Lambda with sorted() in Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the sorted() function do in Python?

It returns a new list containing all items from the iterable in ascending order by default.

beginner

What is a lambda function in Python?

A lambda function is a small anonymous function defined with the lambda keyword, which can take any number of arguments but has only one expression.

intermediate

How do you use a lambda function with sorted()?

You pass the lambda function to the key parameter of sorted() to specify the value to sort by for each item.

intermediate

Example: Sort a list of tuples by the second item using sorted() and lambda.

sorted_list = sorted([(1, 3), (2, 1), (3, 2)], key=lambda x: x[1]) # Result: [(2, 1), (3, 2), (1, 3)]

intermediate

How to sort a list of strings by their length using sorted() and lambda?

Use sorted(strings, key=lambda s: len(s)) to sort strings by length from shortest to longest.

What does the key parameter in sorted() do?

A Specifies a function to extract a comparison key from each element.

B Changes the original list in place.

C Reverses the list before sorting.

D Filters elements before sorting.

The key parameter takes a function that returns a value used to compare elements during sorting.

Which of these is a valid lambda function to sort a list of numbers in descending order using sorted()?

A sorted(numbers, key=lambda x: -x)

B sorted(numbers, key=lambda x: x)

C sorted(numbers, reverse=False)

D sorted(numbers, key=lambda x: x+1)

Using key=lambda x: -x sorts numbers by their negative value, effectively sorting descending.

What will this code output? sorted(['apple', 'banana', 'cherry'], key=lambda x: x[0])

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

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

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

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

Sorting by the first letter keeps the list in the same order because 'a' < 'b' < 'c'.

How can you sort a list of dictionaries by the value of the key 'age' using lambda?

A sorted(list_of_dicts, key=lambda d: age)

B sorted(list_of_dicts, key=lambda d: d.age)

C sorted(list_of_dicts, key=lambda d: d['age'])

D sorted(list_of_dicts, key=lambda d: d.get('name'))

Use lambda d: d['age'] to access the 'age' value in each dictionary for sorting.

What is the output of sorted([3, 1, 2], key=lambda x: x % 2)?

A [1, 2, 3]

B [2, 3, 1]

C [3, 1, 2]

D [1, 3, 2]

Sorting by x % 2 groups even numbers (0) before odd numbers (1), so 2 comes before 3 and 1.

Explain how to use a lambda function with sorted() to sort a list of complex items.

Think about how to tell Python what part of each item to sort by.

Describe a real-life example where sorting with a lambda function would be useful.

Consider sorting a list of people by age or sorting products by price.

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