为不同的功能分离**kwargs

Question

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

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不是关键字arg,eat那么skit如何才能传递与其调用的函数相关的关键字args？

Answer 1

您可以kwargs根据func_code.co_varnames函数过滤字典:

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})

另请参阅:您是否可以列出Python函数接收的关键字参数？

Answer 2

如果添加**kwargs到所有定义,则可以传递整个批次:

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)

任何内容**kwargs都不是一个明确的关键字参数,只会留下来kwargs并被例如忽略eat.

例:

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