How to Sort a List in Python

What you’ll build or solve

You’ll sort a Python list in ascending or descending order, including sorting text, numbers, and lists of dictionaries.

When this approach works best

This approach works best when you:

• Need to display items in a predictable order, like prices low-to-high or names A-to-Z.
• Want to rank or pick top results, like highest scores or most recent dates.
• Have structured data and need to sort by a specific field, like sorting a list of users by age.

Avoid this approach when:

• You need to preserve the original order and only filter items. Sorting changes order, so filter first or copy the list.

Prerequisites

• Python installed
• You know what a list is

Step-by-step instructions

1) Pick sort() or sorted() based on whether you want to change the list

Python gives you two main ways to sort:

• my_list.sort() sorts the list in place and returns None.
• sorted(my_list) returns a new sorted list and keeps the original unchanged.

numbers= [3,1,2]

numbers.sort()
print(numbers)

Using sorted():

numbers= [3,1,2]

sorted_numbers=sorted(numbers)
print(sorted_numbers)
print(numbers)

What to look for: if you do result = numbers.sort(), result becomes None.

2) Sort in descending order

Use reverse=True to sort from largest to smallest, or Z-to-A for text.

scores= [10,5,25,7]
scores.sort(reverse=True)
print(scores)

With sorted():

scores= [10,5,25,7]
top_first=sorted(scores,reverse=True)
print(top_first)

3) Sort text with consistent casing

String sorting is case-sensitive, so uppercase letters can come before lowercase letters. For a more natural alphabetical sort, use key=str.lower.

names= ["mina","Sam","ada","Zoe"]

names_sorted=sorted(names)
print(names_sorted)

Case-insensitive sort:

names= ["mina","Sam","ada","Zoe"]

names_sorted=sorted(names,key=str.lower)
print(names_sorted)

What to look for: using key=str.lower does not change the strings, it only changes how Python compares them.

4) Sort by a custom rule with key=

The key function tells Python what value to sort by.

Sort by length:

words= ["python","go","javascript","c"]
print(sorted(words,key=len))

Sort by absolute value:

values= [-10,3,-2,7]
print(sorted(values,key=abs))

5) Sort lists of dictionaries by a field

A common real-world case is sorting records by a specific key like age, price, or created_at.

users= [
    {"name":"Mina","age":34},
    {"name":"Sam","age":27},
    {"name":"Ada","age":30},
]

by_age=sorted(users,key=lambdau:u["age"])
print(by_age)

If a key might be missing, sort more safely with get() and a default:

users= [
    {"name":"Mina","age":34},
    {"name":"Sam"},
    {"name":"Ada","age":30},
]

by_age=sorted(users,key=lambdau:u.get("age",0))
print(by_age)

What to look for: if you sort with u["age"] and one item has no age, you get a KeyError.

Examples you can copy

Example 1: Sort numbers and keep the original list

numbers= [12,3,9,1]
sorted_numbers=sorted(numbers)

print("sorted:",sorted_numbers)
print("original:",numbers)

Example 2: Sort names case-insensitively

names= ["mina","Sam","ada","Zoe"]
print(sorted(names,key=str.lower))

Example 3: Sort products by price, highest first

products= [
    {"name":"Notebook","price":4.5},
    {"name":"Pen","price":1.2},
    {"name":"Backpack","price":29.0},
]

by_price_desc=sorted(products,key=lambdap:p["price"],reverse=True)
print(by_price_desc)

Example 4: Sort by two fields (primary, then secondary)

This uses a tuple to sort by city first, then by name within each city.

people= [
    {"name":"Naomi","city":"Boston"},
    {"name":"Sam","city":"New York"},
    {"name":"Ada","city":"Los Angeles"},
]

sorted_people=sorted(
people,
key=lambdap: (p["city"],p["name"].lower())
)
print(sorted_people)

Common mistakes and how to fix them

Mistake 1: Saving the result of .sort()

You might write:

numbers= [3,1,2]
result=numbers.sort()
print(result)

Why it breaks: .sort() sorts in place and returns None.

Correct approach:

numbers= [3,1,2]
numbers.sort()
print(numbers)

numbers= [3,1,2]
result=sorted(numbers)
print(result)

Mistake 2: Sorting a list with mixed types

You might write:

items= [3,"2",1]
items.sort()

Why it breaks: Python cannot compare numbers and strings during sorting.

Correct approach: convert all items to the same type before sorting.

items= [3,"2",1]
items_clean= [int(x)ifisinstance(x,str)elsexforxinitems]
items_clean.sort()
print(items_clean)

Troubleshooting

• If you see TypeError: '<' not supported between instances of ..., check for mixed types and convert them before sorting.
• If you see None after sorting, you saved the return value of .sort(). Use sorted() or sort the list directly.
• If you see KeyError while sorting dictionaries, use dict.get("key", default) in your key= function.
• If your sort looks wrong for text, use key=str.lower for case-insensitive ordering.
• If numbers sort like strings, for example "10" before "2", convert them to int or float first.

Quick recap

• Use list.sort() to sort in place, or sorted(list) to get a new list.
• Add reverse=True for descending order.
• Use key= to control how items are compared.
• For dictionaries, sort with key=lambda d: d["field"] or d.get("field", default).
• Fix mixed-type issues by converting values before sorting.