Group_by for categorization in Ruby - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the group_by method do in Ruby?

It groups elements of a collection based on the result of a block, returning a hash where keys are the group criteria and values are arrays of elements in each group.

beginner

How do you use group_by to categorize numbers as even or odd?

You pass a block that returns :even or :odd based on each number. For example:
numbers.group_by { |n| n.even? ? :even : :odd }

beginner

What type of object does group_by return?

It returns a Hash where each key is a group label and each value is an array of elements belonging to that group.

intermediate

Can group_by be used on arrays of strings? Give an example.

Yes! For example, grouping words by their first letter:
words.group_by { |word| word[0] } groups words by their starting character.

beginner

What happens if the block passed to group_by returns the same value for multiple elements?

All those elements are grouped together under the same key in the resulting hash.

What does group_by return when called on an array?

A An array sorted by the block's return value

B A hash grouping elements by the block's return value

C A single element from the array

D A boolean indicating if grouping was successful

group_by returns a hash where keys are group labels and values are arrays of grouped elements.

How would you group an array of numbers into 'positive' and 'negative' using group_by?

A numbers.group_by { |n| n > 0 ? :positive : :negative }

B numbers.group_by(&:positive?)

C numbers.group_by { |n| n.abs }

D numbers.group_by { |n| n < 0 }

Option A correctly uses a block returning symbols :positive or :negative based on the number's sign.

If you group words by their length using group_by, what will the keys of the resulting hash be?

A The length of each word

B The first letter of each word

C The words themselves

D The last letter of each word

Grouping by length means the keys are integers representing word lengths.

What happens if the block passed to group_by returns nil for some elements?

A Those elements are ignored

B Those elements are grouped under an empty string key

C An error is raised

D Those elements are grouped under the <code>nil</code> key

Elements with block result nil are grouped under the nil key in the hash.

Which of these is a valid use of group_by?

A [1,2,3,4].group_by

B [1,2,3,4].group_by(2)

C [1,2,3,4].group_by { |n| n % 2 }

D [1,2,3,4].group_by(&:to_s)

Options C and D are valid uses of group_by: C uses a block to group numbers by remainder when divided by 2 (even/odd), and D uses a symbol to proc to group by string representation.

Explain how the group_by method works in Ruby and give a simple example.

Think about how you can sort items into labeled boxes based on a rule.

Describe a real-life situation where you could use group_by to organize data.

Imagine sorting your books or clothes into categories.

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