How to get the nth element of an Enumerable in Ruby

Question

For example, to return the 10,000th prime number I could write:

require \'prime\'
Prime.first(10000).last #=> 104729

But creating a huge

Answer 1

Based on sawa's solution, here's a more succinct alternative:

Prime.find.with_index(1) { |_, i| i == 10000 }
#=> 104729

or

Prime.find.with_index { |_, i| i == 9999 }
#=> 104723

Answer 2

The closest thing I can think of to a hypothetical at method is drop, which skips the indicated number of elements. It tries to return an actual array though so you need to combine it with lazy if you are using with infinite sequences, e.g.

Prime.lazy.drop(9999).first

Answer 3

What about this?

Prime.to_enum.with_index(1){|e, i| break e if i == 10000}
# => 104729

For enumerators that are not lazy, you probably want to put lazy on the enumerator.

Answer 4

module Enumerable
  def at(n)
    enum = is_a?(Enumerator) ? self : each
    (n-1).times { enum.next }
    enum.next
  end
end

(0..4).at(3)
  #=> 2 
{ a:1, b:2, c:3, d:4, e:5 }.at(3)
  #=> [:c, 3] 
'0ab_1ab_2ab_3ab_4ab'.gsub(/.(?=ab)/).at(3)
  #=> "2" 
require 'prime'; Prime.at(10000)
  #=> 104729 
e = 1.step; e.at(10)
  #=> 10 
[0,1,2,3,4].at(6)
  #=> nil 
'0ab_1ab_2ab_3ab_4ab'.gsub(/.(?=ab)/).at(6)
  #=> StopIteration (iteration reached an end)

Note that '0ab_1ab_2ab_3ab_4ab'.gsub(/.(?=ab)/).class #=> Enumerator.

Some refinements would be needed, such as checking that n is an integer greater than zero and improving the exception handling, such as dealing with the case when self is not an enumerator but has no method each.