把List变为map,并遇到重复值时自动过滤、并返回有序map

发布时间 2023-12-14 11:54:05作者: 漠孤烟

Student:

    @Data
    @AllArgsConstructor
    public class Student {
        private String name;
        private Integer age;
        private Integer score;
    }

把list转成map

        List<Student> students = List.of(
                new Student("zhangsan", 20, 80),
                new Student("zhangsan", 20, 81),
                new Student("lisi", 21, 82),
                new Student("wangwu", 20, 83),
                new Student("zhaoliu", 20, 83)
        );

        Map<String, Integer> scoreMap = students.stream()
                .collect(Collectors.toMap(
                        Student::getName,
                        Student::getScore
                ));
        System.out.println("scoreMap = " + scoreMap); // 报 `java.lang.IllegalStateException: Duplicate key zhangsan (attempted merging values 80 and 81)`

遇重复key时,两个value就会导致报错。

遇到list里有重复key时报错的解决方法:

        // 遇重复key, 取第一个value
        Map<String, Integer> scoreMap2 = students.stream()
                .collect(Collectors.toMap(
                        Student::getName,
                        Student::getScore,
                        (score1, score2) -> score1
                ));
        System.out.println("scoreMap2 = " + scoreMap2);

运行正常,打印结果:scoreMap2 = {lisi=82, zhaoliu=83, zhangsan=80, wangwu=83}

扩展:如果想让map有序,即map key序列与list一致,如何实现?

如下:

        // 遇重复key, 取第一个value,并按list里的序列生成map
        Map<String, Integer> scoreMap3 = students.stream()
                .collect(Collectors.toMap(
                        Student::getName,
                        Student::getScore,
                        (score1, score2) -> score1,
                        LinkedHashMap::new
                ));
        System.out.println("scoreMap3 = " + scoreMap3);

打印结果:scoreMap3 = {zhangsan=80, lisi=82, wangwu=83, zhaoliu=83}