How to properly override the default factory of defaultdict?
ian.g.kelly at gmail.com
Fri Feb 19 19:24:40 EST 2016
On Thu, Feb 18, 2016 at 10:41 AM, Herman <sorsorday at gmail.com> wrote:
> From: Ben Finney <ben+python at benfinney.id.au>
>> you are using the inheritance hierarchy but thwarting it by not using
>> ‘super’. Instead::
>> super().__init__(self, default_factory, *a, **kw)
>> super().__getitem__(self, key)
>> \ "Those who will not reason, are bigots, those who cannot, are |
>> `\ fools, and those who dare not, are slaves." —“Lord” George |
>> _o__) Gordon Noel Byron |
>> Ben Finney
> super does not work for defaultdict. I am using python 2.7. If I use
> super(defaultdict, self).__init__(default_factory, *a, **kw), I get the
> super(defaultdict, self).__init__(default_factory, *a, **kw)
> TypeError: 'function' object is not iterable
You're using it incorrectly. If your class is named
DefaultDictWithEnhancedFactory, then the super call would be:
super(DefaultDictWithEnhancedFactory, self).__init__(default_factory, *a, **kw)
You pass in the current class so that super can look up the next
class. If you pass defaultdict instead, super will think that it's
being called *by* defaultdict and call the __init__ method on its own
superclass, dict, which has a different signature.
defaultdict.__init__ is effectively skipped.
> Look like inheriting from defaultdict is easier. I don't even have to
> override the constructor as suggested by Chris Angelico above. Thanks.
True, although there's a faint code smell as this technically violates
the Liskov Substitution Principle; the default_factory attribute on
defaultdict instances is expected to be a function of zero arguments,
More information about the Python-list