505

What is the easiest way to convert

[x1, x2, x3, ... , xN]

to

[[x1, 2], [x2, 3], [x3, 4], ... , [xN, N+1]]

10 Answers 10

926

If you're using ruby 1.8.7 or 1.9, you can use the fact that iterator methods like each_with_index, when called without a block, return an Enumerator object, which you can call Enumerable methods like map on. So you can do:

arr.each_with_index.map { |x,i| [x, i+2] }

In 1.8.6 you can do:

require 'enumerator'
arr.enum_for(:each_with_index).map { |x,i| [x, i+2] }
9
  • Thanks! Could you give me a pointer to documentation for .each_with_index.map ? Commented Jan 15, 2011 at 1:41
  • 2
    @Misha: map is a method of Enumerable as always. each_with_index, when called without a block, returns an Enumerator object (in 1.8.7+), which mixes in Enumerable, so you can call map, select, reject etc. on it just like on an array, hash, range etc.
    – sepp2k
    Commented Jan 15, 2011 at 1:45
  • 14
    IMO this is simpler and better-reading in 1.8.7+: arr.map.with_index{ |o,i| [o,i+2] }
    – Phrogz
    Commented Jan 15, 2011 at 2:43
  • 4
    @Phrogz: map.with_index doesn't work in 1.8.7 (map returns an array when called without a block in 1.8).
    – sepp2k
    Commented Jan 15, 2011 at 2:50
  • 3
    Important to note this doesn't work with .map! if you want to directly affect the array you're looping on.
    – Ash Blue
    Commented Jul 25, 2013 at 17:38
307

Ruby has Enumerator#with_index(offset = 0), so first convert the array to an enumerator using Object#to_enum or Array#map:

[:a, :b, :c].map.with_index(2).to_a
#=> [[:a, 2], [:b, 3], [:c, 4]]
1
  • 15
    I believe this is the better answer, because it will work with map! foo = ['d'] * 5; foo.map!.with_index { |x,i| x * i }; foo #=> ["", "d", "dd", "ddd", "dddd"]
    – Connor
    Commented Feb 27, 2014 at 21:47
204

In ruby 1.9.3 there is a chainable method called with_index which can be chained to map.

For example:

array.map.with_index { |item, index| ... }
0
19

Over the top obfuscation:

arr = ('a'..'g').to_a
indexes = arr.each_index.map(&2.method(:+))
arr.zip(indexes)
1
  • 21
    Andrew must have great job security! :)
    – David J.
    Commented Jul 19, 2012 at 6:36
14

I have always enjoyed the syntax of this style:

a = [1, 2, 3, 4]
a.each_with_index.map { |el, index| el + index }
# => [1, 3, 5, 7]

Invoking each_with_index gets you an enumerator you can easily map over with your index available.

1
9

Here are two more options for 1.8.6 (or 1.9) without using enumerator:

# Fun with functional
arr = ('a'..'g').to_a
arr.zip( (2..(arr.length+2)).to_a )
#=> [["a", 2], ["b", 3], ["c", 4], ["d", 5], ["e", 6], ["f", 7], ["g", 8]]

# The simplest
n = 1
arr.map{ |c| [c, n+=1 ] }
#=> [["a", 2], ["b", 3], ["c", 4], ["d", 5], ["e", 6], ["f", 7], ["g", 8]]
4

A fun, but useless way to do this:

az  = ('a'..'z').to_a
azz = az.map{|e| [e, az.index(e)+2]}
4
  • 1
    Why the hate? It is a functioning way of doing this AND I even say that is is a silly way to achieve the results.
    – Automatico
    Commented Sep 9, 2014 at 11:13
  • 1
    the call to #index means this is now an O(N^2) loop also why the +2 ? :)
    – rogerdpack
    Commented Mar 13, 2017 at 16:40
  • 3
    As I write A fun, but useless way. +2 is to create the output the OP asks for
    – Automatico
    Commented Mar 13, 2017 at 18:04
  • Fun and useless is different from "really inefficient and poor programming practice" Commented Oct 23, 2021 at 23:30
3
a = [1, 2, 3]
p [a, (2...a.size+2).to_a].transpose
2
module Enumerable
  def map_with_index(&block)
    i = 0
    self.map { |val|
      val = block.call(val, i)
      i += 1
      val
    }
  end
end

["foo", "bar"].map_with_index {|item, index| [item, index] } => [["foo", 0], ["bar", 1]]
1
  • This might be an easier way to go for 1.8.6 and 1.8.7 (yes some of us still use it) instead of having to use weirder stuff like each_with_index.map etc. and even those of us on newer versions might prefer it to having to use map.with_index FWIW :)
    – rogerdpack
    Commented Mar 13, 2017 at 16:42
1

I often do this:

arr = ["a", "b", "c"]

(0...arr.length).map do |int|
  [arr[int], int + 2]
end

#=> [["a", 2], ["b", 3], ["c", 4]]

Instead of directly iterating over the elements of the array, you're iterating over a range of integers and using them as the indices to retrieve the elements of the array.

2
  • 1
    If you read the other answers, I hope you now realise there are better approaches. So not sure why you needed to add this.
    – nathanvda
    Commented Nov 28, 2014 at 12:05
  • If Andrew Grimm's answer deserves ten votes, then this one deserves at least one! Commented May 18, 2017 at 20:20

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