Lambda with sorted() in Python - Mini Project: Build & Apply

Sorting a List of Dictionaries Using Lambda with sorted()

📖 Scenario: You work in a small bookstore. You have a list of books, each with a title and a price. You want to organize the books by their price from lowest to highest so customers can easily see the cheapest books first.

🎯 Goal: Build a Python program that sorts a list of dictionaries representing books by their price using sorted() with a lambda function.

📋 What You'll Learn

Create a list called books with dictionaries containing 'title' and 'price' keys

Create a variable called sorted_books that sorts books by the 'price' key using sorted() and a lambda

Print the sorted_books list

💡 Why This Matters

🌍 Real World

Sorting lists of dictionaries is common when working with data like products, books, or records where you want to order items by a specific detail.

💼 Career

Many jobs require sorting data efficiently. Using lambda with sorted() is a quick way to organize data without writing complex code.

1

Create the list of books

Create a list called books with these exact dictionaries: {'title': 'Book A', 'price': 12.99}, {'title': 'Book B', 'price': 8.99}, and {'title': 'Book C', 'price': 15.50}.

Python

# Create the list called books with the given dictionaries
# Your code here

Hint

Use square brackets [] to create a list and curly braces {} for each dictionary.

2

Prepare to sort the books

Create a variable called sorted_books and set it to None for now. This will hold the sorted list later.

Python

books = [
    {'title': 'Book A', 'price': 12.99},
    {'title': 'Book B', 'price': 8.99},
    {'title': 'Book C', 'price': 15.50}
]

# Create a variable sorted_books and set it to None
# Your code here

Hint

Just write sorted_books = None for now.

3

Sort the books by price using lambda with sorted()

Use sorted() with a lambda function to sort the books list by the 'price' key. Assign the result to sorted_books. Use lambda book: book['price'] as the key function.

Python

books = [
    {'title': 'Book A', 'price': 12.99},
    {'title': 'Book B', 'price': 8.99},
    {'title': 'Book C', 'price': 15.50}
]

# Sort books by price using sorted() and lambda, assign to sorted_books
# Your code here

Hint

Use sorted(books, key=lambda book: book['price']) to sort by price.

4

Print the sorted list of books

Print the sorted_books list to show the books sorted by price.

Python

books = [
    {'title': 'Book A', 'price': 12.99},
    {'title': 'Book B', 'price': 8.99},
    {'title': 'Book C', 'price': 15.50}
]

sorted_books = sorted(books, key=lambda book: book['price'])

# Print the sorted_books list
# Your code here

Hint

Use print(sorted_books) to show the sorted list.

Practice

1. What does the key argument do when used with sorted() and a lambda function in Python?

easy

A. It reverses the order of the sorted list automatically.

B. It changes the original list to a sorted list in place.

C. It tells sorted() how to compare items by extracting a value from each item.

D. It filters out items that do not match the lambda condition.

Solution

1. Step 1: Understand the role of key in sorted()

   The key argument takes a function that extracts a value from each item to use for sorting.

2. Step 2: Understand how lambda works with key

   The lambda function defines this extraction rule inline, telling sorted() what to compare.

3. Final Answer:

   It tells sorted() how to compare items by extracting a value from each item. -> Option C

4. Quick Check:

   key=lambda x: value means compare by value [OK]

Hint: Remember: key=lambda extracts sort value from each item [OK]

Common Mistakes:

• Thinking key changes the original list
• Confusing key with reverse sorting
• Assuming lambda filters items

2. Which of the following is the correct syntax to sort a list of tuples data = [(2, 'b'), (1, 'a'), (3, 'c')] by the first element using sorted() and a lambda?

easy

A. sorted(data, key=lambda x: x[0])

B. sorted(data, lambda x: x[0])

C. sorted(data, key=lambda x: x[1])

D. sorted(data, key=x[0])

Solution

1. Step 1: Identify the correct use of key argument

   The key argument must be named and assigned a function, here a lambda.

2. Step 2: Check lambda syntax and index

   lambda x: x[0] correctly extracts the first element of each tuple for sorting.

3. Final Answer:

   sorted(data, key=lambda x: x[0]) -> Option A

4. Quick Check:

   Correct syntax uses key= and lambda with index [OK]

Hint: Always use key= with lambda for sorting rules [OK]

Common Mistakes:

• Omitting key= argument name
• Using wrong tuple index
• Passing lambda as second positional argument

3. What is the output of this code?

data = ['apple', 'banana', 'cherry']
sorted_list = sorted(data, key=lambda x: len(x))
print(sorted_list)

medium

A. ['apple', 'banana', 'cherry']

B. ['banana', 'cherry', 'apple']

C. ['cherry', 'banana', 'apple']

D. ['apple', 'cherry', 'banana']

Solution

1. Step 1: Understand sorting by length

   The lambda extracts the length of each string: apple(5), banana(6), cherry(6).

2. Step 2: Sort strings by their length

   Sorted order by length is: apple(5), banana(6), cherry(6). Since banana and cherry have same length, original order among them is preserved (banana before cherry).

3. Final Answer:

   ['apple', 'banana', 'cherry'] -> Option A

4. Quick Check:

   Sort by len(x) = ['apple', 'banana', 'cherry'] [OK]

Hint: Sort by length with key=lambda x: len(x) [OK]

Common Mistakes:

• Assuming alphabetical sort instead of length
• Mixing order of equal length items
• Forgetting sorted returns new list

4. Find the error in this code snippet:

data = [{'name': 'Alice', 'age': 30}, {'name': 'Bob', 'age': 25}]
sorted_data = sorted(data, key=lambda x: x['age'])
print(sorted_data)

medium

A. The code will raise a KeyError because 'age' is missing.

B. There is no error; the code sorts by age correctly.

C. The lambda function should use double quotes for keys.

D. sorted() cannot sort dictionaries.

Solution

1. Step 1: Check dictionary keys and lambda syntax

   Each dictionary has the key 'age', and lambda accesses it correctly with single quotes.

2. Step 2: Confirm sorted() usage on list of dicts

   sorted() can sort list of dictionaries using key function; no error occurs.

3. Final Answer:

   There is no error; the code sorts by age correctly. -> Option B

4. Quick Check:

   Access dict keys with quotes; sorted works on list of dicts [OK]

Hint: Dictionaries can be sorted by key with lambda [OK]

Common Mistakes:

• Thinking single quotes cause error
• Assuming KeyError without missing keys
• Believing sorted() can't handle dicts

5. You have a list of tuples representing products and their prices:

products = [('pen', 1.5), ('notebook', 2.0), ('eraser', 0.5), ('pencil', 1.5)]

How would you sort this list first by price ascending, then by product name alphabetically using sorted() and lambda?

hard

A. sorted(products, key=lambda x: x[1] and x[0])

B. sorted(products, key=lambda x: x[1] or x[0])

C. sorted(products, key=lambda x: x[0], x[1])

D. sorted(products, key=lambda x: (x[1], x[0]))

Solution

1. Step 1: Understand sorting by multiple criteria

   To sort by price then name, use a tuple in key: (price, name).

2. Step 2: Write lambda returning tuple for sorting

   lambda x: (x[1], x[0]) returns price first, then product name.

3. Final Answer:

   sorted(products, key=lambda x: (x[1], x[0])) -> Option D

4. Quick Check:

   Use tuple in key=lambda for multi-level sort [OK]

Hint: Use tuple in lambda key for multi-level sorting [OK]

Common Mistakes:

• Using logical operators instead of tuple
• Passing multiple key arguments
• Reversing order of sort keys