Java 8 Stream 中的 group by 分组

Java About 2,134 words

需求

使用CollectorsgroupingBy完成各种分组操作。

Bean 类

@Data
class Student {
    private String name;
    private int score;
    private int age;
}

集合

Student zs1 = new Student("zs", 100, 20);
Student ls = new Student("ls", 90, 25);
Student ww = new Student("ww", 90, 23);
Student zs2 = new Student("zs", 85, 30);

List<Student> list = List.of(zs1, ls, ww, zs2);

学生姓名分组

Map<String, List<Student>> nameStudentsMap = list.stream().collect(Collectors.groupingBy(Student::getName));

输出

{ww=[Student{name='ww', score=90, age=23}], ls=[Student{name='ls', score=90, age=25}], zs=[Student{name='zs', score=100, age=20}, Student{name='zs', score=85, age=30}]}

按分数分组取学生对象集合

Map<Integer, List<Student>> scoreStudentsMap = list.stream().collect(Collectors.groupingBy(Student::getScore));

输出:可以看到90分有2个学生,其余都是1个学生。

{100=[Student{name='zs', score=100, age=20}], 85=[Student{name='zs', score=85, age=30}], 90=[Student{name='ls', score=90, age=25}, Student{name='ww', score=90, age=23}]}

统计分组中包含的个数

Map<Integer, Long> scoreCountMap = list.stream().collect(Collectors.groupingBy(Student::getScore, Collectors.counting()));

输出:可以看到90分有2个人,其余都是1个人。

{100=1, 85=1, 90=2}

按分数分组取学生姓名集合

Collectors.groupingBy分组中使用Collectorsmapping映射学生到姓名的关系。

Map<Integer, List<String>> scoreStudentNamesMap = list.stream().collect(Collectors.groupingBy(Student::getScore, Collectors.mapping(Student::getName, Collectors.toList())));

输出

{100=[zs], 85=[zs], 90=[ls, ww]}

按分数分组取学生年龄大于 20 的学生集合

Map<Integer, Set<Student>> scoreStudentsAgeGt20Map = list.stream().collect(Collectors.groupingBy(Student::getScore, Collectors.filtering(student -> student.getAge() > 20, Collectors.toSet())));

输出

{100=[], 85=[Student{name='zs', score=85, age=30}], 90=[Student{name='ww', score=90, age=23}, Student{name='ls', score=90, age=25}]}

按分数分组取学生姓名以 z 开始的姓名集合

Map<Integer, Set<String>> scoreStudentNameStartWithMap = list.stream().collect(Collectors.groupingBy(Student::getScore, Collectors.mapping(Student::getName, Collectors.filtering(name -> name.startsWith("z"), Collectors.toSet()))));

输出

{100=[zs], 85=[zs], 90=[]}
Views: 925 · Posted: 2023-06-09

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