How can I get Python to automatically create missing key/value pairs in a dictionary?

Question

I'm creating a dictionary structure that is several levels deep. I'm trying to do something like the following:

dict = {}
dict['a']['b'] = True

At the moment the above fails because key 'a' does not exist. At the moment I have to check at every level of nesting and manually insert an empty dictionary. Is there some type of syntactic sugar to be able to do something like the above can produce:

{'a': {'b': True}}

Without having to create an empty dictionary at each level of nesting?

http://stackoverflow.com/questions/4178249/infinitely-nested-dictionary-in-python — Joran Beasley, May 23 '13 at 22:20

score 35 · Accepted Answer · edited Sep 10 '21 at 07:07

35

As others have said, use defaultdict. This is the idiom I prefer for arbitrarily-deep nesting of dictionaries:

def nested_dict():
    return collections.defaultdict(nested_dict)

d = nested_dict()
d[1][2][3] = 'Hello, dictionary!'
print(d[1][2][3]) # Prints Hello, dictionary!

This also makes checking whether an element exists a little nicer, too, since you may no longer need to use get:

if not d[2][3][4][5]:
    print('That element is empty!')

This has been edited to use a def rather than a lambda for pep8 compliance. The original lambda form looked like this below, which has the drawback of being called <lambda> everywhere instead of getting a proper function name.

>>> nested_dict = lambda: collections.defaultdict(nested_dict)
>>> d = nested_dict()
>>> d[1][2][3]
defaultdict(<function <lambda> at 0x037E7540>, {})

edited Sep 10 '21 at 07:07

Craig McQueen

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answered May 23 '13 at 22:28

Henry Keiter

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That's an elegant implementation. – Peter DeGlopper May 23 '13 at 22:30
Oh, wow, I've never considered a recursive factory before. I'd use `def` to define the `nested_dict` function though, rather than a lambda. – Blckknght May 23 '13 at 22:32
Great answer. Just one hint on this: `pickle` can not handle it by default: `AttributeError: Can't pickle local object 'd..'` – gies0r Aug 30 '18 at 22:01
@gies0r you mean this problem happens when "saving the dict as pickle" or "loading the pickle into dict"? – Luk Aron Nov 28 '19 at 09:25
@LukAron saving the dict as pickle – gies0r Nov 29 '19 at 18:49
[Named lambdas are bad practice](https://stackoverflow.com/a/38381663/4518341). Use a `def` instead. – wjandrea May 25 '20 at 13:09

score 2 · Answer 2 · edited May 23 '17 at 12:10

2

Use defaultdict.

Python: defaultdict of defaultdict?

edited May 23 '17 at 12:10

Community

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answered May 23 '13 at 22:19

Matt Ball

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score 0 · Answer 3 · answered May 23 '13 at 22:24

0

Or you can do this, since dict() function can handle **kwargs:

http://docs.python.org/2/library/functions.html#func-dict

print dict(a=dict(b=True))
# {'a': {'b' : True}}

answered May 23 '13 at 22:24

Peter Varo

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score 0 · Answer 4 · answered May 23 '13 at 22:30

If the depth of your data structure is fixed (that is, you know in advance that you need mydict[a][b][c] but not mydict[a][b][c][d]), you can build a nested defaultdict structure using lambda expressions to create the inner structures:

two_level = defaultdict(dict)
three_level = defaultdict(lambda: defaultdict(dict))
four_level = defaultdict(lamda: defaultdict(lambda: defaultdict(dict)))

How can I get Python to automatically create missing key/value pairs in a dictionary?

4 Answers4

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