Duncan Booth wrote:
> Ron Adam wrote:
>
>>James Stroud wrote:
>>
>>>Here it goes with a little less overhead:
>>>
>>>
>
> <example snipped>
>
>>But it's not a dictionary anymore so you can't use it in the same places
>>you would use a dictionary.
>>
>> foo(**n)
>>
>>Would raise an error.
>>
>>So I couldn't do:
>>
>> def foo(**kwds):
>> kwds = namespace(kwds)
>> kwds.bob = 3
>> kwds.alice = 5
>> ...
>> bar(**kwds) #<--- do something with changed items
>>
>
> I agree with Steven D'Aprano's reply, but if you are set on it you could
> try this:
>
>
>>>>class namespace(dict):
>
> def __init__(self, *args, **kw):
> dict.__init__(self, *args, **kw)
> self.__dict__ = self
>
>
>
>>>>n = namespace({'bob':1, 'carol':2, 'ted':3, 'alice':4})
>>>>n.bob
>
> 1
>
>>>>n.bob = 3
>>>>n['bob']
>
> 3
>
> The big problem of course with this sort of approach is that you cannot
> then safely use any of the methods of the underlying dict as they could be
> masked by values.
>
> P.S. James, *please* could you avoid top-quoting.
Or worse, the dictionary would become not functional depending on what
methods were masked.
And this approach reverses that, The dict values will be masked by the
methods, so the values can't effect the dictionary methods. But those
specific values are only retrievable with the standard dictionary notation.
class namespace(dict):
__getattr__ = dict.__getitem__
__setattr__ = dict.__setitem__
__delattr__ = dict.__delitem__
n = namespace()
n.__getattr__ = 'yes' # doesn't mask __getattr__ method.
print n['__getattr__'] -> 'yes'
The value is there and __getattr__() still works. But n.__getattr__
returns the method not the value.
So is there a way to keep the functionality without loosing the methods?
BTW, I agree with Steven concerning data structures. This really isn't
a substitute for a data structure. Many keys will not work with this.
n.my name = 'Ron'
n.(1,2) = 25
n.John's = [ ... ]
The use case I'm thinking of is not as a shortcut for data structures,
but instead, as a way to keep names as names, and maintaining those
names in a group. Thus the namespace association.
def foo(**kwds):
kwds = namespace(kwds)
print kwds.name
print kwds.value
...
name = 'ron'
value = 25
foo( name=name, position=position )
Cheers,
Ron
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