Sorting a dictionary based on its child dictionary

Question

I want to sort this dictionary according to age

I have tried some ways to sort this like create a list of the age the sort that but then i won't able to assign the age to its corresonding index


students = {
    'a': {
        "age":18,
        "class": "IV"
    },
    'b': {
        "age":19,
        "class": "IV"
    },
    'c': {
        "age":17,
        "class": "IV"
    },
    'd': {
        "age":17.5,
        "class": "IV"
    },
    'e': {
        "age":19.5,
        "class": "IV"
    },
}

you can't guarantee order with regular dicts, use `OrderedDict` instead. — eagr, Apr 30 '22 at 11:05
@eagr only for versions 3.6 and below https://stackoverflow.com/questions/39980323/are-dictionaries-ordered-in-python-3-6 — Nin17, Apr 30 '22 at 11:07

score 1 · Answer 1 · answered Apr 30 '22 at 11:03

You can do it with a dictionary comprehension:

{k: v for k, v in sorted(students.items(), key=lambda i: i[1]['age'])}

Output:

{'c': {'age': 17, 'class': 'IV'},
 'd': {'age': 17.5, 'class': 'IV'},
 'a': {'age': 18, 'class': 'IV'},
 'b': {'age': 19, 'class': 'IV'},
 'e': {'age': 19.5, 'class': 'IV'}}

Alexander · Answer 2 · 2022-04-30T12:48:20.050

1

Try using the builtin sorted function.

students_sorted = dict(sorted(students.items(), key=lambda x: x[1]["age"]))

students_sorted:

{
  'c': {'age': 17, 'class': 'IV'}, 
  'd': {'age': 17.5, 'class': 'IV'},
  'a': {'age': 18, 'class': 'IV'}, 
  'b': {'age': 19, 'class': 'IV'}, 
  'e': {'age': 19.5, 'class': 'IV'}
}

edited Apr 30 '22 at 12:48

answered Apr 30 '22 at 11:05

Alexander

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score 1 · Answer 3 · answered Apr 30 '22 at 11:05

1

first you can to sort the key/value pairs of your dict by a custom sort key function and then put it back together

sorted_students = {
    key: value for key, value in sorted(
        students.items(), key=lambda item: item[1]['age']
    )
}

answered Apr 30 '22 at 11:05

Raphael

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Sorting a dictionary based on its child dictionary

3 Answers3