Ruby select by index

Question

I am trying to select elements out of an array:

arr = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n']

whose index is a Fibonacci number. I want the result:

['a', 'b', 'c', 'd', 'f', 'i', 'n']

My code returns both the element and the index.

def is_fibonacci?(i, x = 1, y = 0)
  return true if i == x || i == 0
  return false if x > i
  is_fibonacci?(i, x + y, x)
end

arr.each_with_index.select do |val, index|
  is_fibonacci?(index)
end

This code returns:

[["a", 0], ["b", 1], ["c", 2], ["d", 3], ["f", 5], ["i", 8], ["n", 13]]

Please help me understand how I can still iterate through the array and evaluate the index but only return the element.

Answer 1

You can change the last bit of your code to

arr.select.with_index do |val, index|
  is_fibonacci?(index)
end

This works because if you call a method such as select without a block, you get an Enumerator object, on which you can then chain more Enumerable methods.

In this case I've used with_index, which is very similar to calling each_with_index on the original array. However because this happens after the select instead of before, select returns items from the original array, without the indices appended

Answer 2

Your code seems great so far, I wouldn't change it. You can go over your results after the fact and change the [element, index] pairs to only contain the element by mapping over each pair and only taking the first:

>> results = [["a", 0], ["b", 1], ["c", 2], ["d", 3], ["f", 5], ["i", 8], ["n", 13]]
>> results.map(&:first)
=> ["a", "b", "c", "d", "f", "i", "n"]