为不同的功能分离 **kwargs

Question

给定一个将多个函数作为参数的高阶函数，该函数如何将关键字参数传递给函数参数？

例子

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，所以skit 如何只传递与其调用的函数相关的关键字args？

Answer 1

您可以根据func_code.co_varnames （在 python 2 中）过滤kwargs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， __code__应该用来代替func_code 。 所以函数将是：

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

另请参阅： 您能否列出函数接收的关键字参数？

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将留在kwargs并被例如eat忽略。

例子：

>>> 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，在参数之后是 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