Question

我有一个学生列表，我从数据库中获取它。我希望，根据学生的年龄来构建该列表，并最终得到这样的列表：

[
  30 => [
    "id1" => "Name1",
    "id2" => "Name2",
  ],
  31 => [
    "id5" => "Name3",
    "id6" => "Name4",
  ]
]

在我的代码中，我收到了这样的学生名单：

List<ApplicantModel> applicants = applicantDao.getApplicantsByApplicationStatus(applicantTypeId);

我想要在“for”中创建集合：

for(int i=0;i<=applicants.size()-1;i++){
    //code here to create the collection with the above structure 
}

或者，如果您有其他更好的建议？谢谢！

Answer 1

您可以使用TreeMap（根据年龄和ID的顺序）：它可能看起来像这样:(我不知道获取年龄和id的方法的确切名称 - 所以你可能需要调整getAge和getId方法调用）：

import java.util.LinkedList;
import java.util.List;
import java.util.TreeMap;

public class ApplicantStructureTest {

    public static void main(String[] args) {
        List<ApplicantModel> applicants = new LinkedList<>();
        applicants.add(new ApplicantModel("id1", 30, "name1"));
        applicants.add(new ApplicantModel("id2", 30, "name2"));
        applicants.add(new ApplicantModel("id5", 31, "name3"));
        applicants.add(new ApplicantModel("id6", 31, "name4"));

        // here is your for loop
        TreeMap<Integer, TreeMap<String, String>> structured = new TreeMap<Integer, TreeMap<String, String>>();
        for (int i = 0; i <= applicants.size() - 1; i++) {
            ApplicantModel applicant = applicants.get(i);
            Integer age = applicant.getAge();
            TreeMap<String, String> ageMap = structured.get(age);
            if (ageMap == null) {
                ageMap = new TreeMap<String, String>();
                structured.put(age, ageMap);
            }
            ageMap.put(applicant.getId(), applicant.getName());
        }


        System.out.println(structured);
    }

    public static class ApplicantModel {
        private Integer age;
        private String id;
        private String name;

        public ApplicantModel(String id, Integer age, String name) {
            this.id = id;
            this.age = age;
            this.name = name;
        }

        public String getName() {
            return name;
        }

        public Integer getAge() {
            return age;
        }

        public String getId() {
            return id;
        }

    }

}

此代码将打印：

{30 = {id1 = name1，id2 = name2}，31 = {id5 = name3，id6 = name4}}

Answer 2

您还可以在地图中创建结构并列出：

Map<Integer,List<ApplicantModel>> structured = new HashMap<Integer, List<ApplicantModel>>();
for(int i = 0; i <= applicants.size() - 1; i++){
  ApplicantModel applicant = applicants.get(i);

  Integer age = applicant.getAge();
  List<ApplicantModel> list = structured.get(age);
  if (list == null) {
    list = new  List<ApplicantModel>();
    structured.put(age, list);
  }
  list.add(applicant);
}

Answer 3

类似于Krzysztof Cichocki获得的结果也可以通过流（Java 8或更高版本）获得：

    Map<Integer, Map<String, String>> byAge = studentList.stream()
            .collect(Collectors.groupingBy(ApplicantModel::getAge, 
                                           Collectors.toMap(ApplicantModel::getId, ApplicantModel::getName)));

这会产生（我承认不是读者友好的格式）：

{30={id2=name2, id1=name1}, 31={id6=name4, id5=name3}}

如果您特别想要TreeMap，请使用三参数groupingBy()和/或四参数toMap()：

    Map<Integer, Map<String, String>> byAge = studentList.stream()
            .collect(Collectors.groupingBy(ApplicantModel::getAge, 
                                           TreeMap::new, 
                                           Collectors.toMap(ApplicantModel::getId,
                                                            ApplicantModel::getName,
                                                            (u, v) -> { throw new IllegalArgumentException(); },
                                                            TreeMap::new)));

现在元素从toString()：

排序

{30={id1=name1, id2=name2}, 31={id5=name3, id6=name4}}

集合 - 向集合添加元素

3 个答案: