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?

{"0": {"_ref": "ipam:stat/ZGkMTAuMTQ4LjEyLjAvMjIvMA:default/10.158.2.0/22",
"comment": "VLAN 0",
"dhcp_utilization": 0,
"disable": false,
"enable_ddns": false,
"extattrs": {"CITY": {"value": "city" },
      "COUNTRY": {
        "value": "ABC"
      },
      "Helpnow ID": {
        "value": "TA"
            "value": 0
      }
    },
    "members": [],
    "network": "10.158.2.0",
    "utilization": 4
  },{"0": {"_ref": "ipam:stat/ZGkMQ4LjEyLjAvMjIvMA:default/10.109.2.0/22",
    "comment": "VLAN 0",
    "dhcp_utilization": 0,
    "disable": false,
    "enable_ddns": false,
    "extattrs": {
      "CITY": {
        "value": "city"
      },
      "COUNTRY": {
        "value": "CDS"
      },
      "Helpnow ID": {
        "value": "TA"
            "value": 0
      }
    },
    "members": [],
    "network": "10.109.2.0",
    "utilization": 9
  }]

`Is there any possible way to convert it to a list of dictionaries?` Yes, there is. Any other questions? How about starting with a [mcve]? — cs95, Mar 13 '18 at 04:56
Possible duplicate of [Converting Python Dictionary to List](https://stackoverflow.com/questions/1679384/converting-python-dictionary-to-list) — 4castle, Mar 13 '18 at 05:02

score 14 · Accepted Answer · edited Mar 13 '18 at 05:09

14

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'}]

edited Mar 13 '18 at 05:09

Van Peer

2,127
2
25
35

answered Mar 13 '18 at 05:01

NewNewton

1,015
1
10
22

score 2 · Answer 2 · answered Aug 21 '20 at 13:24

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}]

This is exactly what I wanted. – A Yashwanth Jun 09 '21 at 20:03 — A Yashwanth, Jun 09 '21 at 20:03

DarkMuesli · Answer 3 · 2022-01-19T08:20:02.083

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.

score 0 · Answer 4 · answered Apr 17 '22 at 20:49

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'}}]

score 0 · Answer 5 · answered Jan 11 '23 at 04:19

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)

Converting a dictionary of dictionaries to a List of dictionaries

5 Answers5

Linked