CODEWITHSUNDEEP
pythonpython-3.xdictionary
Vandhana
Vandhana

Reputation: 333

Swap keys of nested dictionaries

I have a dictionary as follows:

Each key has a dictionary associated with it.

dict_sample = {'a': {'d0': '1', 'd1': '2', 'd2': '3'}, 'b': {'d0': '1'}, 'c': {'d1': '1'}}

I need the output as follows:

output_dict = {'d0': {'a': 1, 'b': 1}, 'd1': {'a': 2, 'c': 1}, 'd2': {'a': 3}}

I'd appreciate any help on the pythonic way to achieve this. Thank You !

Upvotes: 4

Views: 1241

Answers (2)

blhsing
blhsing

Reputation: 106792

You can use dict.setdefault on a new dict with a nested loop:

d = {}
# for each key and sub-dict in the main dict
for k1, s in dict_sample.items():
    # for each key and value in the sub-dict
    for k2, v in s.items():
        # this is equivalent to d[k2][k1] = int(v), except that when k2 is not yet in d,
        # setdefault will initialize d[k2] with {} (a new dict)
        d.setdefault(k2, {})[k1] = int(v)

d would become:

{'d0': {'a': 1, 'b': 1}, 'd1': {'a': 2, 'c': 1}, 'd2': {'a': 3}}

Upvotes: 3

avayert
avayert

Reputation: 684

I believe this produces the desired output

>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>>
>>> dict_sample = {'a': {'d0': '1', 'd1': '2', 'd2': '3'}, 'b': {'d0': '1'}, 'c': {'d1': '1'}}
>>>
>>> for key, value in dict_sample.items():
...     for k, v in value.items():
...         d[k][key] = v
...
>>> d
defaultdict(<class 'dict'>, {'d0': {'a': '1', 'b': '1'}, 'd1': {'a': '2', 'c': '1'}, 'd2': {'a': '3'}})

Upvotes: 3

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