How to switch the nesting structure of a dictionary of dictionaries in Python

问题

If I have the following dictionary:

dict_a = {'a': {'x': 0, 'y': 1, 'z': 2},
          'b': {'x': 3, 'y': 4, 'z': 5}}

What is the best way to switch the structure of the dictionary to:

dict_b = {'x': {'a': 0, 'b': 3},
          'y': {'a': 1, 'b': 4},
          'z': {'a': 2, 'b': 5}}

回答1:

With default dictionary

Given the dictionary is always two levels deep, you can do this with a defaultdict:

from collections import defaultdict

dict_b = defaultdict(dict)

for k,v in dict_a.items():
    for k2,v2 in v.items():
        dict_b[k2][k] = v2

Which gives:

>>> dict_b
defaultdict(<class 'dict'>, {'x': {'a': 0, 'b': 3}, 'z': {'a': 2, 'b': 5}, 'y': {'a': 1, 'b': 4}})
>>> dict(dict_b)
{'x': {'a': 0, 'b': 3}, 'z': {'a': 2, 'b': 5}, 'y': {'a': 1, 'b': 4}}

defaultdict is a subclass of dict, you can turn the result again in a vanilla dict with dict_b = dict(dict_b) (as is demonstrated in the second query).

With pandas

You can also use pandas for this:

from pandas import DataFrame

dict_b = DataFrame(dict_a).transpose().to_dict()

This gives:

>>> DataFrame(dict_a).transpose().to_dict()
{'y': {'a': 1, 'b': 4}, 'x': {'a': 0, 'b': 3}, 'z': {'a': 2, 'b': 5}}

回答2:

We don't need defaultdict to transpose the nested dict if we use setdefault:

def transpose(nested_dict):
    result = dict()

    for x, nested in nested_dict.items():
        for a, val in nested.items():
            result.setdefault(a, dict())[x] = val

    return result

来源：https://stackoverflow.com/questions/43391661/how-to-switch-the-nesting-structure-of-a-dictionary-of-dictionaries-in-python