Originals of indexes on sorted elements in Ruby

Question

arr = [1,3,2,4] arr.sort #=> [1,2,3,4] 

I need an array [0, 2, 1, 3] (source indexes in arr.sort order)

Is there an easy way to do this with Ruby 1.9.3?

Thank you

Answer 1

 xs = [1, 3, 2, 4] original_indexes = xs.map.with_index.sort.map(&:last) #=> [0, 2, 1, 3] 

Answer 2

 arr=[1,3,2,4] p arr.map{|e| arr.sort.index(e)} 

To avoid sorting every time, it is better:

 arr=[1,3,2,4] arr_s = arr.sort p arr.map{|e| arr_s.index(e)} 

---

UPDATED

 arr=[1,3,2,4] start_time = Time.now (1..100000).each do |i| arr.map{|e| arr.sort.index(e)} end elapsed = Time.now - start_time p elapsed xs = [1, 3, 2, 4] start_time = Time.now (1..100000).each do |i| xs.map.with_index.sort.map(&:last) end elapsed = Time.now - start_time p elapsed 

and got the result:

 0.281736 0.504314 

Answer 3

I tested MRI Ruby 2.2.1p85 (on both Mac and CentOS), the tokland solution returns the wrong result:

 xs = [8,3,2,7,5] xs.map.with_index.sort.map(&:last) #=> [2, 1, 4, 3, 0] # wrong 

Evgeny Anfilofeev's solution works, but does not support an unhistorical array:

 arr = [8,3,2,7,5] arr_s = arr.sort arr.map{|e| arr_s.index(e)} #=> [4, 1, 0, 3, 2] # correct arr = [8,3,5,2,8,8,7,5] arr_s = arr.sort arr.map{|e| arr_s.index(e)} #=> [5, 1, 2, 0, 5, 5, 4, 2] 

I approach this:

 arr = [8,3,5,2,8,8,7,5] index_order = [] arr.uniq.sort.each do |a| index_order += arr.each_index.select{|i| arr[i] == a } end r = [] index_order.each_with_index do |a, i| r[a] = i end r #=> [5, 1, 2, 0, 6, 7, 4, 3] 

Answer 4

 array = [6, 20, 12, 2, 9, 22, 17] sorted = array.sort indices = [] array.each do |n| index = (0...sorted.length).bsearch { |x| n <=> sorted[x] } indices << index end indices 

This solution works with O (nlogn)