defaultdict of defaultdict?
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 x.a
and x.b
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()
Answers:
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()
.
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
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.
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.
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')
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})})
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 x.a
and x.b
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()
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()
.
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
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.
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.
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')
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})})