'Converting a dictionary of dictionaries to a List of dictionaries

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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Solution 1:[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'}]

Solution 2:[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}]

Solution 3:[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 dict without having them as a value in the inner dict, the following code produces a 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 networkx graph through the add_nodes_from() function from a dict of dicts produced by the to_dict('index') function from pandas and it worked like a charm.

Solution 4:[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'}}]

Sources

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Source: Stack Overflow

Solution Source
Solution 1 Van Peer
Solution 2 Kirby
Solution 3
Solution 4 Citizen_7