Converting a dictionary of dictionaries to a List of dictionaries

Question

I have a dictionary of dictionaries.Is there any possible way to convert it to a list of dictionaries? And if not, then how is the filter() method applied to filter data in a dictionary of dictionaries?

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Answer 1

A list comprehension should do the job, no need for filter() .

>>> dic_of_dics = {1: {"a": "A"}, 2: {"b": "B"}, 3: {"c": "C"}}

>>> list_of_dics = [value for value in dic_of_dics.values()]

>>>list_of_dics
[{'a': 'A'}, {'b': 'B'}, {'c': 'C'}]

Answer 2

Just to mention a solution with keeping first-level "key" associated with "id" inside next level of dictionaries.

to rephrase it "Converting a dictionary of dictionaries to a List of dictionaries with keeping key inside".

>>> dic_of_dics = {1: {"a": "A"}, 2: {"b": "B"}, 3: {"c": "C"}}

>>> [(lambda d: d.update(id=key) or d)(val) for (key, val) in dic_of_dics.items()]
[{'a': 'A', 'id': 1}, {'b': 'B', 'id': 2}, {'c': 'C', 'id': 3}]

Answer 3

I'm aware the question is somewhat old, but just in case someone needs the solution I was looking for:

If you want to keep the keys of the outer dicts without having them as a value in the inner dict, the following code produces list of tuples (key, dict) for every inner dict.

>>> dic_of_dics = {1: {"a": "A"}, 2: {"b": "B"}, 3: {"c": "C"}}
>>> [(k,v) for k, v in dic_of_dics.items()]
[(1, {'a': 'A'}), (2, {'b': 'B'}), (3, {'c': 'C'})]

I used it to define a graph through the add_nodes_from method in the networkx lib from a dict of dicts produced by the to_dict('index') method from pandas and it worked like a charm.

Answer 4

If I understood the question very well, the task is to turn a dictionary of dictionaries into a list of dictionaries without losing the structure of the data.

A list comprehension would achieve this neatly where every item in the list is essentially a dictionary of a dictionary.

>>> dic_of_dicts = {1: {"a": "A"}, 2: {"b": "B"}, 3: {"c": "C"}}
>>> [{k: v} for k, v in dic_of_dicts.items()]
[{1: {'a': 'A'}}, {2: {'b': 'B'}}, {3: {'c': 'C'}}]

Answer 5

To pass a dictionary of dictionaries to list() is more concise and faster:

dic_of_dicts = {1: {"a": "A"}, 2: {"b": "B"}, 3: {"c": "C"}}
list_of_dicts = list(dic_of_dics.values())
list_of_dicts
[{'a': 'A'}, {'b': 'B'}, {'c': 'C'}]

Benchmark:

%timeit [value for value in dic_of_dics.values()]
153 ns ± 1.13 ns per loop (mean ± std. dev. of 7 runs, 10,000,000 loops each)

%timeit list(dic_of_dics.values())
95.3 ns ± 0.639 ns per loop (mean ± std. dev. of 7 runs, 10,000,000 loops each)