sort nested dictionary keys numerically

Question

if I have a dictionary like this

{'even': {'1': 'odd', '0': 'even'}, 'odd': {'1': 'even', '0': 'odd'}}"

I want to sort this dictionary and make it look like this

{'even': {'0': 'even', '1': 'odd'}, 'odd': {'0': 'odd', '1': 'even'}}"

How could this be done?

Dictionaries aren't meant to be an ordered data type, why did you want to do this? What have you done so far to try and do this? — Sayse, Oct 19 '20 at 12:34

Stef · Accepted Answer · 2020-10-19T13:06:39.913

Since python 3.6/3.7, dictionaries are ordered by insertion order, so it is possible to sort a dict. If you are using an older version of python, then it is not possible: dicts are not ordered.

Here I give a function to sort a dict by keys, and use it with one level of nesting.

d = {'even': {'1': 'odd', '0': 'even'}, 'odd': {'1': 'even', '0': 'odd'}}

def sorted_simple_dict(d):
  return {k: v for k, v in sorted(d.items())}

def sorted_once_nested_dict(d):
  return {k: sorted_simple_dict(v) for k, v in sorted(d.items())}

sorted_d = sorted_once_nested_dict(d)
# {'even': {'0': 'even', '1': 'odd'}, 'odd': {'0': 'odd', '1': 'even'}}

If you want to sort arbitrarily deep nested dicts, you can use recursion and isinstance( , dict):

d = {'even': {'even': {'even': {'1': 'odd', '0': 'even'}, 'odd': {'1': 'even', '0': 'odd'}}, '1': 'odd', '0': 'even'}, 'odd': {'1': 'even', '0': 'odd'}}

def sorted_nested_dict(d):
  if isinstance(d, dict):
    return {k: sorted_nested_dict(v) for k, v in sorted(d.items())}
  else:
    return d

sorted_d = sorted_nested_dict(d)
# {'even': {'0': 'even', '1': 'odd', 'even': {'even': {'0': 'even', '1': 'odd'}, 'odd': {'0': 'odd', '1': 'even'}}}, 'odd': {'0': 'odd', '1': 'even'}}

See also:

How do I sort a dictionary by key?

sort nested dictionary keys numerically

1 Answers1