views:

226

answers:

2

Thanks to some great folks on SO, I discovered the possibilities offered by collections.defaultdict, notably in readability and speed. I have put them to use with success.

Now I would like to implement three levels of dictionaries, the two top ones being defaultdict and the lowest one being int. I don't find the appropriate way to do this. Here is my attempt:

from collections import defaultdict
d = defaultdict(defaultdict)
a = [("key1", {"a1":22, "a2":33}),
     ("key2", {"a1":32, "a2":55}),
     ("key3", {"a1":43, "a2":44})]
for i in a:
    d[i[0]] = i[1]

Now this works, but the following, which is the desired behavior, doesn't:

d["key4"]["a1"] + 1

I suspect that I should have declared somewhere that the second level defaultdict is of type int, but I didn't find where or how to do so.

The reason I am using defaultdict in the first place is to avoid having to initialize the dictionary for each new key.

Any more elegant suggestion?

Thanks pythoneers!

+6  A: 

Use:

d = defaultdict(lambda: defaultdict(int))

This will create a new defaultdict(int) whenever a new key is accessed in d.

interjay
exactly what I came in to say.
job
Totally does the trick :) Thx
Morlock
+1  A: 

Look at nosklo's answer here for a more general solution.

class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

Testing:

a = AutoVivification()

a[1][2][3] = 4
a[1][3][3] = 5
a[1][2]['test'] = 6

print a

Output:

{1: {2: {'test': 6, 3: 4}, 3: {3: 5}}}
miles82
Thanks for the link @miles82 (and the edit, @voyager). How pythonesque and safe is this approach?
Morlock