112

If I have d=dict(zip(range(1,10),range(50,61))) how can I build a collections.defaultdict out of the dict?

The only argument defaultdict seems to take is the factory function, will I have to initialize and then go through the original d and update the defaultdict?

Salvador Dali
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Karthick
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3 Answers3

123

Read the docs:

The first argument provides the initial value for the default_factory attribute; it defaults to None. All remaining arguments are treated the same as if they were passed to the dict constructor, including keyword arguments.

from collections import defaultdict
d=defaultdict(int, zip(range(1,10),range(50,61)))

Or given a dictionary d: 

from collections import defaultdict
d=dict(zip(range(1,10),range(50,61)))
my_default_dict = defaultdict(int,d)
Oded BD
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Jochen Ritzel
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    Is it possible to extend this approach for nested dicts? The following doesn't work as intended: `default_dict = defaultdict(None,{"a":1,"b":{"c":3}})` for example `default_dict["e"]` raises a KeyError instead returning None – alancalvitti Dec 28 '18 at 16:13
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    @alancalvitti that's because passing `None` doesn't mean that it will generate None for unknown keys, it will not use one at all. You need to do `defaultdict(lambda: None, {....})` – Eric Darchis Jan 15 '19 at 10:47
16

You can construct a defaultdict from dict, by passing the dict as the second argument.

from collections import defaultdict

d1 = {'foo': 17}
d2 = defaultdict(int, d1)

print(d2['foo'])  ## should print 17
print(d2['bar'])  ## should print 1 (default int val )
muhsin mohammed
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-1

You can create a defaultdict with a dictionary by using a callable.

from collections import defaultdict

def dict_():
    return {'foo': 1}

defaultdict_with_dict = defaultdict(dict_)
hoi ja
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