How to implement an ordered, default dict? [duplicate]

Answer 1

Here is another possibility, inspired by Raymond Hettinger's super() Considered Super, tested on Python 2.7.X and 3.4.X:

from collections import OrderedDict, defaultdict

class OrderedDefaultDict(OrderedDict, defaultdict):
    def __init__(self, default_factory=None, *args, **kwargs):
        #in python3 you can omit the args to super
        super(OrderedDefaultDict, self).__init__(*args, **kwargs)
        self.default_factory = default_factory

If you check out the class's MRO (aka, help(OrderedDefaultDict)), you'll see this:

class OrderedDefaultDict(collections.OrderedDict, collections.defaultdict)
 |  Method resolution order:
 |      OrderedDefaultDict
 |      collections.OrderedDict
 |      collections.defaultdict
 |      __builtin__.dict
 |      __builtin__.object

meaning that when an instance of OrderedDefaultDict is initialized, it defers to the OrderedDict's init, but this one in turn will call the defaultdict's methods before calling __builtin__.dict, which is precisely what we want.

Answer 2

Here's another solution to think about if your use case is simple like mine and you don't necessarily want to add the complexity of a DefaultOrderedDict class implementation to your code.

from collections import OrderedDict

keys = ['a', 'b', 'c']
items = [(key, None) for key in keys]
od = OrderedDict(items)

(None is my desired default value.)

Note that this solution won't work if one of your requirements is to dynamically insert new keys with the default value. A tradeoff of simplicity.

Update 3/13/17 - I learned of a convenience function for this use case. Same as above but you can omit the line items = ... and just:

od = OrderedDict.fromkeys(keys)

Output:

OrderedDict([('a', None), ('b', None), ('c', None)])

And if your keys are single characters, you can just pass one string:

OrderedDict.fromkeys('abc')

This has the same output as the two examples above.

You can also pass a default value as the second arg to OrderedDict.fromkeys(...).

Answer 3

Another simple approach would be to use dictionary get method

>>> from collections import OrderedDict
>>> d = OrderedDict()
>>> d['key'] = d.get('key', 0) + 1
>>> d['key'] = d.get('key', 0) + 1
>>> d
OrderedDict([('key', 2)])
>>>