Java - 按规则从HashMap中获得前15名 [英] Java - get top 15 from HashMap by rule

问题描述

我想请求帮助。假设我有HashMap，其键值如下所示：

  String1 0.99 
 String2 0.98 
 String3 0.97 
 String4 0.98 
 String5 0.5 
 String6 0.4 
 String7 0.3 
等
  

我想用这个公式保存前15个条目： Math.abs（value - 0.5）。

这些数据的计算值（按公式计算）为：

  String1 0.49 
 String2 0.48 
 String3 0.47 
 String4 0.48 
 String5 0 
 String6 0.1 
 String7 0.2 
  

  String1 0.49 
 String4 0.48 
 String2 0.48 
 String3 0.47 
 String7 0.2 
 String6 0.1 
 String5 0 
  

现在我想拥有一个数组，其中索引将是原始值的顺序和值，如下所示：

  array [0] = 0.99 
 array [1] = 0.98 
 array [2] = 0.98 
 arra y [3] = 0.97 
 array [4] = 0.3 
 array [5] = 0.4 
 array [6] = 0.5 
  

感谢大家的帮助

任何帮助将不胜感激。谢谢

解决方案

第一个选项是迭代整个地图并找到您的值 - 可能将其放入数组中并使用O（n * log n）排序算法或15 * O（n）遍历它。

你可以放弃密钥唯一的条件吗？然后使用有序对（String，float）并编写一个客户 Comparator ，它在float值上实现您的规则，并使用 SortedSet 。

可能有涉及 BiMap 的解决方案，我不确定。 / p>

I would like to ask for help. Let's say I have HashMap with key-value like this:

String1 0.99
String2 0.98
String3 0.97
String4 0.98
String5 0.5
String6 0.4
String7 0.3
etc.

And I would like to save to array the top 15 entries by this formula: Math.abs(value - 0.5).

The counted values (by the formula) for this data would be:

String1 0.49
String2 0.48
String3 0.47
String4 0.48
String5 0
String6 0.1
String7 0.2

The values sorted

String1 0.49
String4 0.48
String2 0.48
String3 0.47
String7 0.2
String6 0.1
String5 0

And now I would like to have array, where index would be the order and value the original value, like this:

array[0] = 0.99
array[1] = 0.98
array[2] = 0.98
array[3] = 0.97
array[4] = 0.3
array[5] = 0.4
array[6] = 0.5 

Thanks everyone for help

Any help would be appreciated. Thank You

解决方案

First option is to iterate over the whole map and find your values - perhaps throw it into an array and use an O(n * log n) sort algorithm or 15*O(n) walks over it.

Can you drop the condition that the keys are unique? Then work with ordered pairs (String, float) and write a customer Comparator that implements your rule on the "float" value, and use a SortedSet.

There may be a solution involving BiMap, I'm not sure.

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