Group_by for categorization in Ruby - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to group numbers by their parity (even or odd).

Ruby

numbers = [1, 2, 3, 4, 5]
groups = numbers.group_by { |n| n.[1] }
puts groups

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

Ato_s

Bodd?

Cabs

Deven?

Attempts:

3 left

💡 Hint

Common Mistakes

Using odd? groups by odd numbers, which is not what the question asks.

Using to_s or abs does not categorize by parity.

The even? method returns true for even numbers, so grouping by n.even? categorizes numbers into even and odd groups.

2fill in blank

medium

Complete the code to group words by their first letter.

Ruby

words = ['apple', 'banana', 'apricot', 'blueberry']
groups = words.group_by { |word| word.[1] }
puts groups

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

Afirst

Blength

Cupcase

Dreverse

Attempts:

3 left

💡 Hint

Common Mistakes

Using length groups by word length, not first letter.

Using upcase returns the whole word in uppercase, not the first letter.

The first method returns the first character of a string, so grouping by word.first groups words by their first letter.

3fill in blank

hard

Fix the error in the code to group numbers by their remainder when divided by 3.

Ruby

numbers = [1, 2, 3, 4, 5, 6]
groups = numbers.group_by { |n| n [1] 3 }
puts groups

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

A/

B%

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using / divides numbers but does not give remainder.

Using * or - does not relate to remainder.

The modulo operator % gives the remainder of division, so n % 3 groups numbers by their remainder when divided by 3.

4fill in blank

hard

Fill both blanks to group words by their length and select only words longer than 3 letters.

Ruby

words = ['cat', 'dog', 'elephant', 'bee', 'ant']
groups = words.group_by { |word| word.[1] }
filtered = groups.select { |length, words| length [2] 3 }
puts filtered

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

Alength

B>

C<

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using size is also valid for length but only one correct answer is accepted here.

Using < selects shorter words, not longer.

Grouping by word.length groups words by their length. Selecting where length is greater than 3 filters only longer words.

5fill in blank

hard

Fill all three blanks to group numbers by their sign and select only positive numbers.

Ruby

numbers = [-2, -1, 0, 1, 2]
groups = numbers.group_by { |n| n.[1] }
positive = groups.select { |sign, nums| sign == [2] }
puts positive[[3]]

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

Anegative?

Btrue

Cfalse

Dpositive?

Attempts:

3 left

💡 Hint

Common Mistakes

Using negative? groups negative numbers, not positive.

Using false selects non-positive numbers.

The positive? method returns true for positive numbers. Selecting groups where sign is true filters positive numbers. Accessing positive[true] gets the array of positive numbers.

Practice

1. What does the Ruby method group_by do when called on an array?

easy

A. It converts the array into a string.

B. It groups elements into a hash based on a given condition.

C. It removes duplicate elements from the array.

D. It sorts the array in ascending order.

Solution

1. Step 1: Understand the purpose of group_by

   The group_by method organizes elements by a condition and returns a hash.

2. Step 2: Compare with other array methods

   Sorting, removing duplicates, or converting to string are different methods like sort, uniq, or join.

3. Final Answer:

   It groups elements into a hash based on a given condition. -> Option B

4. Quick Check:

   group_by = groups hash [OK]

Hint: Remember: group_by returns a hash of grouped elements [OK]

Common Mistakes:

• Confusing group_by with sort or uniq
• Expecting an array instead of a hash
• Thinking it changes element order only

2. Which of the following is the correct syntax to group an array nums by even or odd using group_by?

easy

A. nums.group_by(n % 2 == 0)

B. nums.group_by do n % 2 == 0 end

C. nums.group_by { |n| n % 2 == 0 }

D. nums.group_by ->(n) { n % 2 == 0 }

Solution

1. Step 1: Recall block syntax for group_by

   The method requires a block with curly braces or do-end and a block variable.

2. Step 2: Check each option's syntax

   nums.group_by { |n| n % 2 == 0 } uses correct block with { |n| ... }. nums.group_by(n % 2 == 0) misses block. nums.group_by do n % 2 == 0 end lacks block variable. nums.group_by ->(n) { n % 2 == 0 } uses lambda incorrectly.

3. Final Answer:

   nums.group_by { |n| n % 2 == 0 } -> Option C

4. Quick Check:

   Correct block syntax = nums.group_by { |n| n % 2 == 0 } [OK]

Hint: Use curly braces with block variable for group_by [OK]

Common Mistakes:

• Omitting block variable in block
• Passing condition as argument instead of block
• Using lambda syntax incorrectly

3. What is the output of this Ruby code?

words = ['apple', 'banana', 'avocado', 'blueberry']
groups = words.group_by { |w| w[0] }
puts groups

medium

A. [["a", ["apple", "avocado"]], ["b", ["banana", "blueberry"]]]

B. ["apple", "avocado", "banana", "blueberry"]

C. {"apple"=>"a", "banana"=>"b"}

D. {"a"=>["apple", "avocado"], "b"=>["banana", "blueberry"]}

Solution

1. Step 1: Understand group_by with first letter

   The block groups words by their first character, so keys are 'a' and 'b'.

2. Step 2: Check output format of group_by

   It returns a hash with keys as letters and values as arrays of words starting with that letter.

3. Final Answer:

   {"a"=>["apple", "avocado"], "b"=>["banana", "blueberry"]} -> Option D

4. Quick Check:

   group_by first char = hash with arrays [OK]

Hint: group_by returns a hash, not an array or string [OK]

Common Mistakes:

• Expecting an array instead of a hash
• Confusing keys and values in output
• Misreading output as string instead of hash

4. Identify the error in this Ruby code snippet:

numbers = [1, 2, 3, 4]
groups = numbers.group_by do n
  n.even?
end
puts groups

medium

A. Missing block variable in the do-end block.

B. group_by cannot be used with do-end blocks.

C. The method even? does not exist for integers.

D. The array must be sorted before using group_by.

Solution

1. Step 1: Check block syntax for group_by

   The block must have a variable to represent each element, like do |n| ... end.

2. Step 2: Identify the missing block variable

   The code uses do n without vertical bars, so n is treated as a local variable, causing error.

3. Final Answer:

   Missing block variable in the do-end block. -> Option A

4. Quick Check:

   Block variable required in do-end [OK]

Hint: Always use |variable| in do-end blocks [OK]

Common Mistakes:

• Omitting vertical bars around block variable
• Thinking even? is undefined for integers
• Believing group_by needs sorted array

5. Given an array of hashes representing people:

people = [
  {name: 'Alice', age: 30},
  {name: 'Bob', age: 20},
  {name: 'Carol', age: 30},
  {name: 'Dave', age: 20}
]

Which code correctly groups people by their age using group_by?

hard

A. people.group_by { |person| person[:age] }

B. people.group_by { |person| person.name }

C. people.group_by { |person| person['age'] }

D. people.group_by { |person| person.age }

Solution

1. Step 1: Understand data structure and access

   Each element is a hash with symbol keys like :age, so access with person[:age].

2. Step 2: Check each option's key access

   people.group_by { |person| person[:age] } uses correct symbol key access. people.group_by { |person| person.name } and D use dot notation which doesn't work on hashes. people.group_by { |person| person['age'] } uses string key which is incorrect here.

3. Final Answer:

   people.group_by { |person| person[:age] } -> Option A

4. Quick Check:

   Hash keys accessed by symbols = people.group_by { |person| person[:age] } [OK]

Hint: Use symbol keys with [:key] to access hash values [OK]

Common Mistakes:

• Using dot notation on hashes
• Using string keys instead of symbols
• Confusing array of hashes with objects