Lazy enumerators in Ruby - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a lazy enumerator from an array.

Ruby

numbers = [1, 2, 3, 4, 5]
lazy_enum = numbers.[1]

Drag options to blanks, or click blank then click option'

Alazy

Bmap

Cselect

Deach

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'map' or 'select' directly without 'lazy' creates immediate enumerables.

Using 'each' returns the original enumerable, not lazy.

The lazy method converts an enumerable into a lazy enumerator, allowing deferred computation.

2fill in blank

medium

Complete the code to get the first 3 even numbers lazily.

Ruby

numbers = (1..Float::INFINITY).lazy
result = numbers.select { |n| n.even? }.[1](3).force

Drag options to blanks, or click blank then click option'

Atake

Bmap

Cdrop

Dfirst

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'first' returns an element immediately, not a lazy enumerator.

Using 'drop' skips elements instead of taking them.

The take method selects the first N elements lazily before forcing evaluation.

3fill in blank

hard

Fix the error in the code to lazily map and select odd numbers.

Ruby

numbers = (1..10).lazy
result = numbers.[1] { |n| n * 2 }.select(&:odd?).force

Drag options to blanks, or click blank then click option'

Aselect

Beach

Cmap

Dreject

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'select' instead of 'map' applies a filter, not a transformation.

Using 'each' does not return a lazy enumerator.

The map method transforms each element lazily before selection.

4fill in blank

hard

Fill both blanks to create a lazy enumerator that drops the first 5 elements and takes the next 3.

Ruby

numbers = (1..100).lazy
result = numbers.[1](5).[2](3).force

Drag options to blanks, or click blank then click option'

Adrop

Btake

Cselect

Dmap

Attempts:

3 left

💡 Hint

Common Mistakes

Reversing the order of 'take' and 'drop' changes the result.

Using 'select' or 'map' instead of 'drop' or 'take' changes behavior.

drop skips the first N elements lazily, and take selects the next N elements lazily.

5fill in blank

hard

Fill all three blanks to create a lazy enumerator that maps numbers to their squares and selects those greater than 10.

Ruby

numbers = (1..10).lazy
result = numbers.[1] { |n| n * n }.[2] { |n| n [3] 10 }.force

Drag options to blanks, or click blank then click option'

Amap

Bselect

C>

Ddrop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'drop' instead of 'select' changes filtering behavior.

Using '<' instead of '>' changes the filter condition.

map transforms numbers to their squares, select filters those greater than 10.

Practice

1. What is the main benefit of using .lazy enumerators in Ruby?

easy

A. They sort elements in ascending order by default.

B. They immediately compute all elements for faster access.

C. They convert arrays into hashes automatically.

D. They delay computation until results are needed, saving memory and time.

Solution

1. Step 1: Understand what lazy enumerators do

   Lazy enumerators delay the processing of elements until you actually ask for them, which helps save memory and time especially with large or infinite collections.

2. Step 2: Compare options with this behavior

   Only They delay computation until results are needed, saving memory and time. describes this delaying behavior. Other options describe immediate computation, conversion, or sorting, which are not the main features of lazy enumerators.

3. Final Answer:

   They delay computation until results are needed, saving memory and time. -> Option D

4. Quick Check:

   Lazy enumerators delay work = A [OK]

Hint: Lazy means delay work until needed, not immediate [OK]

Common Mistakes:

• Thinking lazy enumerators compute all elements immediately
• Confusing lazy enumerators with sorting or conversion methods
• Assuming lazy enumerators change data structure type

2. Which of the following is the correct way to create a lazy enumerator from an array [1, 2, 3] in Ruby?

easy

A. [1, 2, 3].to_lazy

B. lazy([1, 2, 3])

C. [1, 2, 3].lazy

D. Array.lazy([1, 2, 3])

Solution

1. Step 1: Recall the syntax for creating lazy enumerators

   In Ruby, calling .lazy on an enumerable creates a lazy enumerator.

2. Step 2: Check each option's syntax

   Only [1, 2, 3].lazy is valid Ruby syntax. The others are not valid methods or functions.

