How to subclass an OrderedDict?

Question

Subclassing a Python dict works as expected:

>>> class DictSub(dict):
...     def __init__(self):
...         self[1] = 10
...         
>>> DictSub()
{1: 10}

However, doing the same thing with a collections.OrderedDict does not work:

>>> import collections
>>> class OrdDictSub(collections.OrderedDict):
...     def __init__(self):
...         self[1] = 10
...         
>>> OrdDictSub()
(…)
AttributeError: 'OrdDictSub' object has no attribute '_OrderedDict__root'

Thus, the OrderedDict implementation uses a private __root atribute, which prevents the subclass OrdDictSub from behaving like the DictSub subclass. Why? How can one inherit from an OrderedDict?

Asked By: Eric O Lebigot

||

Source

Answer 1

Try initializing the superclass in the __init__ method:

def __init__(self):
    collections.OrderedDict.__init__(self)
    self[1] = 10

This is the normal way to initialize a subclass. You don’t have to call the superclass’s __init__ method in general, but if you have no knowledge of the superclass’s implementation you really should call __init__.

Answered By: Dietrich Epp

Answer 2

You need to invoke OrderedDict.__init__ from your __init__:

class OrdDictSub(collections.OrderedDict):
    def __init__(self):
        super(OrdDictSub, self).__init__()

You haven’t given OrderedDict a chance to initialize itself. Technically, you want to do this for your dict subclass as well, since you want a fully initialized dict. The fact that dict works without it is just luck.

Answered By: Ned Batchelder

How to subclass an OrderedDict?

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