如何将kwargs传递给python中的另一个kwargs?

时间:2014-07-23 11:34:43

标签: python

def a(**akwargs):
    def b(bkwargs=akwargs):
        # how to not only use akwargs defaultly,but also define bkwargs by
        # myself?
        print bkwargs
    return b

如果我想实现以下功能。我怎么能用上面的代码呢?

>>>a(u='test')()
{'u': test}

>>>a(u='test')(u='test2')
{'u': test2}

3 个答案:

答案 0 :(得分:4)

有点不清楚你想要什么,但我认为这就是诀窍:

def a(**akwargs):
    def b(bkwargs=akwargs, **kwargs):
        # how to not only use akwargs defaultly,but also define bkwargs by
        # myself?
        if not bkwargs:
            bkwargs = kwargs
        else:
            # it depends what you want here (merge or replace?), but probably
            # something like bkwargs.update(kwargs) or kwargs.update(bkwargs)
    return b

答案 1 :(得分:2)

使用外部函数中的kwargs作为默认值,并更新传递给内部函数的kwargs。

from functools import partial

def outer(**kwargs):
    def inner(**kwargs):
        return(kwargs)
    return partial(inner, **kwargs)

如果没有为内部分配kwargs,下一个使用外部函数的kwargs。

def outer(**outer_kwargs):
    def inner(**inner_kwargs):
        kwargs = inner_kwargs or outer_kwargs
        return kwargs
    return inner

答案 2 :(得分:1)

为什么不这样做?

def a(**akwargs):
    def b(**bkwargs):
        allkwargs = {}
        allkwargs.update(akwargs)
        allkwargs.update(bkwargs)
        print allkwargs
    return b

这使用a中的值,但允许您通过向b传递更多内容来覆盖它。所以:

>>> a(u='test')()
{'u': 'test'}
>>> a(u='test')(u='test2')
{'u': 'test2'}