Question

考虑以下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 + "]";
    }
}

让我们说我希望按年份按照我的技能进行分组，这就是我们每年groupBy的做法：

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？</ strong>

Answer 1

仅使用Java 8功能的flatMapping替代

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

你可以稍微优化一下

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 2

我们可以使用 Java-9中添加的Collectors.flatMapping收藏家。使用flatMapping，我们可以将中间列表展平为一个容器。 flatMapping可用于原始流的元素可转换为流的情况。

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

API说明 ：

flatMapping（）收集器在多级缩减中使用时非常有用，例如groupingBy或partitioningBy的下游。

Answer 3

实现目标的另一种方法是使用静态工厂方法Collector.of()来实现自己的收集器：

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。

Answer 4

尝试更加面向对象。所以我想创建一个新的小对象

public static class AsGroup {
    private final Integer year;
    private final Collection<Skill> skillSet;

    public AsGroup(Integer year, Collection<Skill> skillSet) {

        this.year = year;
        this.skillSet = skillSet;
    }

    public AsGroup addSkills(AsGroup asGroupSkills) {
        this.skillSet.addAll(asGroupSkills.skillSet);
        return this;
    }
}

然后你可以用以下方法解决问题：

Map<Integer, Optional<com.company.Main.AsGroup>> groupedByYear = workExperienceList.stream()
                .map(workExperience ->
                        new AsGroup(workExperience.getYear(), new HashSet<>(workExperience.getSkill()))
                ).collect(groupingBy((asGroup) -> asGroup.year,
                          reducing((group1, group2) -> (group1.addSkills(group2))))
                );

您可以按照以下方式使用它：

groupedByYear.forEach(((year, groupedSkills) -> System.out.println(year + " " + groupedSkills.get().skillSet)));

它打印如下：

2017 [Skill [skill=Skill-1], Skill [skill=Skill-1], Skill [skill=Skill-4], Skill [skill=Skill-2], Skill [skill=Skill-2], Skill [skill=Skill-3]]
2018 [Skill [skill=Skill-1], Skill [skill=Skill-2], Skill [skill=Skill-9]]

将列表流转换为单个Container

4 个答案: