Multi-level defaultdict with variable depth?

I'd do it with a subclass of dict that defines __missing__:

>>> class NestedDict(dict):
...     def __missing__(self, key):
...             self[key] = NestedDict()
...             return self[key]
...
>>> table = NestedDict()
>>> table['A']['B1']['C1'] = 1
>>> table
{'A': {'B1': {'C1': 1}}}

You can't do it directly with defaultdict because defaultdict expects the factory function at initialization time, but at initialization time, there's no way to describe the same defaultdict. The above construct does the same thing that default dict does, but since it's a named class (NestedDict), it can reference itself as missing keys are encountered. It is also possible to subclass defaultdict and override __init__.

Your example says that at any level there can be a value, and also a dictionary of sub-elements. That is called a tree, and there are many implementations available for them. This is one:

from collections import defaultdict
class Tree(defaultdict):
    def __init__(self, value=None):
        super(Tree, self).__init__(Tree)
        self.value = value

root = Tree()
root.value = 1
root['a']['b'].value = 3
print root.value
print root['a']['b'].value
print root['c']['d']['f'].value

Outputs:

1
3
None

You could do something similar by writing the input in JSON and using json.load to read it as a structure of nested dictionaries.

I think the simplest implementation of a recursive dictionary is this. Only leaf nodes can contain values.

# Define recursive dictionary
from collections import defaultdict
tree = lambda: defaultdict(tree)

Usage:

# Create instance
mydict = tree()

mydict['a'] = 1
mydict['b']['a'] = 2
mydict['c']
mydict['d']['a']['b'] = 0

# Print
import prettyprint
prettyprint.pp(mydict)

Output:

{
  "a": 1, 
  "b": {
    "a": 1
  }, 
  "c": {},
  "d": {
    "a": {
      "b": 0
    }
  }
}

You can do it without even defining a class:

from collections import defaultdict

nested_dict = lambda: defaultdict(nested_dict)
nest = nested_dict()

nest[0][1][2][3][4][5] = 6