Sorting a Dictionary by Nested Key

Question

Consider a dict of the following form:

dic = { "First": { 3: "Three" }, "Second": { 1: "One" }, "Third": { 2:"Two" } }

I would like to sort it by the nested dic key (3, 1, 2)

I tried using the lambda function in the following manner but it returns a "KeyError: 0"

dic = sorted(dic.items(), key=lambda x: x[1][0])

The expected output would be:

{ "Second": { 1: "One" }, "Third": { 2: "Two" }, "First": { 3:"Three" } }

In essence what I want to know is how to designate a nested key independently from the main dictionary key.

What would the expected output be if the nested dicts contained more than one key each? For example, {"first": {3: "three", 1:"one"}, "second": {2:"two"}}? — Kevin
– Kevin, Commented Jul 9, 2019 at 19:56
Do you mean you want to sort the dictionary's nested keys, then iterate over / use those keys? — duhaime
– duhaime, Commented Jul 9, 2019 at 19:56
@ScottHunter, As of Python 3.7, dictionaries are ordered, so in principle they can be sorted. Now as for whether it's a good idea to sort them, is another question entirely... — Kevin
– Kevin, Commented Jul 9, 2019 at 20:05

John Coleman · Accepted Answer · 2019-07-09 20:00:43Z

In the lambda function, x is a key-value pair, x[1] is the value, which is itself a dictionary. x[1].keys() is its keys, but it needs to be turned into a list if you want to get its one and only item by its index. Thus:

sorted(dic.items(), key = lambda x: list(x[1].keys())[0])

which evaluates to:

[('Second', {1: 'One'}), ('Third', {2: 'Two'}), ('First', {3: 'Three'})]

Ammar · Accepted Answer · 2019-07-09 20:28:40Z

dic = {'First': {3: 'Three'}, 'Second': {1: 'One'}, 'Third': {2: 'Two'}} sorted_list = sorted(dic.items(), key=lambda x:list(x[1].keys())[0]) sorted_dict = dict(sorted_list) print(sorted_dict)

You need to get the keys for the nested dictionary first and then convert them into list and sort over its first index. You will get a sorted list. All you need to convert this list to dictionary using dict(). I hope that helps. This snippet works for python3.

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