defaultdict of defaultdict?
Solution 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 i.e. d[Key_doesnt_exist][Key_doesnt_exist]
it will return 0, which is the return value of the argument of the last defaultdict i.e. int()
.
Solution 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
Solution 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.
Solution 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})})