python - 为不同的函数分离 **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 的关键字参数，那么 skit 怎么能只传递与其调用的函数相关的关键字参数呢？

Answer 1

您可以根据函数的 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 中，应该使用 __code__ 而不是 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})

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