" + exposed_func_1.__name__)
print(exposed_func_1.__doc__)

结果是：

^{pr2}$

调用inspect.signature(exposed_func_1).parameters返回所需的签名。但是，使用inspect.getfullargspec(exposed_func_1)仍然返回template的签名。至少，如果您在template的定义中放置所有要生成的函数的公共参数，这些参数将出现。在

如果因为某种原因这是个坏主意，请告诉我！在

更复杂的例子

通过在内部函数中分层并定义更多不同的行为，可以得到比这更复杂的结果：

import functools

def wrapper(inner_func, outer_arg, outer_kwarg=None):
    def wrapped_func(f):
        @functools.wraps(f)
        def template(common_exposed_arg, *other_args, common_exposed_kwarg=None, **other_kwargs):
            print("\nstart of template.")
            print("outer_arg: ", outer_arg, " outer_kwarg: ", outer_kwarg)
            inner_arg = outer_arg * 10 + common_exposed_arg
            inner_func(inner_arg, *other_args, common_exposed_kwarg=common_exposed_kwarg, **other_kwargs)
            print("template done")
        return template
    return wrapped_func

# Build two examples.
def inner_fcn_1(hidden_arg, exposed_arg, common_exposed_kwarg=None):
    print("inner_fcn, hidden_arg: ", hidden_arg, ", exposed_arg: ", exposed_arg, ", common_exposed_kwarg: ", common_exposed_kwarg)

def inner_fcn_2(hidden_arg, common_exposed_kwarg=None, other_exposed_kwarg=None):
    print("inner_fcn_2, hidden_arg: ", hidden_arg, ", common_exposed_kwarg: ", common_exposed_kwarg, ", other_exposed_kwarg: ", other_exposed_kwarg)

@wrapper(inner_fcn_1, 1)
def exposed_function_1(common_exposed_arg, other_exposed_arg, common_exposed_kwarg=None):
    """exposed_function_1 docstring: this dummy function exposes the right signature """
    print("this won't get printed")

@wrapper(inner_fcn_2, 2, outer_kwarg="outer")
def exposed_function_2(common_exposed_arg, common_exposed_kwarg=None, other_exposed_kwarg=None):
    """ exposed_2 doc """
    pass

这有点冗长，但关键是，在使用它创建函数时，来自您（程序员）的动态输入有很大的灵活性，在使用公开的输入（来自函数的用户）时也有很大的灵活性。在

Answer 3

使用类的一种可能的解决方案：

def make_fun(args_list):
    args_list = args_list[:]

    class MyFunc(object):
        def __call__(self, *args, **kwargs):
            if len(args) > len(args_list):
                raise ValueError('Too many arguments passed.')

            # At this point all positional arguments are fine.
            for arg in args_list[len(args):]:
                if arg not in kwargs:
                    raise ValueError('Missing value for argument {}.'.format(arg))

            # At this point, all arguments have been passed either as
            # positional or keyword.
            if len(args_list) - len(args) != len(kwargs):
                raise ValueError('Too many arguments passed.')

            for arg in args:
                print(arg)

            for arg in args_list[len(args):]:
                print(kwargs[arg])

    return MyFunc()

functionA = make_fun(['paramA', 'paramB'])
functionB = make_fun(['arg1', 'arg2', 'arg3'])

functionA(3, paramB=1)       # Works
try:
    functionA(3, 2, 1)           # Fails
except ValueError as e:
    print(e)

try:
    functionB(0)                 # Fails
except ValueError as e:
    print(e)

try:
    functionB(arg1=1, arg2=2, arg3=3, paramC=1)                 # Fails
except ValueError as e:
    print(e)