3. Final Answer:

   [1, 2, 3].lazy -> Option C

4. Quick Check:

   Use .lazy on enumerable = A [OK]

Hint: Use .lazy method on enumerable to create lazy enumerator [OK]

Common Mistakes:

• Trying to call non-existent methods like to_lazy
• Using lazy as a standalone function
• Confusing class methods with instance methods

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

result = (1..Float::INFINITY).lazy.select { |x| x % 3 == 0 }.map { |x| x * 2 }.first(5)
puts result.inspect

medium

A. [3, 6, 9, 12, 15]

B. [6, 12, 18, 24, 30]

C. [2, 4, 6, 8, 10]

D. RuntimeError: infinite loop

Solution

1. Step 1: Understand the lazy chain

   The code creates an infinite range from 1 upwards, then lazily selects numbers divisible by 3, then doubles them, and finally takes the first 5 results.

2. Step 2: Calculate the first 5 values

   Numbers divisible by 3 starting from 1 are 3, 6, 9, 12, 15. Doubling each gives 6, 12, 18, 24, 30.

3. Final Answer:

   [6, 12, 18, 24, 30] -> Option B

4. Quick Check:

   Lazy select multiples of 3, map *2, first 5 = B [OK]

Hint: Trace lazy steps and apply first(n) to get output [OK]

Common Mistakes:

• Forgetting to double the numbers after selection
• Confusing selected numbers with original range
• Expecting an error from infinite range without lazy

4. Identify the error in this Ruby code using lazy enumerators:

numbers = (1..10).lazy.map { |n| n * 2 }
puts numbers.take(5)

medium

A. The take method should be called with .to_a to get results.

B. The code will run fine and print the first 5 doubled numbers.

C. The code will raise a syntax error because lazy is not defined.

D. Calling take returns an Enumerator, not an Array, causing puts to raise an error.

Solution

1. Step 1: Understand what take returns on a lazy enumerator

   Calling take(n) on a lazy enumerator returns another lazy enumerator, not an array.

2. Step 2: How to get actual values

   To get an array of results, you must call to_a after take(5), like numbers.take(5).to_a.

3. Final Answer:

   The take method should be called with .to_a to get results. -> Option A

4. Quick Check:

   Use take(n).to_a to get array from lazy enumerator [OK]

Hint: Add .to_a after take(n) to get array output [OK]

Common Mistakes:

• Expecting take(n) alone to return an array
• Confusing lazy enumerator with normal enumerator
• Assuming lazy is undefined or causes syntax error

5. Given a large array arr = (1..1000000).to_a, which Ruby code snippet efficiently finds the first 3 even numbers greater than 500000 using lazy enumerators?

hard

A. arr.lazy.select { |x| x > 500000 && x.even? }.first(3)

B. arr.select { |x| x > 500000 && x.even? }.first(3)

C. arr.lazy.map { |x| x * 2 }.first(3)

D. arr.take(3).lazy.select(&:even?)

Solution

1. Step 1: Understand the goal and data size

   We want the first 3 even numbers greater than 500,000 from a large array. Using lazy enumerators avoids processing the entire array.

2. Step 2: Analyze each option

   arr.lazy.select { |x| x > 500000 && x.even? }.first(3) uses lazy, filters with select for the condition, then takes first 3 - efficient and correct.
   arr.select { |x| x > 500000 && x.even? }.first(3) processes entire array eagerly, inefficient.
   arr.lazy.map { |x| x * 2 }.first(3) maps all elements, not filtering by condition.
   arr.take(3).lazy.select(&:even?) takes first 3 elements then filters, which misses numbers > 500,000.

3. Final Answer:

   arr.lazy.select { |x| x > 500000 && x.even? }.first(3) -> Option A

4. Quick Check:

   Lazy select with condition then first(3) = C [OK]

Hint: Use lazy select with condition, then first(n) for efficiency [OK]

Common Mistakes:

• Using eager select causing slow processing
• Filtering after taking first 3 elements incorrectly
• Using map instead of select for filtering