不同函数的分离**

Question

给定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，那么短剧怎么能只传递与它正在调用的函数相关的关键字args呢？

Answer 1

您可以根据函数的kwargs (在python 2中)筛选func_code.co_varnames字典：

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})

在python 3中，应该使用__code__而不是func_code。因此，其功能是：

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

另见：Can you list the keyword arguments a function receives?

Answer 2

如果将**kwargs添加到所有定义中，则可以传递全部内容：

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中任何不是显式关键字参数的内容都只会留在kwargs中，而被eat忽略。

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)

* ->在参数之前是args，在参数后面是kwargs(关键字参数)

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