Remove adjacent identical elements in a Ruby Array?

Question

Ruby 1.8.6

I have an array containing numerical values. I want to reduce it such that sequences of the same value are reduced to a single instance of that value.

Answer 1

Unless you are very concerned with the speed that block will calculate at, I would suggest you simply add this line to the end of your block to get the desired output:

a.compact!

That will just remove all the nil elements you introduced to the array earlier (the would-be duplicates), forming your desired output: [1, 2, 3, 2, 3]

If you want another algorithm, here is something far uglier than yours. :-)

require "pp"

a = [1, 1, 1, 2, 2, 3, 3, 3, 3, 2, 2, 2, 3, 3, 3]

i = 0

while i < a.size do
  e = a[i]
  j = i

  begin
    j += 1
  end while e == a[j]

  for k in i+1..j-1 do
    a[k] = nil
  end

  i = j
end

pp a
a.compact!
pp a

Gives you the output:

[1, nil, nil, 2, nil, 3, nil, nil, nil, 2, nil, nil, 3, nil, nil]
[1, 2, 3, 2, 3]

In my opinion, your code is fine. Just add the a.compact! call and you are sorted.