Nested, infinite dictionary in python

Question

I want set the keys and values in the dictionary. Here is an example of what I do.

class NestedDict(dict):
    def __getitem__(self, key):
        if key in self: return self.get(key)
        return self.setdefault(key, NestedDict())

>>> c = NestedDict()
>>> c
{}
>>> c['a']['b'] = 'test'
>>> c['a']['c'] = 2
>>> c
{'a': {'c': 2, 'b': 'test'}}
>>> c['a']['c'] += 1
>>> c
{'a': {'c': 3, 'b': 'test'}}
>>> c['a']['d'] += 1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +=: 'NestedDict' and 'int'

Any ideas how to solve this problem? I want be able to use += and -=. Of course if value doesn't exist then += 1 is these same as = 1. Maybe there's a better solution?

Thanks.

it's because your c['a']['d'] doesn't have a value, so your initiate it and want to increment something. juste do c['a']['d'] = 0 then you can to what you want — lpostula, Mar 20 '12 at 09:21
when you write `c[someNonExistingKey] +=1`, what result would you want? — georg, Mar 20 '12 at 09:26
i know, but i want write something that would automatically set 0 first if no value and is used =+ — eshlox, Mar 20 '12 at 09:57
If I say `c[non_existing_key]`, should it return a `NestedDict` or `0`? — Noufal Ibrahim, Mar 20 '12 at 10:39
Also see http://stackoverflow.com/questions/2435989/how-do-i-do-advanced-python-hash-autovivification — ncoghlan, Aug 23 '12 at 12:24

Noufal Ibrahim · Answer 1 · 2012-08-24T18:14:48.400

Since d is non existent in c['a'], what kind of behaviour do you expect when you try to add 1 to it? It will first call __getitem__, not find the key and then return a NestedDict which doesn't support in place addition with an int.

On a side note, It looks to me as though you're trying to implement a dictionary with a "default" value. I would use the defaultdict class available in the collections module like so

from collections import defaultdict

def create_nested_dict():
   return defaultdict(create_nested_dict)

c = create_nested_dict()

score 2 · Accepted Answer · answered Mar 20 '12 at 11:40

Since you have control over your objects placed inside the nested dict, just define addition and subtraction to them in a way that when added to an object, it simply returns the other object (effectvely behaving as a numerical 0 when added to numbers):

class NestedDict(dict):
    def __getitem__(self, key):
        if key in self: return self.get(key)
        return self.setdefault(key, NestedDict())
    def __add__(self, other):
        return other
    def __sub__(self, other):
        return other

And voilá:

>>> n = NestedDict()
>>> n["a"]["b"] += 1
>>> n["a"]["b"]
1

Perfect. Exactly what I wanted. Thank you. – eshlox Mar 20 '12 at 12:52 — eshlox, Mar 20 '12 at 12:52

score 1 · Answer 3 · answered Mar 20 '12 at 09:33

First of all, setdefault() already returns self[key] if it exists, you don't need the first line of your __getitem__().

As for your problem, you can't do it. At least not with your existing code. You're already creating a NestedDict for every node that doesn't exist. Now you want some of them to be numbers?

Perhaps you should create a dynamic node class which could function either as a dict or as a number, depending on the need?

Nested, infinite dictionary in python

3 Answers3