根据他们的名字对他们的分数进行排名

时间:2018-05-06 06:33:18

标签: java arrays sorting

我想根据他们的分数对一些替代品进行排名,并打印每个替代品的名称及其排名,我该怎么做?

这是MWE:

import java.util.Arrays;
public class Ranking {
public static void main(String[] args) {

//The score of the alternatives :
double [] score = new double [4] ;
score[0] = 6.75 ;
score[1] = 9.0 ;
score[2] = 6.75 ;
score[3] = 5.50;

//The name of the alternatives :
String [] name = new String [4] ;
name[0] = "a1";
name[1] = "a2";
name[2] = "a3";
name[3] = "a4";

//Descending Sorting the score array like this :
for(int i=0;i<4;i++)
score[i]= - score[i];

Arrays.sort(score);

for(int i=0;i<4;i++)
score[i]= - score[i];

//print the result
for(int i=0;i<4;i++)
System.out.println(score[i] + " rank = " + (i+1));
}
//result :
//9.0 rank = 1
//6.75 rank = 2
//6.75 rank = 3
//5.5 rank = 4

但我想要这样的结果:

name : a2 a1 a3 a4
rank : 1  2  3  4

我该怎么做?

3 个答案:

答案 0 :(得分:1)

您对单个数组进行排序,因此无法在名称 - 等级之间映射结果。

在java中,通常的方法是将数据抽象为对象并对其进行操作:

声明你的模型类:

public YourModel {
  private String name;
  private double score;

  public Rank(String name, double score) {
    this.name = name;
    this.score = score;
  }
  // you create your getters/ setters
}

向数组中添加数据:

YourModel[] models = new YourModel[4];
models[0] = new YourModel("a1", 6.75D);

...添加其他型号......

对数组进行排序:

Arrays.sort(models, Comparator.comparing(YourModel::getScore));

完成此步骤后,您将对数组进行排序:

[{"a2", 9.0} ....]

您可以打印结果:

for(int i=0;i<4;i++)
   System.out.println(models[i].getName() + " rank = " + (i+1));
}

答案 1 :(得分:1)

您应该使用Map<String, Double>

尝试以下示例:

Map<String, Double> map = new HashMap<>();
      map.put("a1", 6.75);
      map.put("a2", 9.0);
      map.put("a3", 6.8);

      map = map.entrySet().stream()
      .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
      .collect(Collectors.toMap(
         Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));

//Iterate through the map

您将以反向排序的顺序得到答案。

.sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))用于按值排序,LinkedHashMap::new用于维护Map的排序

答案 2 :(得分:1)

当您必须对某些内容进行排序时,您应该查看TreeSetTreeMap结构。实际上Array不是排序是主要功能的地方。

您似乎对所有对象都有唯一的names,那么您必须将TreeSet与reverce比较器一起使用。因此,当您遍历Set时,您将获得所需的订单:

final class Model {

    private final String name;
    private final double score;

    public Model(String name, double score) {
        this.name = name;
        this.score = score;
    }
}

final Comparator<Model> reverceScore = (d1, d2) -> {
    if (d1.name.equalsIgnoreCase(d2.name))
        return 0;

    int res = Double.compare(d2.score, d1.score);
    return res != 0 ? res : d1.name.compareTo(d2.name);
};

Set<Model> data = new TreeSet<>(reverceScore);
data.add(new Model("a1", 6.75));
data.add(new Model("a2", 9));
data.add(new Model("a3", 6.75));
data.add(new Model("a4", 5.5));

int rank = 1;

for(Model model : data)
    System.out.println("rank: " + rank++ + " - name: " + model.name + ", score: " + model.score);