Group_by for categorization in Ruby - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of group_by with string length

What is the output of this Ruby code using group_by to categorize words by their length?

Ruby

words = ['cat', 'dog', 'bird', 'fish']
groups = words.group_by { |word| word.length }
puts groups

A {"cat"=>[3], "dog"=>[3], "bird"=>[4], "fish"=>[4]}

B {"cat"=>3, "dog"=>3, "bird"=>4, "fish"=>4}

C [[3, ["cat", "dog"]], [4, ["bird", "fish"]]]

D {3=>["cat", "dog"], 4=>["bird", "fish"]}

💡 Hint

Remember, group_by returns a hash where keys are the result of the block and values are arrays of elements.

The group_by method groups elements by the value returned from the block. Here, words are grouped by their length, so keys are lengths and values are arrays of words with that length.

🧠 Conceptual

intermediate

Understanding group_by return type

What type of object does Ruby's group_by method return when called on an array?

A A hash with keys as group criteria and values as arrays of grouped elements

B A single array with grouped elements concatenated

C An array of arrays

D A string representation of grouped elements

💡 Hint

Think about how you would access groups by their category after using group_by.

group_by returns a hash where each key is the result of the block for grouping, and each value is an array of elements that belong to that group.

🔧 Debug

advanced

Fix the error in group_by usage

This Ruby code tries to group numbers by even or odd but raises an error. What is the cause?

Ruby

numbers = [1, 2, 3, 4]
groups = numbers.group_by do |n|
  if n % 2 == 0
    'even'
  else
    'odd'
  end
end
puts groups

A No error, code runs correctly

B TypeError because group_by expects a symbol argument

C SyntaxError due to missing 'end' for the block

D NameError because 'groups' is undefined

💡 Hint

Check if all blocks and conditionals are properly closed with end.

The block passed to group_by has an if statement but is missing the closing end for the block itself, causing a syntax error.

❓ Predict Output

advanced

Output of group_by with nested arrays

What is the output of this Ruby code that groups nested arrays by their first element?

Ruby

pairs = [[1, 'a'], [2, 'b'], [1, 'c'], [2, 'd']]
groups = pairs.group_by { |pair| pair[0] }
puts groups

A {1=>[[1, "a"], [1, "c"]], 2=>[[2, "b"], [2, "d"]]}

B {[1, "a"]=>1, [2, "b"]=>2, [1, "c"]=>1, [2, "d"]=>2}

C [[1, [[1, "a"], [1, "c"]]], [2, [[2, "b"], [2, "d"]]]]

D {1=>["a", "c"], 2=>["b", "d"]}

💡 Hint

Remember that group_by keeps the original elements grouped by the key returned from the block.

The code groups the pairs by the first element of each pair. The result is a hash with keys 1 and 2, and values are arrays of pairs starting with those keys.

🚀 Application

expert

Count words by their first letter using group_by

Given an array of words, which code correctly uses group_by to count how many words start with each letter?

A words.group_by(&:first).map { |k, v| [k, v.size] }

B words.group_by { |w| w[0] }.map { |k, v| [k, v.length] }.to_h

C words.group_by { |w| w[0] }.transform_values(&:count)

D words.group_by { |w| w[0] }.count

💡 Hint

After grouping, you need to convert the grouped hash to a hash of counts.

Option B groups words by their first letter, then maps each group to a pair of letter and count, finally converts the array of pairs back to a hash.

Option B uses transform_values(&:count) but transform_values requires Ruby 2.4 or later.

Option B uses &:first which works for strings, but returns an array of [letter, count] pairs instead of a hash.

Option B calls count on the grouped hash, which returns the number of keys, not counts per group.

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