Order an array so that elements can be floated across columns

column_order.rb

# There is definitely an easier way to do this, but I'm spacing on what it is. >.<

# Takes an ordered list and re-orders it so elements can be css-floated and still appear in order.
# e.g. helper.column_order([1,2,3,4], 2) would produce [1,3,2,4] so elements appear on the page as |1 3|
#                                                                                                  |2 4|
def column_order(list, number_of_columns)
  remainder = list.length % number_of_columns
  number_of_rows = list.length / number_of_columns + [remainder, 1].min
  reordered_list = []

  if remainder > 0
    1.upto(number_of_columns - remainder - 1) { |i| list.insert(-i * number_of_columns, nil) } # pad the original array with nils
  end

  list.each_with_index do |item, index|
    row = index % number_of_rows
    reordered_list[row] ||= []
    reordered_list[row] << item
  end

  reordered_list.flatten.compact
end

#====== SPECS

describe '#column_order' do
  it 're-orders a list so it can be flowed horizontally but appear as vertical columns' do
    # | 1 3 |
    # | 2 4 |
    expect(helper.column_order(['1','2','3','4'], 2)).to eq(['1','3','2','4'])
  end

  it 'also works with 3 columns' do
    # | 1 3 5 |
    # | 2 4 6 |
    expect(helper.column_order(['1','2','3','4','5','6'], 3)).to eq(['1','3','5','2','4','6'])
  end

  it 'also works as a 3 x 3' do
    # | 1 4 7 |
    # | 2 5 8 |
    # | 3 6 9 |
    expect(helper.column_order(['1','2','3','4','5','6','7','8','9'], 3)).to eq(['1','4','7','2','5','8','3','6','9'])
  end

  it 'works with list lengths that do not divide into even columns (remainder 1)' do
    # | 1 4 6 |
    # | 2 5 7 |
    # | 3     |
    expect(helper.column_order(['1','2','3','4','5','6','7'], 3)).to eq(['1','4','6','2','5','7','3'])
  end

  it 'works with list lengths that do not divide into even columns (remainder 2)' do
    # | 1 4 7 |
    # | 2 5 8 |
    # | 3 6   |
    expect(helper.column_order(['1','2','3','4','5','6','7','8'], 3)).to eq(['1','4','7','2','5','8','3', '6'])
  end
end