將列表流轉換為單個Container

Question

考慮下面的WorkExperience類：

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 + "]";
    }
}     

假設我希望按年份分組我的技能，這就是我們如何按年份groupBy ：

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返回： Map<Integer, Set<List<Skills>>>
但我需要的是： Map<Integer, Set<Skills>>

如何將List流轉換為單個Container？

Answer 1

使用Java 8功能的flatMapping的替代方案將是

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

你可以稍微優化一下

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

我們可以使用Java-9中添加的Collectors.flatMapping器。 通過使用flatMapping ， 我們可以將中間列表展平為單個容器。 flatMapping可用於原始流的元素可轉換為流的情況。

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

API注意 ：

flatMapping（）收集器在多級縮減中使用時非常有用，例如groupingBy或partitioningBy的下游。

Answer 3

另一種實現所需方法的方法是使用靜態工廠方法Collector.of()實現自己的收集Collector.of() ：

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;
        })
    );

與其他答案相比，這是相當混亂和臃腫。

Answer 4

嘗試更加面向對象。 所以我想創建一個新的小對象

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決你的問題：

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))))
                );

您可以按如下方式使用它：

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]]