How to create ordered dictionary of dictionary of list in Python

Question

With the following class

from collections import defaultdict, OrderedDict
class NestedOrderedDict(OrderedDict):
    def __missing__(self,k):
        val = self[k] = NestedOrderedDict()
        return self[k]

I tried to make ordered dictionary of dictionary of list

orddodol = NestedOrderedDict()
orddodol["foo"]["a"].append(1)
orddodol["foo"]["b"].append(1)
orddodol["foo"]["b"].append(12)
orddodol["bar"]["a"].append(2)
orddodol["bar"]["a"].append(3)

But it failed giving

AttributeError: 'NestedOrderedDict' object has no attribute 'append'

What's the right way to do it?

Answer 1

You have to initialize the second level dictionary values as lists:

from collections import defaultdict, OrderedDict
class NestedOrderedDict(OrderedDict):
    def __missing__(self,k):
        val = self[k] = NestedOrderedDict()
        return self[k]

orddodol = NestedOrderedDict()
orddodol["foo"]["a"] = []           # these
orddodol["foo"]["b"] = []           # three
orddodol["bar"]["a"] = []           # lines
orddodol["foo"]["a"].append(1)
orddodol["foo"]["b"].append(1)
orddodol["foo"]["b"].append(12)
orddodol["bar"]["a"].append(2)
orddodol["bar"]["a"].append(3)

Answer 2

You almost did it right :) If you want to create ordered dictionary of dictionary of list , then you'd better use defaultdict(list) for that purpose:

from collections import defaultdict, OrderedDict
class NestedOrderedDict(OrderedDict):
    def __missing__(self, k):
   >>>> self[k] = defaultdict(list)
        return self[k]

orddodol = NestedOrderedDict()
orddodol["foo"]["a"].append(1)
orddodol["foo"]["b"].append(1)
orddodol["foo"]["b"].append(12)
orddodol["bar"]["a"].append(2)
orddodol["bar"]["a"].append(3)

Answer 3

from collections import defaultdict, OrderedDict
class NestedOrderedDict(OrderedDict):
    def __missing__(self,k):
        val = self[k] = NestedOrderedDict()
        return self[k]

orddodol = NestedOrderedDict()
orddodol["foo"]["a"] = []
orddodol["foo"]["b"] = []
orddodol["bar"]["a"] = []
orddodol["foo"]["a"].append(1)
orddodol["foo"]["b"].append(1)
orddodol["foo"]["b"].append(12)
orddodol["bar"]["a"].append(2)
orddodol["bar"]["a"].append(3)

or you can implement a defaultdict with ordered like this post Can I do an ordered, default dict in Python

from collections import OrderedDict, Callable
class DefaultOrderedDict(OrderedDict):
    # Source: https://stackoverflow.com/a/6190500/562769
    def __init__(self, default_factory=None, *a, **kw):
        if (default_factory is not None and
           not isinstance(default_factory, Callable)):
            raise TypeError('first argument must be callable')
        OrderedDict.__init__(self, *a, **kw)
        self.default_factory = default_factory

    def __getitem__(self, key):
        try:
            return OrderedDict.__getitem__(self, key)
        except KeyError:
            return self.__missing__(key)

    def __missing__(self, key):
        if self.default_factory is None:
            raise KeyError(key)
        self[key] = value = self.default_factory()
        return value

    def __reduce__(self):
        if self.default_factory is None:
            args = tuple()
        else:
            args = self.default_factory,
        return type(self), args, None, None, self.items()

    def copy(self):
        return self.__copy__()

    def __copy__(self):
        return type(self)(self.default_factory, self)

    def __deepcopy__(self, memo):
        import copy
        return type(self)(self.default_factory,
                          copy.deepcopy(self.items()))

    def __repr__(self):
        return 'OrderedDefaultDict(%s, %s)' % (self.default_factory,
                                               OrderedDict.__repr__(self))