根据另一个数组的元素对数组进行排序 [英] Sort an array according to the elements of another array

问题描述

我有一组 id

a1 = [1, 2, 3, 4, 5]  

我有另一个随机排列的 id 对象数组

and I have another array of objects with ids in random order

a2 = [(obj_with_id_5), (obj_with_id_2), (obj_with_id_1), (obj_with_id_3), (obj_with_id_4)]  

现在我需要根据a1中id的顺序对a2进行排序.所以 a2 现在应该变成:

Now I need to sort a2 according to the order of ids in a1. So a2 should now become:

[(obj_with_id_1), (id_2), (id_3), (id_4), (id_5)]  

a1 可能是 [3, 2, 5, 4, 1] 或任何顺序，但 a2 应该对应于 a1 中 id 的顺序.

a1 might be [3, 2, 5, 4, 1] or in any order but a2 should correspond to the order of ids in a1.

我喜欢这个:

a1.each_with_index do |id, idx|
  found_idx = a1.find_index { |c| c.id == id }
  replace_elem = a2[found_idx]
  a2[found_idx] = a2[idx]
  a2[idx] = replace_elem
end  

但是如果 a2 的元素顺序与 a1 完全相反，这仍然可能会遇到 O(n^2) 时间.有人可以告诉我排序 a2 的最有效方法吗?

But this still might run into an O(n^2) time if order of elements of a2 is exactly reverse of a1. Can someone please tell me the most efficient way of sorting a2?

推荐答案

hash_object = objects.each_with_object({}) do |obj, hash| 
  hash[obj.object_id] = obj
end

[1, 2, 3, 4, 5].map { |index| hash_object[index] }
#=> array of objects in id's order

我相信运行时间将是 O(n)

I believe that the run time will be O(n)

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