class 的自定义双星运算符？

Question

如何实现自定义双星运算符 (**) 来解包，类似于 __iter__ 与单星运算符 (*) 的工作方式？

例如：

class PlayerManager(object):

    def __init__(self, players=None):
        self.players = players or []

    # Made up method to support ** operator
    def __dict_iter__(self):
        for player in self.players:
            yield get_steamid(player), player

def print_players(**players):
    print(players)

player_manager = PlayerManager([list, of, players])
print_players(**player_manager)

输出：

{
    'STEAM_0:0:02201': <Player object at 0x0000000000>,
    'STEAM_0:0:10232': <Player object at 0x0000000064>,
    'STEAM_0:0:73602': <Player object at 0x0000000128>
}

Answer 1

实施 Mapping ABC. Technically, the language docs don't specify which Mapping methods are used, so assuming you only need some subset used by the current implementation is a bad idea. All it says is:

If the syntax **expression appears in the function call, expression must evaluate to a mapping, the contents of which are treated as additional keyword arguments. In the case of a keyword appearing in both expression and as an explicit keyword argument, a TypeError exception is raised.

所以如果你实现Mapping ABC，你肯定有正确的接口，不管它是否依赖.items()，直接迭代和__getitem__调用等

仅供参考，在检查时，CPython 3.5 中的行为绝对取决于如何你实现 Mapping （如果你继承自 dict，它使用直接访问 dict 内部的优化路径，如果你不这样做，它会迭代 .keys() 并查找每个键）。所以，是的，不要偷工减料，实施整个 ABC。多亏了从 Mapping ABC 及其父类继承的默认实现，只需以下代码即可完成此操作：

class MyMapping(Mapping):
    def __getitem__(self, key):
        ...
    def __iter__(self):
        ...
    def __len__(self):
        ...

您继承的默认实现在某些情况下可能不是最优的（例如 items 和 values 会做涉及迭代和查找的半邪恶的事情，其中直接访问器可能更快，具体取决于内部结构)，所以如果您将它用于其他目的，我建议使用优化版本覆盖那些。

Answer 2

正如@ShadowRanger 所说，实现映射。这是一个例子：

from collections.abc import Mapping

class Foo(Mapping):
    def __iter__(self):
        yield "a"
        yield "b"

    def __len__(self):
        return 2

    def __getitem__(self, item):
        return ord(item)

f = Foo()

print(*f)
print(dict(**f))

程序输出：

a b
{'a': 97, 'b': 98}

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