python 参数的默认值是否被 args 或 kwargs 接受？

Question

我在尝试制作测试装饰器时发现了一个奇怪的实例。代码如下：

def doublefunc(fn):
    def warpped(*arg, **kwargs):
        arg = (i if type(i) is not int else i*2 for i in arg)
        for k, v in kwargs.iteritems():
            kwargs[k] = v *2 if type(v) is int else v
        return fn(*arg, **kwargs)
    return warpped

@doublefunc
def add(x, y, m, z=10, v='bbaa'):
    print 'x=%s, y=%s, m=%s, z=%s, v=%s' % (x, y, m, z, v)
    return x + y + z

当我这样调用函数时：

print add(1, 1, 'teststring')

返回值是14而不是24，也就是说函数add中参数z的默认值，在装饰函数内部不被接受。为什么？

Answer 1

这很棘手，但您可以通过使用 inspect 模块

获得未传递的默认参数

import inspect

def get_unpassed_defaults(fn, args, kwargs):
        args_len = len(args) + len(kwargs)
        argspec = inspect.getargspec(fn)

        defaults = argspec.defaults
        default_names = argspec.args[-len(defaults):]

        number_of_args = len(argspec.args)
        number_of_positional_args = number_of_args - len(argspec.defaults)
        number_of_kwargs = number_of_args - number_of_positional_args
        number_of_kwargs_passed_as_positional = len(args) - number_of_positional_args

        default_values = defaults[number_of_kwargs_passed_as_positional:]
        default_names = default_names[number_of_kwargs_passed_as_positional:]
        defaults_dict = dict(zip(default_names, default_values))

        for kwarg_name in kwargs:
            if kwarg_name in defaults_dict:
                defaults_dict.pop(kwarg_name)

        return defaults_dict

然后在 doublefunc 中利用它：

def doublefunc(fn):
    def wrapped(*arg, **kwargs):
        unpassed_defaults = get_unpassed_defaults(fn, arg, kwargs)
        kwargs.update(unpassed_defaults)

        arg = (i if type(i) is not int else i*2 for i in arg)
        for k, v in kwargs.iteritems():
            kwargs[k] = v *2 if type(v) is int else v

        return fn(*arg, **kwargs)
    return wrapped

@doublefunc
def add(x, y, m, z=10, v='bbaa'):
    print 'x=%s, y=%s, m=%s, z=%s, v=%s' % (x, y, m, z, v)
    return x + y + z

print add(1, 1, "teststring", v=3)

给出：

x=2, y=2, m=teststring, z=20, v=6
24

Answer 2

原因是装饰器函数不查找包装函数的内部属性（在本例中为默认 kwargs）。如果您仍然想这样做，那么您只需要使用字典即可。

Won't work with Python 2 because there is no getfullargspec method, so for that follow @Nolen Royalty's answer

为Python3

>>> import inspect
>>> def doublefunc(fn):
        def wrapped(*arg, **kwargs):
            defaults = (inspect.getfullargspec(fn).kwonlydefaults)
            defaults.update(kwargs)
            kwargs = defaults
            arg = (i if type(i) is not int else i*2 for i in arg)
            for k, v in kwargs.items():
                kwargs[k] = v *2 if type(v) is int else v
            return fn(*arg, **kwargs)
        return wrapped

>>> @doublefunc
def add(x, y, m, *args, z=10, v='bbaa'):
    print('x={0}, y={1}, m= {2}, z={3}, v={4}'.format(x,y,m,z,v))
    return x + y + z

>>> print(add(1, 1, 'teststring', v=1))

结果：

x=2, y=2, m= teststring, z=20, v=2
24

python 参数的默认值是否被 args 或 kwargs 接受？

Is python parameters' default value accepted by args or kwargs?

python

parameters

default-value