python class that acts as mapping for **unpacking

问题

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.

回答1:

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'}

回答2:

If you're trying to create a Mapping — not just satisfy the requirements for passing to a function — then you really should inherit from collections.Mapping. As described in the documentation, you need to implement just:

__getitem__
__len__
__iter__

The Mixin will implement everything else for you: __contains__, keys, items, values, get, __eq__, and __ne__.

来源：https://stackoverflow.com/questions/26909472/which-of-dicts-methods-is-called-when-pythons-double-asterisk-operator-is-used