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.