Python Basics — Tuples: Practice Exercises

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Tuples — Practice Exercises

Test your understanding of tuple creation, unpacking, named tuples, and tuple vs list usage.

Exercise 1 — Tuple Operations

Given t = (10, 20, 30, 40, 50, 20, 30, 20), write code to:

  1. Count how many times 20 appears
  2. Find the index of first 30
  3. Get the last 3 elements as a new tuple
  4. Check if 60 is in the tuple
Show Solution 
t = (10, 20, 30, 40, 50, 20, 30, 20)

# 1. Count occurrences of 20
print(t.count(20))  # 3

# 2. Index of first 30
print(t.index(30))  # 2

# 3. Last 3 elements
print(t[-3:])  # (20, 30, 20)

# 4. Check membership
print(60 in t)  # False

Exercise 2 — Unpacking Practice

Write a function head_tail(lst) that takes a list and returns the first element and a list of the remaining elements using star unpacking.

# Expected:
head_tail([1, 2, 3, 4, 5])  # → (1, [2, 3, 4, 5])
head_tail([10])              # → (10, [])
Show Solution 
def head_tail(lst):
    first, *rest = lst
    return first, rest

print(head_tail([1, 2, 3, 4, 5]))  # (1, [2, 3, 4, 5])
print(head_tail([10]))              # (10, [])
print(head_tail(["a", "b", "c"]))   # ('a', ['b', 'c'])

Exercise 3 — Named Tuple Student Record

Create a named tuple Student with fields name, age, and gpa. Then:

  1. Create 3 students
  2. Find the student with the highest GPA
  3. Convert the top student to a dictionary
Show Solution 
from collections import namedtuple

Student = namedtuple("Student", "name age gpa")

students = [
    Student("Alice", 20, 3.9),
    Student("Bob", 22, 3.7),
    Student("Charlie", 21, 3.95),
]

# Find highest GPA
top = max(students, key=lambda s: s.gpa)
print(top)  # Student(name='Charlie', age=21, gpa=3.95)

# Convert to dict
print(top._asdict())
# {'name': 'Charlie', 'age': 21, 'gpa': 3.95}

Exercise 4 — Swap Multiple Variables

Without using any temporary variables, swap the values of three variables so that a gets b’s value, b gets c’s value, and c gets a’s value.

a, b, c = 1, 2, 3
# After swap: a=2, b=3, c=1
Show Solution 
a, b, c = 1, 2, 3

a, b, c = b, c, a

print(a)  # 2
print(b)  # 3
print(c)  # 1

This works because Python evaluates the entire right side (b, c, a) as a tuple first, then unpacks it.

Exercise 5 — Tuple as Dictionary Key

Create a distance lookup table using (city1, city2) tuples as keys. Then write a function that returns the distance between any two cities regardless of the order they’re given.

Show Solution 
distances = {
    ("New York", "London"): 5570,
    ("New York", "Tokyo"): 10838,
    ("London", "Tokyo"): 9566,
    ("Paris", "London"): 344,
}

def get_distance(city1, city2):
    # Check both orderings
    if (city1, city2) in distances:
        return distances[(city1, city2)]
    elif (city2, city1) in distances:
        return distances[(city2, city1)]
    else:
        return None

print(get_distance("New York", "London"))  # 5570
print(get_distance("London", "New York"))  # 5570 (reverse order works)
print(get_distance("Tokyo", "London"))     # 9566
print(get_distance("Paris", "Tokyo"))      # None (not in table)

Scoring Guide

Score Level
5/5 Excellent — tuples mastered
3-4/5 Good — review unpacking and named tuples
1-2/5 Keep practicing — re-read the chapter

Next chapter: Dictionaries →

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