将列表流转换为单个容器

Question

考虑下面的WorkExperience类：

public class WorkExperience {
    private int year;
    private List<Skills> skill;

    public WorkExperience(int year, List<Skills> skill) {
        this.year = year;
        this.skill = skill;
    }   
    //getter setter         
}

public class Skills {
    private String skills;

    public Skills(String skills) {
        this.skills = skills;
    }

    @Override
    public String toString() {
        return "Skills [skills=" + skills + "]";
    }
}     

假设我想按年按我的技能分组，这就是我们按年做groupBy的方法：

public static void main(String[] args) {

    List<Skills> skillSet1 = new  ArrayList<>();
    skillSet1.add(new Skills("Skill-1"));
    skillSet1.add(new Skills("Skill-2"));
    skillSet1.add(new Skills("Skill-3"));

    List<Skills> skillSet2 = new  ArrayList<>();
    skillSet2.add(new Skills("Skill-1"));
    skillSet2.add(new Skills("Skill-4"));
    skillSet2.add(new Skills("Skill-2"));


    List<Skills> skillSet3 = new  ArrayList<>();
    skillSet3.add(new Skills("Skill-1"));
    skillSet3.add(new Skills("Skill-9"));
    skillSet3.add(new Skills("Skill-2"));

    List<WorkExperience> workExperienceList = new ArrayList<>();
    workExperienceList.add(new WorkExperience(2017,skillSet1));
    workExperienceList.add(new WorkExperience(2017,skillSet2));
    workExperienceList.add(new WorkExperience(2018,skillSet3));

    Map<Integer, Set<List<Skills>>> collect = workExperienceList.stream().collect(
        Collectors.groupingBy(
            WorkExperience::getYear,
            Collectors.mapping(WorkExperience::getSkill, Collectors.toSet())
        )
    );
}

groupBy正在返回：Map<Integer, Set<List<Skills>>>

但我需要的是：Map<Integer, Set<Skills>>

如何将列表流转换为单个容器？

Answer 1

我们可以使用在Collectors.flatMapping Java-9中添加的收集器。通过使用flatMapping、，我们可以将中间列表扁平化成一个容器。在原始流的元素可以转换为流的情况下，可以使用 flatMapping。

workExperienceList.stream().collect(Collectors.groupingBy(
                              WorkExperience::getYear, 
                              Collectors.flatMapping(workexp -> workexp.getSkill().stream(), 
                                             Collectors.toSet())));

API Note

flatMapping()收集器在多级还原(如groupingBy或partitioningBy的下游)中最有用。

Answer 2

使用Java 8特性替代flatMapping的方法将是

Map<Integer, Set<Skills>> map = workExperienceList.stream()
    .collect(Collectors.toMap(
        WorkExperience::getYear,
        we -> new HashSet<>(we.getSkill()),
        (s1, s2)-> { s1.addAll(s2); return s1; }));

您可以对此进行一些优化

Map<Integer, Set<Skills>> map = workExperienceList.stream()
    .collect(Collectors.toMap(
        WorkExperience::getYear,
        we -> new HashSet<>(we.getSkill()),
        (s1, s2) -> {
            if(s1.size() > s2.size()) { s1.addAll(s2); return s1; }
            else { s2.addAll(s1); return s2; }
        }));

Answer 3

实现所需的另一种方法是使用静态工厂方法Collector.of()实现自己的收集器。

Map<Integer, Set<Skills>> collect = workExperienceList.stream()
    .collect(Collector.of(
        HashMap::new,
        ( map, e ) -> map.computeIfAbsent(e.getYear(), k -> new HashSet<>()).addAll(e.getSkill()),
        ( left, right ) -> {
            right.forEach(( year, set ) -> left.computeIfAbsent(year, k -> new HashSet<>()).addAll(set));
            return left;
        })
    );

与其他答案相比，这是相当混乱和臃肿的。