为不同的功能分隔**

Question

给出一个以多个函数为参数的高阶函数，该函数如何将关键字参数传递给函数参数？

示例

def eat(food='eggs', how_much=1):
    print(food * how_much)


def parrot_is(state='dead'):
    print("This parrot is %s." % state)


def skit(*lines, **kwargs):
    for line in lines:
        line(**kwargs)

skit(eat, parrot_is)  # eggs \n This parrot is dead.
skit(eat, parrot_is, food='spam', how_much=50, state='an ex-parrot') # error

[state不是eat的关键字arg，那么如何跳过仅传递与其正在调用的功能相关的关键字args？

Answer 1

您可以基于函数的kwargs过滤func_code.co_varnames词典：

def skit(*lines, **kwargs):
    for line in lines:
        line(**{key: value for key, value in kwargs.iteritems() 
                if key in line.func_code.co_varnames})

另请参见：Can you list the keyword arguments a Python function receives?

Answer 2

如果将**kwargs添加到所有定义中，则可以通过全部：

def eat(food='eggs', how_much=1, **kwargs):
    print(food * how_much)


def parrot_is(state='dead', **kwargs):
    print("This parrot is %s." % state)


def skit(*lines, **kwargs):
    for line in lines:
        line(**kwargs)

**kwargs中的任何内容也不是明确的关键字参数，只会留在kwargs中，并被例如eat。

示例：

>>> skit(eat, parrot_is, food='spam', how_much=50, state='an ex-parrot')
spamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspam
This parrot is an ex-parrot.

Answer 3

def sample(a,b,*,c=0):
    print(a,b,c)

*->在参数为args之前和参数为kwargs（关键字参数）之后

sample(1,2,c=10)
sample(a=1,b=2,c=1) # this also work 
sample(1,2,c) # TypeError: sample() takes 2 positional arguments but 3 were given