Generating dictionary keys on the fly [duplicate]

Answer 1

class D(dict):
    def __missing__(self, key):
        self[key] = D()
        return self[key]

d = D()
d['a']['b']['c'] = 3

Answer 2

I like Dave's answer better, but here's an alternative.

from collections import defaultdict
d = defaultdict(lambda : defaultdict(int))
>>> d['a']['b'] += 1
>>> d
defaultdict(<function <lambda> at 0x652f0>, {'a': defaultdict(<type 'int'>, {'b': 1})})
>>> d['a']['b']
1

http://tumble.philadams.net/post/85269428/python-nested-defaultdicts

It's definitely not pretty to have to use lambdas to implements the inner defaulted collections, but apparently necessary.

Answer 3

You could use a tuple as the key for the dict and then you don't have to worry about subdictionaries at all:

mydict[(key,subkey,subkey2)] = "value"

Alternatively, if you really need to have subdictionaries for some reason you could use collections.defaultdict.

For two levels this is straightforward:

>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>> d['key']['subkey'] = 'value'
>>> d['key']['subkey']
'value'

For three it's slightly more complex:

>>> d = defaultdict(lambda: defaultdict(dict))
>>> d['key']['subkey']['subkey2'] = 'value'
>>> d['key']['subkey']['subkey2']
'value'

Four and more levels are left as an exercise for the reader. :-)