从ruby哈希值数组中获取前n个元素

Hey I have an array where each element is a hash containing a few values and a count.

嘿，我有一个数组，每个元素都是一个哈希，包含一些值和一个计数。

result = [
           {"count" => 3,"name" => "user1"}, 
           {"count" => 10,"name" => "user2"}, 
           {"count" => 10, "user3"},
           {"count" => 2, "user4"}
         ]

I can sort the array by count as follows:

我可以按照如下方式对数组进行排序:

result = result.sort_by do |r|
  r["count"]
end

Now I want to be able to retrieve the top n entries based on count (not just first(n)) Is there an elegant way to do this? So as an example, let n = 1 I would expect a result set of.

现在，我希望能够根据count(不只是第一个(n))检索*n项，是否有一种优雅的方法来实现这一点?作为一个例子，让n = 1我期望结果集。

[{"count" => 10,"name" => "user2"}, {"count" => 10, "user3"}]

since I asked for all entries with the highest score.. if I asked for top 2 highest scores I'd get

因为我要求所有的分数都是最高的。如果我要求最高的2分，我将得到

 [{"count" => 10,"name" => "user2"}, {"count" => 10, "user3"}, {"count" => 3, "user1"}]

4 个解决方案

#1

Enumerable#group_by to the rescue (as usual):

可列举的#group_by救援(一如既往):

result.group_by { |r| r["count"] }
      .sort_by  { |k, v| -k }
      .first(2)
      .map(&:last)
      .flatten

Most of the work is done by the group_by. The sort_by simply lines things up so that first(2) will pick off the groups you want. Then map with last will extract the count/name hashes that you started with and the final flatten will clean up the extra left over arrays.

大部分工作是由group_by完成的。sort_by简单地将其排列起来，这样第一个(2)就会选出你想要的组。然后使用last的map将提取开始时的计数/名称散列，最后的flatten将清理剩下的数组。

#2

This solution is not elegant in terms of being concise, but it has better time complexity. In other words, it should execute a lot faster for a very large number of hashes.

这个解决方案在简洁方面并不优雅，但它具有更好的时间复杂度。换句话说，对于大量的散列，它应该执行得更快。

You will need to install the "algorithms" gem in order to use the Heap data structure:

需要安装“算法”gem才能使用堆数据结构:

Heaps are an efficient data structure when you need to find the largest or smallest elements in a group. This particular type of heap is optimal if the value of "n" is much smaller than the total number of pairs.

当您需要查找组中最大或最小的元素时，堆是一种有效的数据结构。如果“n”的值远远小于对的总数，那么这种特殊类型的堆是最优的。

require 'algorithms'
def take_highest(result,n)
  max_heap = Containers::Heap.new(result){|x,y| (x["count"] <=> y["count"]) == 1}
  last = max_heap.pop
  count = 0
  highest = [last]
  loop do   
    top = max_heap.pop
    break if top.nil?
    count += (top["count"] == last["count"] ? 0 : 1)
    break if count == n
    highest << top
    last = top
  end
  highest
end

#3

Starting in Ruby 2.2.0, max_by takes an extra argument that lets you ask for a certain number of top elements instead of just getting one. Using this, we can improve on mu is too short's answer

从Ruby 2.2.0开始，max_by采用了一个额外的参数，允许您请求一定数量的*元素，而不是只获得一个。利用这个，我们可以改进mu是太短的答案。

result = [
           {count: 3, name: 'user1'},
           {count: 10, name: 'user2'},
           {count: 10, name: 'user3'},
           {count: 2, name: 'user4'}
         ]
p result.group_by { |r| r[:count] }
      .max_by(2, &:first)
      .flat_map(&:last)
      .sort_by { |r| -r[:count] }
# => [{:count=>10, :name=>"user2"}, {:count=>10, :name=>"user3"}, {:count=>3, :name=>"user1"}]

The docs don't say if the array returned by max_by is sorted. If that turns out to be true though, we could just use reverse in the last step instead of sorting.

文档没有说明max_by返回的数组是否已排序。如果这是真的，我们可以在最后一步使用逆向排序而不是排序。

#4

new_result = result.
  sort_by { |r| -r["count"] }.
  chunk { |r| r["count"] }.
  take(2).
  flat_map(&:last)

#=> [{"count"=>10, "name"=>"user3"}, 
#    {"count"=>10, "name"=>"user2"}, 
#    {"count"=> 3  "name"=>"user1"}]

#1