How to Sort a List of Dictionaries in Python

If you have a list where each item is a dictionary, you can sort it by one dictionary key such as age, name, price, or city.

This is common when working with real data from JSON files, APIs, or form input.

Quick answer

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": 22},
    {"name": "Cara", "age": 27}
]

sorted_people = sorted(people, key=lambda person: person["age"])
print(sorted_people)

Output:

[{'name': 'Ben', 'age': 22}, {'name': 'Cara', 'age': 27}, {'name': 'Ana', 'age': 30}]

Use sorted() when you want a new sorted list. Use a key function to choose which dictionary value to sort by.

What this page helps you do

Sort a list of dictionaries by a specific key
Sort numbers like age or price
Sort text like name or city
Reverse the order when needed
Avoid common beginner mistakes

Use `sorted()` with a key

A list of dictionaries looks like this:

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": 22},
    {"name": "Cara", "age": 27}
]

Each item in the list is a dictionary.

To sort this kind of data, use sorted() and tell Python which value to compare.

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": 22},
    {"name": "Cara", "age": 27}
]

sorted_people = sorted(people, key=lambda item: item["age"])

print(sorted_people)

Output:

[{'name': 'Ben', 'age': 22}, {'name': 'Cara', 'age': 27}, {'name': 'Ana', 'age': 30}]

What each part does

sorted(people, ...) sorts the list
key= tells Python what value to use for sorting
lambda item: item["age"] gets the age from each dictionary

If lambda is new to you, see how to use lambda functions in Python.

Sort by a number value

This is useful for keys like age, score, price, or quantity.

products = [
    {"name": "Notebook", "price": 4.50},
    {"name": "Pen", "price": 1.20},
    {"name": "Bag", "price": 18.00}
]

sorted_products = sorted(products, key=lambda item: item["price"])
print(sorted_products)

Output:

[{'name': 'Pen', 'price': 1.2}, {'name': 'Notebook', 'price': 4.5}, {'name': 'Bag', 'price': 18.0}]

Python sorts number values from lowest to highest by default.

Sort numbers in descending order

Use reverse=True if you want highest to lowest.

products = [
    {"name": "Notebook", "price": 4.50},
    {"name": "Pen", "price": 1.20},
    {"name": "Bag", "price": 18.00}
]

sorted_products = sorted(products, key=lambda item: item["price"], reverse=True)
print(sorted_products)

Output:

[{'name': 'Bag', 'price': 18.0}, {'name': 'Notebook', 'price': 4.5}, {'name': 'Pen', 'price': 1.2}]

If you only need to sort a normal list of values, see how to sort a list in Python.

Sort by a string value

You can also sort alphabetically by a string key like name, title, or city.

people = [
    {"name": "Cara", "age": 27},
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": 22}
]

sorted_people = sorted(people, key=lambda item: item["name"])
print(sorted_people)

Output:

[{'name': 'Ana', 'age': 30}, {'name': 'Ben', 'age': 22}, {'name': 'Cara', 'age': 27}]

Case-sensitive sorting

Uppercase and lowercase letters can change the result.

people = [
    {"name": "ana"},
    {"name": "Ben"},
    {"name": "cara"}
]

sorted_people = sorted(people, key=lambda item: item["name"])
print(sorted_people)

This may not sort the way you expect, because Python compares strings by their character values.

Case-insensitive sorting

Use .lower() if you want a more natural alphabetical order.

people = [
    {"name": "ana"},
    {"name": "Ben"},
    {"name": "cara"}
]

sorted_people = sorted(people, key=lambda item: item["name"].lower())
print(sorted_people)

Output:

[{'name': 'ana'}, {'name': 'Ben'}, {'name': 'cara'}]

Sort the original list in place

If you want to change the original list instead of creating a new one, use .sort().

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": 22},
    {"name": "Cara", "age": 27}
]

people.sort(key=lambda item: item["age"])
print(people)

Output:

[{'name': 'Ben', 'age': 22}, {'name': 'Cara', 'age': 27}, {'name': 'Ana', 'age': 30}]

Important difference

sorted() returns a new list
.sort() changes the original list
.sort() returns None

This is a common beginner mistake:

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": 22}
]

result = people.sort(key=lambda item: item["age"])
print(result)

Output:

None

If you want a new list, use sorted() instead.

Handle missing keys safely

Using item["age"] works only if every dictionary has an age key.

If one dictionary is missing that key, Python raises a KeyError.

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben"},
    {"name": "Cara", "age": 27}
]

sorted_people = sorted(people, key=lambda item: item["age"])
print(sorted_people)

That fails because {"name": "Ben"} does not have an age.

Safer approach with `get()`

Use .get() when keys may be missing.

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben"},
    {"name": "Cara", "age": 27}
]

sorted_people = sorted(people, key=lambda item: item.get("age", 0))
print(sorted_people)

Output:

[{'name': 'Ben'}, {'name': 'Cara', 'age': 27}, {'name': 'Ana', 'age': 30}]

Here, missing age values are treated as 0.

Choose a default value that makes sense for your data.

Use 0 if missing numbers should come first
Use a large number if missing values should come last
Use "" for missing text values in some cases

To learn more about this method, see Python dictionary get() method. If you are seeing a missing-key error, read KeyError in Python: causes and fixes.

When sorting may fail

Sorting can fail if the values are not consistent.

Mixed types

This usually causes a TypeError:

people = [
    {"name": "Ana", "age": 30},
    {"name": "Ben", "age": "22"},
    {"name": "Cara", "age": 27}
]

sorted_people = sorted(people, key=lambda item: item["age"])
print(sorted_people)

One age is an integer, but another is a string. Python cannot always compare them.

Missing keys

This raises a KeyError when you use item["age"] and one dictionary does not have that key.

`None` values

If some values are None, sorting may also fail or give unexpected results.

Before sorting, check what your data looks like:

print(data)
print(type(data))
print(data[0])
print(data[0].keys())
print([item.get('age') for item in data])
print([type(item.get('age')) for item in data])

These checks help you answer questions like:

Is data really a list?
Is each item a dictionary?
Do all dictionaries have the same keys?
Are all age values the same type?

Common mistakes

Using item["key"] when some dictionaries do not have that key
Trying to sort mixed types like integers and strings together
Using .sort() and expecting it to return a new list
Forgetting reverse=True when descending order is needed
Getting unexpected string order because of uppercase and lowercase letters

FAQ

What is the easiest way to sort a list of dictionaries in Python?

Use sorted() with key=.

sorted(data, key=lambda item: item["age"])

What is the difference between `sorted()` and `sort()`?

sorted() creates a new list
list.sort() changes the original list

How do I sort in descending order?

Add reverse=True.

sorted(data, key=lambda item: item["age"], reverse=True)

How do I sort alphabetically by name?

Use a string key:

sorted(data, key=lambda item: item["name"])

How do I avoid errors when a key is missing?

Use .get() with a default value:

sorted(data, key=lambda item: item.get("age", 0))