How to construct a defaultdict from a dictionary?

Question

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?

Asked By: Karthick

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Source

Answer 1

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)

Answered By: Jochen Ritzel

Answer 2

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 )

Answered By: muhsin mohammed

Answer 3

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_)

Answered By: hoi ja

How to construct a defaultdict from a dictionary?

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