How to Sort a Dictionary in Python (By Key and By Value)

To sort a dictionary in Python, sort its items and build a new dict: dict(sorted(d.items())) sorts by key, and dict(sorted(d.items(), key=lambda kv: kv[1])) sorts by value. Add reverse=True for descending order. Dictionaries themselves have no sort() method, but since Python 3.7 they keep insertion order, so the new dict stays in the sorted order. This guide covers sorting a Python dictionary by key (alphabetically, case-insensitive), by value, by nested values, and the top N items.

All examples were run with Python 3.12.5 in the Windows Command Prompt. Reference: Sorting HOW TO and dict in the Python docs.

Sort a dictionary by key and by value

prices = {"mango": 2.5, "apple": 1.2, "kiwi": 0.8, "banana": 0.5}

by_key = dict(sorted(prices.items()))                              # sort by key
by_value = dict(sorted(prices.items(), key=lambda kv: kv[1]))      # sort by value

print("by key:  ", by_key)
print("by value:", by_value)

Output:

by key:   {'apple': 1.2, 'banana': 0.5, 'kiwi': 0.8, 'mango': 2.5}
by value: {'banana': 0.5, 'kiwi': 0.8, 'apple': 1.2, 'mango': 2.5}

d.items() gives (key, value) pairs. Tuples compare by their first element, so sorted(d.items()) orders by key; a key function that returns kv[1] orders by value.

Sort a dictionary by key alphabetically

Plain sorting compares characters by code point, so all uppercase letters come before lowercase ones. Use str.casefold() for true alphabetical order:

cities = {"boston": 675, "Chicago": 2746, "austin": 979, "Denver": 715}

print(sorted(cities))                                 # just the sorted keys (a list)
print(dict(sorted(cities.items())))                   # uppercase keys come first
print(dict(sorted(cities.items(), key=lambda kv: kv[0].casefold())))   # true alphabetical order
print(dict(sorted(cities.items(), reverse=True)))     # Z to A

Output:

['Chicago', 'Denver', 'austin', 'boston']
{'Chicago': 2746, 'Denver': 715, 'austin': 979, 'boston': 675}
{'austin': 979, 'boston': 675, 'Chicago': 2746, 'Denver': 715}
{'boston': 675, 'austin': 979, 'Denver': 715, 'Chicago': 2746}

Sort a dictionary by value (ascending and descending)

from operator import itemgetter

stock = {"pens": 120, "notebooks": 45, "folders": 80, "staplers": 12}

print(dict(sorted(stock.items(), key=itemgetter(1))))                 # lowest first
print(dict(sorted(stock.items(), key=itemgetter(1), reverse=True)))   # highest first
print(sorted(stock, key=stock.get, reverse=True)[:2])                 # keys of the top 2 values

from collections import Counter
print(Counter(stock).most_common(2))                                  # top 2 as (key, value) pairs

Output:

{'staplers': 12, 'notebooks': 45, 'folders': 80, 'pens': 120}
{'pens': 120, 'folders': 80, 'notebooks': 45, 'staplers': 12}
['pens', 'folders']
[('pens', 120), ('folders', 80)]

A sorted dict does not stay sorted

The result of dict(sorted(...)) is an ordinary dictionary. New keys are added at the end, so sort again when needed (or keep the data in a sorted list):

d = {"c": 3, "a": 1, "b": 2}
d_sorted = dict(sorted(d.items()))

print(d_sorted)
d_sorted["aa"] = 0                  # new keys go to the END: a dict keeps insertion order,
print(d_sorted)                     # it does not stay sorted by itself
print(dict(sorted(d_sorted.items())))   # sort again when you need it

Output:

{'a': 1, 'b': 2, 'c': 3}
{'a': 1, 'b': 2, 'c': 3, 'aa': 0}
{'a': 1, 'aa': 0, 'b': 2, 'c': 3}

Sort by a nested value or by two keys

employees = {
    "E03": {"name": "Cara", "dept": "IT", "salary": 72000},
    "E01": {"name": "Anna", "dept": "HR", "salary": 65000},
    "E02": {"name": "Ben", "dept": "IT", "salary": 81000},
}

by_salary = dict(sorted(employees.items(), key=lambda kv: kv[1]["salary"], reverse=True))
for emp_id, info in by_salary.items():
    print(emp_id, info["name"], info["salary"])

# two keys: department, then salary (highest first)
for emp_id, info in sorted(employees.items(), key=lambda kv: (kv[1]["dept"], -kv[1]["salary"])):
    print(info["dept"], info["name"], info["salary"])

Output:

E02 Ben 81000
E03 Cara 72000
E01 Anna 65000
HR Anna 65000
IT Ben 81000
IT Cara 72000

Sort a list of dictionaries by a key

A related task: when you have a list of dicts, sort the list with a key that reads the field:

people = [{"name": "Dev", "age": 31}, {"name": "Anna", "age": 25}, {"name": "Ben", "age": 31}]

print(sorted(people, key=lambda p: p["age"]))
print(sorted(people, key=lambda p: (-p["age"], p["name"])))     # oldest first, then by name

Output:

[{'name': 'Anna', 'age': 25}, {'name': 'Dev', 'age': 31}, {'name': 'Ben', 'age': 31}]
[{'name': 'Ben', 'age': 31}, {'name': 'Dev', 'age': 31}, {'name': 'Anna', 'age': 25}]

Common errors

mixed = {3: "three", "two": 2, 1: "one"}

try:
    sorted(mixed)
except TypeError as e:
    print("TypeError:", e)

print(sorted(mixed, key=str))           # compare the keys as strings

d = {"b": 2, "a": 1}
try:
    d.sort()                            # dictionaries have no sort() method
except AttributeError as e:
    print("AttributeError:", e)

Output:

TypeError: '<' not supported between instances of 'str' and 'int'
[1, 3, 'two']
AttributeError: 'dict' object has no attribute 'sort'

• TypeError: ‘<‘ not supported between instances of ‘str’ and ‘int’: the keys have mixed types; sort with key=str or make the types consistent.
• AttributeError: ‘dict’ object has no attribute ‘sort’: use sorted(d.items()) and build a new dict.

More Python dictionary tutorials:

• Sort a list of tuples by the second element
• Initialize a dictionary in Python
• Convert a dictionary to a string
• Get multiple keys from a dictionary

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