Dictionary use cases in Python - Mini Project: Build & Apply

Dictionary use cases

📖 Scenario: You are managing a small bookstore's inventory. You want to keep track of book titles and their quantities using a dictionary.

🎯 Goal: Build a Python program that creates a dictionary of books and quantities, sets a minimum stock threshold, finds books below that threshold, and prints them.

📋 What You'll Learn

Create a dictionary with exact book titles and quantities

Create a variable for minimum stock threshold

Use a dictionary comprehension to find books with stock below the threshold

Print the resulting dictionary of low stock books

💡 Why This Matters

🌍 Real World

Managing inventory in stores or warehouses often uses dictionaries to track items and quantities.

💼 Career

Understanding dictionary operations is essential for data handling, filtering, and reporting in many programming jobs.

1

Create the book inventory dictionary

Create a dictionary called inventory with these exact entries: 'Python Basics': 12, 'Data Science 101': 5, 'Machine Learning': 3, 'Deep Learning': 7, 'AI for Beginners': 2

Python

# Create the inventory dictionary with exact book titles and quantities
# Your code here

Hint

Use curly braces {} to create a dictionary with keys as book titles and values as quantities.

2

Set the minimum stock threshold

Create a variable called min_stock and set it to 5 to represent the minimum acceptable stock level.

Python

inventory = {
    'Python Basics': 12,
    'Data Science 101': 5,
    'Machine Learning': 3,
    'Deep Learning': 7,
    'AI for Beginners': 2
}
# Create min_stock variable and set it to 5
# Your code here

Hint

Just assign the number 5 to a variable named min_stock.

3

Find books with stock below the threshold

Use a dictionary comprehension to create a new dictionary called low_stock_books that contains only the books from inventory with quantities less than min_stock. Use for book, qty in inventory.items() in your comprehension.

Python

inventory = {
    'Python Basics': 12,
    'Data Science 101': 5,
    'Machine Learning': 3,
    'Deep Learning': 7,
    'AI for Beginners': 2
}
min_stock = 5
# Create low_stock_books dictionary using comprehension
# Your code here

Hint

Use {key: value for key, value in dict.items() if condition} format for dictionary comprehension.

4

Print the low stock books

Write a print statement to display the low_stock_books dictionary.

Python

inventory = {
    'Python Basics': 12,
    'Data Science 101': 5,
    'Machine Learning': 3,
    'Deep Learning': 7,
    'AI for Beginners': 2
}
min_stock = 5
low_stock_books = {book: qty for book, qty in inventory.items() if qty < min_stock}
# Print the low_stock_books dictionary
# Your code here

Hint

Use print(low_stock_books) to show the dictionary.

Practice

1. What is the main purpose of a dictionary in Python?

easy

A. To store data as key-value pairs for quick access

B. To store data in a fixed order like a list

C. To perform mathematical calculations

D. To create a sequence of numbers

Solution

1. Step 1: Understand dictionary structure

   Dictionaries store data in pairs where each key maps to a value.

2. Step 2: Identify dictionary use case

   This structure allows fast lookup by key, unlike lists which use indexes.

3. Final Answer:

   To store data as key-value pairs for quick access -> Option A

4. Quick Check:

   Dictionaries = key-value pairs [OK]

Hint: Remember: dictionaries use keys to find values fast [OK]

Common Mistakes:

• Confusing dictionaries with lists or sets
• Thinking dictionaries keep order like lists
• Assuming dictionaries store only numbers

2. Which of the following is the correct way to add a new key-value pair to a dictionary my_dict?

easy

A. my_dict.append('key', 'value')

B. my_dict.add('key', 'value')

C. my_dict['key'] = 'value'

D. my_dict.insert('key', 'value')

Solution

1. Step 1: Recall dictionary syntax for adding items

   To add or update a key-value pair, use square brackets with the key and assign a value.

2. Step 2: Check each option

   Only my_dict['key'] = 'value' correctly adds or updates the dictionary.

3. Final Answer:

   my_dict['key'] = 'value' -> Option C

4. Quick Check:

   Use square brackets to add/update dict items [OK]

Hint: Use square brackets and assignment to add dict items [OK]

Common Mistakes:

• Using list methods like append or insert on dictionaries
• Trying to use a non-existent add() method
• Confusing dictionary syntax with list syntax

3. What will be the output of the following code?

my_dict = {'a': 1, 'b': 2}
my_dict['c'] = 3
print(my_dict)

medium

A. Error: Cannot add new key

B. {'a': 1, 'b': 2}

C. {'c': 3}

D. {'a': 1, 'b': 2, 'c': 3}

Solution

1. Step 1: Understand dictionary update

   Adding my_dict['c'] = 3 adds a new key 'c' with value 3.

2. Step 2: Print the updated dictionary

   Printing shows all keys and values including the new one.

3. Final Answer:

   {'a': 1, 'b': 2, 'c': 3} -> Option D

4. Quick Check:

   Adding key updates dict content [OK]

Hint: Adding key-value pairs updates dictionary instantly [OK]

Common Mistakes:

• Expecting original dict without new key
• Thinking print shows only last added key
• Assuming error when adding new keys

4. Find the error in this code snippet:

my_dict = {'x': 10, 'y': 20}
print(my_dict['z'])

medium

A. KeyError because 'z' is not in the dictionary

B. SyntaxError due to wrong brackets

C. TypeError because dictionary keys must be integers

D. No error, prints 0

Solution

1. Step 1: Check dictionary keys

   Keys are 'x' and 'y', but 'z' is not present.

2. Step 2: Accessing missing key causes error

   Trying to print my_dict['z'] raises a KeyError.

3. Final Answer:

   KeyError because 'z' is not in the dictionary -> Option A

4. Quick Check:

   Access missing key = KeyError [OK]

Hint: Accessing missing keys causes KeyError [OK]

Common Mistakes:

• Assuming missing keys return 0 or None
• Confusing KeyError with SyntaxError
• Trying to use wrong bracket types

5. Given a list of tuples representing student names and scores:

students = [('Alice', 85), ('Bob', 90), ('Alice', 95)]

Which code correctly creates a dictionary with the latest score for each student?

hard

A. scores = dict(students)
scores['Alice'] = 85

B. scores = {name: score for name, score in students}

C. scores = {}
for name, score in students:
if name not in scores:
scores[name] = score

D. scores = {name: max(score) for name, score in students}

Solution

1. Step 1: Understand dictionary comprehension behavior

   When keys repeat, the last value overwrites earlier ones.

2. Step 2: Check each option

   scores = {name: score for name, score in students} uses comprehension that keeps the last score for 'Alice' (95). scores = dict(students) scores['Alice'] = 85 wrongly resets 'Alice' score. scores = {} for name, score in students: if name not in scores: scores[name] = score keeps only first score. scores = {name: max(score) for name, score in students} is invalid syntax.

3. Final Answer:

   scores = {name: score for name, score in students} -> Option B

4. Quick Check:

   Dict comprehension overwrites duplicate keys [OK]

Hint: Dict comprehension keeps last value for duplicate keys [OK]

Common Mistakes:

• Assuming dict keeps first value for duplicates
• Using invalid syntax for max(score)
• Manually overwriting values incorrectly