将List集合中相同属性的对象合并

将List集合中相同属性的对象合并

在实际需求中,可能遇到需要将List集合中的对象,按照某个(多个)属性进行合并成一个对象,并将相同对象的某一个属性(例如:数量)进行相加。

Lambda方式

首先创建实体对象
@Data
@AllArgsConstructor
@NoArgsConstructor
@SuppressWarnings("serial")
public class Student {
    private Integer uid;
    private String uname;
    private Double score;
}
合并代码
public static void main(String[] args) {
        List<Student> list=new ArrayList<>();
        list.add(new Student(1,"张三",60.50));
        list.add(new Student(1,"张三",70.25));
        list.add(new Student(1,"张三",80.25));
        list.add(new Student(2,"李四",60D));

        List<Student> studentList=new ArrayList<>();
        list.parallelStream().collect(Collectors.groupingBy(o -> (o.getUid()+o.getUname()),Collectors.toList())).forEach((id,transfer)->{
            transfer.stream().reduce((a,b)->new Student(a.getUid(),a.getUname(),a.getScore()+b.getScore())).ifPresent(studentList::add);
        });
        for (Student student : studentList) {
            System.out.println(student);
        }
    }

Map方式

实体类重写equals和hashCode方法并添加merge方法
@Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Student student = (Student) o;
        return Objects.equals(uid, student.uid) &&
                Objects.equals(uname, student.uname);
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((uid == null) ? 0 : uid.hashCode());
        result = prime * result + ((uname == null) ? 0 : uname.hashCode());
        return result;
    }


    public static Student merge(Student s1, Student s2) {
        if (!s1.equals(s2)) {
            throw new IllegalArgumentException();
        }
        return new Student(s1.uid, s1.uname, s1.score + s2.score);
    }
测试方法
List<Student> list = new ArrayList<>();
list.add(new Student(1, "张三", 60.50));
list.add(new Student(1, "张三", 70.25));
list.add(new Student(1, "张三", 80.25));
list.add(new Student(2, "李四", 60D));

Map<Student, Student> map = new HashMap<Student, Student>();
        for (Student s : list) {
            if (map.containsKey(s)) {
                map.put(s, Student.merge(s, map.get(s)));
            } else {
                map.put(s, s);
            }
        }
Collection<Student> values = map.values();
values.forEach(System.out::println);

引用原文:
https://blog.csdn.net/weixin_55585048/article/details/116793654

posted @   __先森  阅读(2461)  评论(0编辑  收藏  举报
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