Sequence operators (map, filter) in Kotlin - Mini Project: Build & Apply

Using Sequence Operators: map and filter in Kotlin

📖 Scenario: You are working on a simple Kotlin program that processes a list of ages to find which people are adults and then convert their ages to a string format for display.

🎯 Goal: Build a Kotlin program that uses sequence operators map and filter to select adults (age 18 or older) from a list of ages and then convert those ages to strings.

📋 What You'll Learn

Create a list of ages with exact values

Create a sequence from the list

Use filter to keep only ages 18 or older

Use map to convert the filtered ages to strings

Print the resulting list of strings

💡 Why This Matters

🌍 Real World

Filtering and transforming data is common in apps that process user information, like showing only adult users or formatting data for display.

💼 Career

Understanding sequence operators like <code>map</code> and <code>filter</code> is essential for Kotlin developers working on data processing, Android apps, or backend services.

1

Create the list of ages

Create a list called ages with these exact integer values: 15, 22, 17, 30, 14, 19.

Kotlin

# Create the list of ages
# Your code here

Hint

Use listOf to create the list with the exact numbers.

2

Create a sequence from the list

Create a variable called ageSequence by calling asSequence() on the ages list.

Kotlin

val ages = listOf(15, 22, 17, 30, 14, 19)
# Create a sequence called ageSequence from ages
# Your code here

Hint

Use asSequence() on the ages list to create a sequence.

3

Filter adults and map to strings

Create a variable called adultAgesStrings by applying filter on ageSequence to keep ages 18 or older, then apply map to convert each age to a string, and finally convert the result to a list using toList().

Kotlin

val ages = listOf(15, 22, 17, 30, 14, 19)
val ageSequence = ages.asSequence()
# Filter ages 18 or older and map to strings, then convert to list
# Your code here

Hint

Use filter { it >= 18 } to keep adults, then map { it.toString() } to convert to strings, and toList() to get a list.

4

Print the resulting list

Write a println statement to print the adultAgesStrings list.

Kotlin

val ages = listOf(15, 22, 17, 30, 14, 19)
val ageSequence = ages.asSequence()
val adultAgesStrings = ageSequence.filter { it >= 18 }.map { it.toString() }.toList()
# Print the adultAgesStrings list
# Your code here

Hint

Use println(adultAgesStrings) to display the list.

Practice

1. What does the map operator do in a Kotlin sequence?

easy

A. It removes items that do not match a condition.

B. It transforms each item in the sequence to a new value.

C. It sorts the items in the sequence.

D. It counts the number of items in the sequence.

Solution

1. Step 1: Understand the purpose of map

   The map operator applies a function to each item, changing it to a new value.

2. Step 2: Compare with other operators

   filter removes items, sort orders items, and count counts items, so these are different.

3. Final Answer:

   It transforms each item in the sequence to a new value. -> Option B

4. Quick Check:

   map = transform each item [OK]

Hint: Map changes items; filter removes items [OK]

Common Mistakes:

• Confusing map with filter
• Thinking map sorts or counts items
• Assuming map removes items

2. Which of the following is the correct syntax to filter even numbers from a sequence seq in Kotlin?

easy

A. seq.filter { it % 2 == 0 }

B. seq.map { it % 2 == 0 }

C. seq.filter(it % 2 == 0)

D. seq.map { it / 2 == 0 }

Solution

1. Step 1: Identify correct filter syntax

   In Kotlin, filter takes a lambda with condition inside curly braces: { it % 2 == 0 }.

2. Step 2: Check other options

   map changes values, not filters; seq.filter(it % 2 == 0) misses braces; seq.map { it / 2 == 0 } uses wrong condition.

3. Final Answer:

   seq.filter { it % 2 == 0 } -> Option A

4. Quick Check:

   filter uses braces and condition [OK]

Hint: filter uses braces with condition inside [OK]

Common Mistakes:

• Using map instead of filter
• Missing curly braces in lambda
• Wrong condition inside lambda

3. What is the output of this Kotlin code?

val numbers = sequenceOf(1, 2, 3, 4)
val result = numbers.filter { it % 2 == 0 }.map { it * 10 }.toList()
println(result)

medium

A. [10, 20]

B. [2, 4]

C. [20, 40]

D. [1, 3]

Solution

1. Step 1: Apply filter to keep even numbers

   Filter keeps numbers where it % 2 == 0, so 2 and 4 remain.

2. Step 2: Apply map to multiply each by 10

   Mapping multiplies each by 10: 2 * 10 = 20, 4 * 10 = 40.

3. Final Answer:

   [20, 40] -> Option C

4. Quick Check:

   filter evens then map *10 = [20, 40] [OK]

Hint: Filter first, then map transforms filtered items [OK]

Common Mistakes:

• Mapping before filtering
• Confusing filter condition
• Forgetting to convert sequence to list

4. Find the error in this Kotlin code snippet:

val seq = sequenceOf(1, 2, 3, 4)
val filtered = seq.filter(it % 2 == 0)
println(filtered.toList())

medium

A. Missing curly braces around lambda in filter

B. Using map instead of filter

C. Sequence cannot be converted to list

D. Filter condition should be it % 2 != 0

Solution

1. Step 1: Check filter lambda syntax

   Filter requires a lambda with curly braces: { it % 2 == 0 }. The code misses braces.

2. Step 2: Validate other options

   Map is not used here; sequence can convert to list; condition can be any boolean expression.

3. Final Answer:

   Missing curly braces around lambda in filter -> Option A

4. Quick Check:

   filter needs braces for lambda [OK]

Hint: Always use braces {} for lambdas in filter [OK]

Common Mistakes:

• Omitting braces in lambda
• Confusing filter and map
• Wrong filter condition assumed as error

5. Given a sequence of strings seq = sequenceOf("apple", "banana", "cherry", "date"), which Kotlin code correctly filters strings containing letter 'a' and converts them to uppercase?

hard

A. seq.filter { it.contains('a') }.map { it.lowercase() }.toList()

B. seq.map { it.uppercase() }.filter { it.contains('a') }.toList()

C. seq.map { it.lowercase() }.filter { it.contains('a') }.toList()

D. seq.filter { it.contains('a') }.map { it.uppercase() }.toList()

Solution

1. Step 1: Filter strings containing 'a'

   Use filter { it.contains('a') } to keep only strings with 'a'. This keeps "apple", "banana", "date".

2. Step 2: Convert filtered strings to uppercase

   Use map { it.uppercase() } to change these strings to uppercase.

3. Final Answer:

   seq.filter { it.contains('a') }.map { it.uppercase() }.toList() -> Option D

4. Quick Check:

   Filter then map uppercase = seq.filter { it.contains('a') }.map { it.uppercase() }.toList() [OK]

Hint: Filter first, then map to uppercase [OK]

Common Mistakes:

• Mapping before filtering changes all strings
• Using lowercase instead of uppercase
• Filtering after mapping changes string case before filter