java - Java中对象的聚合列表

Question

我们在 Java 中是否有任何聚合器函数来执行以下聚合？

Person {
    String name;
    String subject;
    String department;
    Long mark1;
    Long mark2;
    Long mark3;
}

列表包含如下数据。

Name    |Subject    |Department |Mark1  |Mark2  |Mark3
--------|-----------|-----------|-------|-------|-----
Clark   |English    |DEP1       |7      |8      |6
Michel  |English    |DEP1       |6      |4      |7
Dave    |Maths      |DEP2       |3      |5      |6
Mario   |Maths      |DEP1       |9      |7      |8

The aggregation criteria is Subject & Dep. The resultant object needs to be

Subject     |Department |Mark1  |Mark2  |Mark3
----------- |-----------|-------|-------|-----
English     |DEP1       |13     |12     |13
Maths       |DEP2       |3      |5      |6
Maths       |DEP1       |9      |7      |8

This aggregation can be achieved by manually iterating through the list and create an aggregated list. Example as below.

private static List<Person> getGrouped(List<Person> origList) {
    Map<String, Person> grpMap = new HashMap<String, Person>();

    for (Person person : origList) {
        String key = person.getDepartment() + person.getSubject();
        if (grpMap.containsKey(key)) {
            Person grpdPerson = grpMap.get(key);
            grpdPerson.setMark1(grpdPerson.getMark1() + person.getMark1());
            grpdPerson.setMark2(grpdPerson.getMark2() + person.getMark2());
            grpdPerson.setMark3(grpdPerson.getMark3() + person.getMark3());
        } else {
            grpMap.put(key, person);
        }
    }
    return new ArrayList<Person>(grpMap.values());
}

但是我们可以利用 Java 8 的任何聚合功能或特性吗？

Answer 1

您可以使用 reduction .聚合mark1的样本如下。

public class Test {

    static class Person {
        Person(String name, String subject, String department, Long mark1, Long mark2, Long mark3) {
            this.name = name;
            this.subject = subject;
            this.department = department;
            this.mark1 = mark1;
            this.mark2 = mark2;
            this.mark3= mark3;
        }
            String name;
            String subject;
            String department;
            Long mark1;
            Long mark2;
            Long mark3;

            String group() {
                return subject+department;
            }

            Long getMark1() {
                return mark1;
            }
    }

      public static void main(String[] args)
      {
        List<Person> list = new ArrayList<Test.Person>();
        list.add(new Test.Person("Clark","English","DEP1",7l,8l,6l));
        list.add(new Test.Person("Michel","English","DEP1",6l,4l,7l));
        list.add(new Test.Person("Dave","Maths","DEP2",3l,5l,6l));
        list.add(new Test.Person("Mario","Maths","DEP1",9l,7l,8l));

        Map<String, Long> groups = list.stream().collect(Collectors.groupingBy(Person::group, Collectors.reducing(
                    0l, Person::getMark1, Long::sum)));

        //Or alternatively as suggested by Holger 
        Map<String, Long> groupsNew = list.stream().collect(Collectors.groupingBy(Person::group, Collectors.summingLong(Person::getMark1)));

        System.out.println(groups);

      }

}

仍在研究通过单个函数生成输出。完成后将更新。