Sorting and Key Functions

Sort lists, dicts, and objects with sorted() and the key parameter — from simple to multi-criteria sorting.

"Python's sorted() can sort almost anything — as long as you tell it what to compare. The key parameter is that instruction."

— ShurAI

sorted() vs .sort()

Python has two ways to sort. Know the difference:

sorted(iterable)

Returns a new sorted list
Original is unchanged
Works on any iterable
sorted([3,1,2]) → [1,2,3]

list.sort()

Sorts the list in place
Returns None
Only works on lists
[3,1,2].sort() → list modified

python

nums = [3, 1, 4, 1, 5, 9, 2]

# sorted() — original untouched
asc  = sorted(nums)             # [1, 1, 2, 3, 4, 5, 9]
desc = sorted(nums, reverse=True) # [9, 5, 4, 3, 2, 1, 1]
print(nums)   # [3, 1, 4, 1, 5, 9, 2]  — unchanged!

# .sort() — modifies in place
nums.sort()
print(nums)   # [1, 1, 2, 3, 4, 5, 9]  — changed!

The key Parameter

Pass a function to key= to control what Python compares when sorting. The function is called on each item to get the sort value:

python

words = ["banana", "fig", "apple", "elderberry", "date"]

# Sort by length (shortest first)
print(sorted(words, key=len))
# ['fig', 'date', 'apple', 'banana', 'elderberry']

# Sort alphabetically, ignoring case
print(sorted(["Banana", "apple", "Cherry"], key=str.lower))
# ['apple', 'Banana', 'Cherry']

Sorting Objects with lambda

Use a lambda to sort by a specific attribute or dict key:

python

students = [
    {"name": "Riya",   "score": 92, "age": 22},
    {"name": "Arjun",  "score": 75, "age": 25},
    {"name": "Sneha",  "score": 88, "age": 21},
    {"name": "Vikram", "score": 75, "age": 23},
]

# Sort by score, highest first
by_score = sorted(students, key=lambda s: s["score"], reverse=True)
for s in by_score:
    print(f"{s['name']}: {s['score']}")
# Riya: 92
# Sneha: 88
# Arjun: 75
# Vikram: 75

Multi-Key Sorting

Return a tuple from the key function to sort by multiple criteria at once:

python

# Sort by score DESC, then by name ASC for ties
by_score_name = sorted(
    students,
    key=lambda s: (-s["score"], s["name"])  # negate score for DESC
)
for s in by_score_name:
    print(f"{s['score']:3}  {s['name']}")
#  92  Riya
#  88  Sneha
#  75  Arjun   ← tie at 75: Arjun before Vikram (alphabetical)
#  75  Vikram

Sorting Objects with attrgetter

python

from operator import attrgetter

class Product:
    def __init__(self, name, price):
        self.name  = name
        self.price = price
    def __repr__(self):
        return f"{self.name}(₹{self.price})"

products = [Product("Rice", 80), Product("Oil", 150), Product("Salt", 25)]

# attrgetter is faster than lambda for attribute access
print(sorted(products, key=attrgetter("price")))
# [Salt(₹25), Rice(₹80), Oil(₹150)]

"Python's sort is stable — items with equal keys keep their original order. This makes multi-key sorting by sorting twice work perfectly: sort by secondary key first, then primary key."

— ShurAI

🧠 Quiz — Q1

What is the key difference between sorted() and list.sort()?

🧠 Quiz — Q2

You want to sort a list of strings by their length. Which call is correct?

🧠 Quiz — Q3

How do you sort a list of dicts by the value of key "score", highest first?

🧠 Quiz — Q4

You sort students by key=lambda s: (-s["score"], s["name"]). What does the tuple achieve?