如何使用Java8流将列表列表转换为单个列表

Question

如何使用流转换以下代码而不使用每个循环.

1. getAllSubjects()返回所有List和每个Subject List<Topic>.所有列表应合并为List<Topic>.
2. 需要得到Map<id,topicName> 从List<Topic>

对象模型:

Subject
  id,....
  List<Topic>
Topic
  id,name

public Map<String, String> getSubjectIdAndName(final String subjectId) {

    List<Subject> list = getAllSubjects(); // api method returns all subjects
    //NEEDS TO IMPROVE CODE USING STREAMS
    list = list.stream().filter(e -> e.getId().equals(subjectId)).collect(Collectors.toList());
    List<Topic> topicList = new ArrayList<>();
    for (Subject s : list) {
        List<Topic> tlist = s.getTopics();
        topicList.addAll(tlist);
    }
    return topicList.stream().collect(Collectors.toMap(Topic::getId, Topic::getName));

}

Answer 1

flatMap在这里使用,不再流.请注意,这toMap假设没有重复的键(或空值)

list.stream()
    .filter(e -> subjectId.equals(e.getId()))
    .flatMap(subject -> subject.getTopics().stream())
    .collect(Collectors.toMap(Topic::getId, Topic::getName));