充当 ** 解包映射的类 [英] Class that acts as mapping for **unpacking

问题描述

如果没有子类化 dict，类需要被视为映射以便可以将其传递给具有 ** 的方法.

from abc import ABCMetauobj 类:__元类__ = ABCMetauobj.register(字典)def f(**k): 返回 ko = uobj()f(**o)# 输出:** 之后的 f() 参数必须是映射，而不是 uobj

至少到了它会抛出缺少映射功能的错误的程度，所以我可以开始实施.

我查看了模拟容器类型，但简单地定义魔术方法没有效果，并且使用 ABCMeta 覆盖并将其注册为 dict 将断言验证为子类，但失败 isinstance(o, dict).理想情况下，我什至不想使用 ABCMeta.

解决方案

__getitem__() 和 keys() 方法就足够了:

<预><代码>>>>D类:定义键(自己):返回 ['a', 'b']def __getitem__(self, key):返回 key.upper()>>>def f(**kwds):打印 kwds>>>f(**D()){'a':'A'，'b':'B'}

Without subclassing dict, what would a class need to be considered a mapping so that it can be passed to a method with **.

from abc import ABCMeta

class uobj:
    __metaclass__ = ABCMeta

uobj.register(dict)

def f(**k): return k

o = uobj()
f(**o)

# outputs: f() argument after ** must be a mapping, not uobj

At least to the point where it throws errors of missing functionality of mapping, so I can begin implementing.

I reviewed emulating container types but simply defining magic methods has no effect, and using ABCMeta to override and register it as a dict validates assertions as subclass, but fails isinstance(o, dict). Ideally, I dont even want to use ABCMeta.

解决方案

The __getitem__() and keys() methods will suffice:

>>> class D:
        def keys(self):
            return ['a', 'b']
        def __getitem__(self, key):
            return key.upper()


>>> def f(**kwds):
        print kwds


>>> f(**D())
{'a': 'A', 'b': 'B'}

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