python class that acts as mapping for **unpacking
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.
__getitem__()和keys()方法就足够了：
>>> 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'}
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__ .
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