CompactMap for optional unwrapping in Swift - Mini Project: Build & Apply

CompactMap for optional unwrapping

📖 Scenario: You have a list of strings representing ages of people. Some strings are valid numbers, and some are not. You want to get a list of only the valid ages as integers.

🎯 Goal: Build a Swift program that uses compactMap to convert an array of optional strings into an array of non-optional integers, filtering out invalid entries.

📋 What You'll Learn

Create an array of strings called ageStrings with the exact values: ["25", "thirty", "42", "", "18"]

Create a variable called validAges that uses compactMap on ageStrings to convert strings to Int

Print the validAges array

💡 Why This Matters

🌍 Real World

Filtering and converting user input or data from text sources where some values may be invalid or missing.

💼 Career

Data cleaning and transformation is a common task in app development, especially when handling user input or external data.

1

Create the array of age strings

Create an array called ageStrings with these exact string values: ["25", "thirty", "42", "", "18"]

Swift

# Create the array ageStrings with the given values
# Your code here

Hint

Use square brackets to create an array of strings.

2

Use compactMap to convert strings to integers

Create a variable called validAges that uses compactMap on ageStrings to convert each string to an Int.

Swift

let ageStrings = ["25", "thirty", "42", "", "18"]
# Use compactMap on ageStrings to create validAges
# Your code here

Hint

Use compactMap with a closure that tries to convert each string to Int.

3

Print the valid ages

Write a print statement to display the validAges array.

Swift

let ageStrings = ["25", "thirty", "42", "", "18"]
let validAges = ageStrings.compactMap { Int($0) }
# Print the validAges array
# Your code here

Hint

Use print(validAges) to show the result.

4

Run and check the output

Run the program and observe the output. It should print the array of valid ages as integers.

Swift

let ageStrings = ["25", "thirty", "42", "", "18"]
let validAges = ageStrings.compactMap { Int($0) }
print(validAges)
# Run the program to see the output

Hint

The output should be [25, 42, 18] because only these strings convert to integers.

Practice

1. What does the compactMap function do when used on an array of optionals in Swift?

easy

A. It unwraps optionals and removes all nil values from the array.

B. It converts all values to strings regardless of their type.

C. It sorts the array in ascending order.

D. It replaces all nil values with zero.

Solution

1. Step 1: Understand what compactMap does

   compactMap takes an array of optionals and unwraps each element, removing any nil values.

2. Step 2: Compare with other options

   It does not convert types, sort, or replace nil with zero. It only filters out nil and unwraps values.

3. Final Answer:

   It unwraps optionals and removes all nil values from the array. -> Option A

4. Quick Check:

   compactMap removes nil = B [OK]

Hint: Remember: compactMap unwraps and drops nils [OK]

Common Mistakes:

• Confusing compactMap with map (which keeps nils)
• Thinking compactMap sorts or changes types
• Assuming compactMap replaces nil with default values

2. Which of the following is the correct syntax to use compactMap on an array numbers of type [Int?] to get an array of non-optional Int values?

easy

A. let result = numbers.compactMap { $0 }

B. let result = numbers.compactMap { Int($0) }

C. let result = numbers.compactMap { $0! }

D. let result = numbers.compactMap { String($0) }

Solution

1. Step 1: Understand the input and output types

   numbers is an array of optional integers [Int?]. We want to unwrap and remove nil values to get [Int].

2. Step 2: Check each option's correctness

   let result = numbers.compactMap { $0 } uses compactMap { $0 } which unwraps optionals and removes nil. let result = numbers.compactMap { Int($0) } tries to convert optional Int to Int again, which is unnecessary. let result = numbers.compactMap { $0! } force unwraps, which can crash if nil. let result = numbers.compactMap { String($0) } converts to String, changing type.

3. Final Answer:

   let result = numbers.compactMap { $0 } -> Option A

4. Quick Check:

   Use compactMap with identity closure to unwrap optionals [OK]

Hint: Use compactMap { $0 } to unwrap optionals safely [OK]

Common Mistakes:

• Using force unwrap inside compactMap causing runtime errors
• Trying to convert optionals unnecessarily
• Changing the type unintentionally

3. What is the output of the following Swift code?

let values: [Int?] = [1, nil, 3, nil, 5]
let unwrapped = values.compactMap { $0 }
print(unwrapped)

medium

A. Compilation error

B. [1, 3, 5]

C. [1, nil, 3, nil, 5]

D. [Optional(1), Optional(3), Optional(5)]

Solution

1. Step 1: Understand the input array

   The array values contains optional integers with some nil values.

2. Step 2: Apply compactMap

   compactMap { $0 } unwraps optionals and removes nil values, resulting in [1, 3, 5].

3. Final Answer:

   [1, 3, 5] -> Option B

4. Quick Check:

   compactMap removes nils and unwraps = [1, 3, 5] [OK]

Hint: compactMap removes nils and unwraps optionals [OK]

Common Mistakes:

• Expecting optionals in output instead of unwrapped values
• Thinking nils remain in the output
• Confusing compactMap with map

4. Identify the error in this Swift code snippet and choose the correct fix:

let strings: [String?] = ["apple", nil, "banana"]
let lengths = strings.compactMap { $0.count }
print(lengths)

medium

A. Error: Use force unwrap $0! before accessing count

B. Error: compactMap should be replaced with map

C. No error; code runs correctly

D. Error: Cannot use $0.count on optional; fix by using compactMap { $0?.count }

Solution

1. Step 1: Analyze the closure inside compactMap

   The closure uses $0.count, but $0 is an optional String?. Accessing count directly on optional causes error.

2. Step 2: Correct the closure to safely unwrap

   Using $0?.count returns an optional Int, which compactMap unwraps and removes nils, fixing the error.

3. Final Answer:

   Error: Cannot use $0.count on optional; fix by using compactMap { $0?.count } -> Option D

4. Quick Check:

   Use optional chaining inside compactMap closure [OK]

Hint: Use optional chaining ($0?.property) inside compactMap [OK]

Common Mistakes:

• Trying to access property directly on optional causing error
• Using map instead of compactMap losing nil filtering
• Force unwrapping causing runtime crashes

5. Given a dictionary [String: Int?] representing item stock counts, how can you create a new dictionary with only items that have a non-nil stock count using compactMap or similar methods?

hard

A. Use filter { $0.value != nil } without unwrapping values.

B. Use mapValues { $0! } to force unwrap all values.

C. Use compactMapValues { $0 } to unwrap and filter out nil values.

D. Use compactMap { $0 } directly on the dictionary.

Solution

1. Step 1: Understand dictionary optional values

   The dictionary has optional Int values. We want to remove keys with nil values and unwrap the rest.

2. Step 2: Use compactMapValues to unwrap and filter

   compactMapValues { $0 } unwraps optionals in values and removes entries with nil values, producing a dictionary with non-optional Int values.

3. Final Answer:

   Use compactMapValues { $0 } to unwrap and filter out nil values. -> Option C

4. Quick Check:

   compactMapValues unwraps and filters dictionary values [OK]

Hint: Use compactMapValues to unwrap and remove nil dictionary values [OK]

Common Mistakes:

• Using mapValues with force unwrap risking crashes
• Filtering without unwrapping values
• Trying to use compactMap directly on dictionary