On Tue, 22 Nov 2005 09:30:49 -0500, Mike Meyer <[EMAIL PROTECTED]> wrote:
>[EMAIL PROTECTED] (Bengt Richter) writes:
>> Has anyone found a way besides not deriving from dict?
>> Shouldn't there be a way?
>> TIA
>> (need this for what I hope is an improvement on the Larosa/Foord OrderedDict
>> ;-)
>>
>> I guess I can just document that you have to spell it dict(d.items()), but
>> I'd
>> like to hide the internal shenanigans ;-)
>
>So why not just hide them:
>
>>>> class md(dict):
>... def __new__(cls, arg):
>... return dict.__new__(cls, arg.items())
>
>
>I'm not sure exactly what you mean by "grabbing sublass inst d contens
>directly", but if dict(d.items()) does it, the above class should do
>it as well. Of course, *other* ways of initializing a dict won't work
>for this class.
>
It's not initializing the dict subclass that is a problem, it is passing
it to the plain dict builtin as a constructor argument and being able to
control what is pulled from the subclass object to do the construction
of the ordinary dict.
Here's where I am on this: (odictb is my version of odict ;-)
>>> from odictb import OrderedDict
>>> d = OrderedDict()
>>> d['a'] = 1
>>> d['b'] = 2
>>> d
{'a': 1, 'b': 2}
>>> dict(d.items())
{'a': 1, 'b': 2}
Looks normal, but I am secretly using key:(insertionseqno, value) to represent
key:value
>>> dict(d)
{'a': (0, 1), 'b': (1, 2)}
and I want to control dict(d) that so the result is like dict(d.items()) above.
The dict builtin apparently skips looking for methods in the dict subclass. If
it were
a subclass of object, dict would probably be looking for __iter__ or
__getitem__ or __???___
but with a dict subclass it seems to bypass such a check. I was hoping that
even so
there might be a __???__ that I didn't know about.
[Oop, just discovered a bug in my implementation ;-/ Hope it doesn't ripple...]
Regards,
Bengt Richter
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