Ruby中已排序元素的原始索引

Question

arr = [1,3,2,4]

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

我想要一个数组[0, 2, 1, 3]（arr.sort顺序中的原始索引）

使用Ruby 1.9.3有一种简单的方法吗？

谢谢

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)}

为避免每次排序，更好的是：

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

---

更新

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

得到了结果：

0.281736
0.504314

Answer 3

我测试了MRI Ruby 2.2.1p85（在Mac和CentOS上），tokland的解决方案返回错误的结果：

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

Yevgeniy Anfilofyev解决方案有效，但不支持非唯一数组：

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]

我是这样的：

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

此解决方案适用于O（nlogn）

Answer 5

(0..arr.size - 1).sort_by { |i| arr[i] }