Can I do an ordered, default dict in Python?

Question

I would like to combine OrderedDict() and defaultdict() from collections in one object, which shall be an ordered, default dict. Is this possible?

Answer 1

Inspired by other answers on this thread, you can use something like,

from collections import OrderedDict

class OrderedDefaultDict(OrderedDict):
    def __missing__(self, key):
        value = OrderedDefaultDict()
        self[key] = value
        return value

I would like to know if there're any downsides of initializing another object of the same class in the missing method.

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

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 4

i tested the default dict and discovered it's also sorted! maybe it was just a coincidence but anyway you can use the sorted function:

sorted(s.items())

i think it's simpler

Answer 5

A simple and elegant solution building on @NickBread. Has a slightly different API to set the factory, but good defaults are always nice to have.

class OrderedDefaultDict(OrderedDict):
    factory = list

    def __missing__(self, key):
        self[key] = value = self.factory()
        return value

Answer 6

If you want a simple solution that doesn't require a class, you can just use OrderedDict.setdefault(key, default=None) or OrderedDict.get(key, default=None). If you only get / set from a few places, say in a loop, you can easily just setdefault.

totals = collections.OrderedDict()

for i, x in some_generator():
    totals[i] = totals.get(i, 0) + x

It is even easier for lists with setdefault:

agglomerate = collections.OrderedDict()

for i, x in some_generator():
    agglomerate.setdefault(i, []).append(x)

But if you use it more than a few times, it is probably better to set up a class, like in the other answers.