正如我对List主题的了解"Subjects":[{"subject":"Math","grades":1},{"subject":"Math","grades":2},{"subject":"Math","grades":3},{"subject":"Math","grades":3},{"subject":"Lab","grades":10},{"subject":"Lab","grades":12}]我想像这样分组和减少结果//Expected Result"Subjects":[{"subject":"Math","grades":[1,2,3]},{"subject":"Lab","grades":[10,12]}]我很好奇如何以 java8 风格映射和减少对象。我在下面放置了过时的代码。我的主课public static void main(String[] args) {List<Subject> list = new ArrayList<>();list.add(new Subject("Math",1));list.add(new Subject("Math",2));list.add(new Subject("Math",3));list.add(new Subject("Math",3));list.add(new Subject("Lab",10));list.add(new Subject("Lab",12)); Map<String, Set<Integer>> result = new HashMap<>();list.stream().forEach(subjects-> { if(result.get(subjects.getSubject())==null){ Set<Integer> set = new HashSet<>(); set.add(subjects.getGrades()); result.put(subjects.getSubject(),set ); }else{ Set<Integer> set =result.get(subjects.getSubject()); set.add(subjects.getGrades()); result.put(subjects.getSubject(), set); } }); result.forEach((key,val)->{ System.out.println("KEY:"+key + " RESULT :"+val); });} public class Subject {private String subject;private Integer grades;public Subject(String subject , Integer grade) {this.subject = subject;this.grades = grade;}/** get set **/}
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白衣染霜花
TA贡献1796条经验 获得超10个赞
您可以使用 aCollectors.groupingBy(Subject::getSubject)
和 aCollectors.mapping(Subject::getGrades, Collectors.toSet())
作为下游。
list.stream() .collect(Collectors.groupingBy(Subject::getSubject, Collectors.mapping(Subject::getGrades, Collectors.toSet())));
它会给你一个Map<String, Set<Integer>>
.
{Lab=[10, 12], Math=[1, 2, 3]}
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