From: William James <w_a_x_man@...>
Date: 2007-09-18T00:30:03+09:00
Subject: Re: group array elements in groups of two

On Sep 17, 10:18 am, Emmanuel Oga <oga_emmanuel_...@yahoo.com.ar>
wrote:
> A better way to do this? :
>
> arr= [1, 2, 3, 4, 5, 6, 7, 8]
> new= []
> while !arr.empty?
>   elem1, elem2= arr.pop, arr.pop
>   new << [elem2, elem1];
> end
> new.reverse!
>
> new= [[1, 2], [3, 4], [5, 6], [7, 8]]

This question has come up several times.

require 'enumerator'
    ==>true
[1, 2, 3, 4, 5, 6, 7, 8].enum_slice(2).to_a
    ==>[[1, 2], [3, 4], [5, 6], [7, 8]]

Without 'require':

f=nil
    ==>nil
[1, 2, 3, 4, 5, 6, 7, 8].partition{f=!f}.transpose
    ==>[[1, 2], [3, 4], [5, 6], [7, 8]]