defaultdict 的 defaultdict 吗？

Question

Is there a way to have a defaultdict(defaultdict(int)) in order to make the following code work?

for x in stuff:
    d[x.a][x.b] += x.c_int

d needs to be built ad-hoc, depending on xa and xb elements.

I could use:

for x in stuff:
    d[x.a,x.b] += x.c_int

but then I wouldn't be able to use:

d.keys()
d[x.a].keys()

Answer 1

Yes like this:

defaultdict(lambda: defaultdict(int))

The argument of a defaultdict (in this case is lambda: defaultdict(int) ) will be called when you try to access a key that doesn't exist. The return value of it will be set as the new value of this key, which means in our case the value of d[Key_doesnt_exist] will be defaultdict(int) .

If you try to access a key from this last defaultdict ie d[Key_doesnt_exist][Key_doesnt_exist] it will return 0, which is the return value of the argument of the last defaultdict ie int() .

Answer 2

The parameter to the defaultdict constructor is the function which will be called for building new elements. So let's use a lambda !

>>> from collections import defaultdict
>>> d = defaultdict(lambda : defaultdict(int))
>>> print d[0]
defaultdict(<type 'int'>, {})
>>> print d[0]["x"]
0

Since Python 2.7, there's an even better solution using Counter :

>>> from collections import Counter
>>> c = Counter()
>>> c["goodbye"]+=1
>>> c["and thank you"]=42
>>> c["for the fish"]-=5
>>> c
Counter({'and thank you': 42, 'goodbye': 1, 'for the fish': -5})

Some bonus features

>>> c.most_common()[:2]
[('and thank you', 42), ('goodbye', 1)]

For more information see PyMOTW - Collections - Container data types and Python Documentation - collections

Answer 3

I find it slightly more elegant to use partial :

import functools
dd_int = functools.partial(defaultdict, int)
defaultdict(dd_int)

Of course, this is the same as a lambda.

Answer 4

Previous answers have addressed how to make a two-levels or n-levels defaultdict . In some cases you want an infinite one:

def ddict():
    return defaultdict(ddict)

Usage:

>>> d = ddict()
>>> d[1]['a'][True] = 0.5
>>> d[1]['b'] = 3
>>> import pprint; pprint.pprint(d)
defaultdict(<function ddict at 0x7fcac68bf048>,
            {1: defaultdict(<function ddict at 0x7fcac68bf048>,
                            {'a': defaultdict(<function ddict at 0x7fcac68bf048>,
                                              {True: 0.5}),
                             'b': 3})})

Answer 5

For reference, it's possible to implement a generic nested defaultdict factory method through:

from collections import defaultdict
from functools import partial
from itertools import repeat


def nested_defaultdict(default_factory, depth=1):
    result = partial(defaultdict, default_factory)
    for _ in repeat(None, depth - 1):
        result = partial(defaultdict, result)
    return result()

The depth defines the number of nested dictionary before the type defined in default_factory is used. For example:

my_dict = nested_defaultdict(list, 3)
my_dict['a']['b']['c'].append('e')

Answer 6

Others have answered correctly your question of how to get the following to work:

for x in stuff:
    d[x.a][x.b] += x.c_int

An alternative would be to use tuples for keys:

d = defaultdict(int)
for x in stuff:
    d[x.a,x.b] += x.c_int
    # ^^^^^^^ tuple key

The nice thing about this approach is that it is simple and can be easily expanded. If you need a mapping three levels deep, just use a three item tuple for the key.