Ruby getting the diagonal elements in a 2d Array

Question

I was trying some problems with my 2D ruby array and my LOC reduces a lot when I do array slicing. So for example,

require "test/unit"

class LibraryTest < Test::Unit::TestCase

  def test_box
    array = [[1,2,3,4],[3,4,5,6], [5,6,7,8], [2,3,4,5]]
    puts array[1][2..3] # 5, 6
    puts array[1..2][1] # 5, 6, 7, 8
  end
end

I want to know if there is a way to get a diagonal slice? Lets say I want to start at [0,0] and want a diagonal slice of 3. Then I would get elements from [0,0], [1,1], [2,2] and I will get an array like [1,4,7] for example above. Is there any magic one-liner ruby code that can achieve this? 3.times do {some magic stuff?}

Answer 1

puts (0..2).collect { |i| array[i][i] }

Answer 2

Better might be a one-liner that utilizes the Matrix library:

require 'matrix'
Matrix.rows(array).each(:diagonal).to_a

Answer 3

I'm picking up @Shai's answer and propose to make it more functional.

First we initialize the array:

arr = [[1, 2, 3, 4], [3, 4, 5, 6], [5, 6, 7, 8], [2, 3, 4, 5]]

Then we prepare the array which serves as padding:

padding = [*0..(arr.length - 1)].map { |i| [nil] * i }
=> [[], [nil], [nil, nil], [nil, nil, nil]]

Then we apply the padding to the array. If you reverse the first usage of the padding or the second one depends if you want to get downward or upward diagonals.

padded = padding.reverse.zip(arr).zip(padding).map(&:flatten)
=> [[nil, nil, nil, 1, 2, 3, 4], [nil, nil, 3, 4, 5, 6, nil], [nil, 5, 6, 7, 8, nil, nil], [2, 3, 4, 5, nil, nil, nil]]

Then we transpose as in @Shai's solution:

padded.transpose.map(&:compact)
=> [[2], [5, 3], [3, 6, 4], [1, 4, 7, 5], [2, 5, 8], [3, 6], [4]]