如何按持续时间对对象列表进行分组和缩减

一种方法是使用级联分组和减少以求和持续时间。以下结果将生成一个 2 级映射，其中包含每个日期和类型的总和：

Map<LocalDate, Map<String, Optional<Person>>> perDateAndTypeAggregated = 
    personList.stream().collect(Collectors.groupingBy(
        Person::getStartDate,
        Collectors.groupingBy(
                Person::getType,
                Collectors.reducing((p1, p2) ->
                    // this is where we sum the durations
                    new Person(p1.getStartDate(), p1.getType(), p1.getTime1().plus(p2.getTime1()), p1.getTime2().plus(p2.getTime2()))
                )
        )
    ));

如果您想要一个列表（地图提供了更丰富的信息，但这是您的应用程序），您可以按如下方式对上一个地图进行展平和排序：

Comparator<Person> byDateThenType =
    Comparator.comparing(Person::getStartDate).thenComparing(Person::getType);

List<Person> result =
    perDateAndTypeAggregated.values().stream()
        .flatMap(m -> m.values().stream())
        .filter(Optional::isPresent)
        .map(Optional::get)
        .sorted(byDateThenType)
        .toList();

编辑：根据 Holger 的评论，可以使用 / 组合来简化上述操作，并通过删除组合来简化第二个操作 （）：toMapgroupingByreducingtoList.filter().map()

Map<LocalDate, Map<String, Person>> perDateAndTypeAggregated2 = 
    personList.stream().collect(Collectors.groupingBy(
        Person::getStartDate,
        Collectors.toMap(
                Person::getType,
                Function.identity(),
                (p1, p2) ->
                    // this is where we sum the durations
                    new Person(p1.getStartDate(), p1.getType(), p1.getTime1().plus(p2.getTime1()), p1.getTime2().plus(p2.getTime2()))
        )
    ));

List<Person> result2 =
    perDateAndTypeAggregated2.values().stream()
        .flatMap(m -> m.values().stream())
        .sorted(byDateThenType)
        .toList();

编辑2：我使用了以下类：Person

public static class Person {
    private LocalDate startDate;
    private String type;
    private Duration time1;
    private Duration time2;

    public Person() {}

    public Person(String startDate, String type, String time1, String time2) {
        this.startDate = LocalDate.parse(startDate);
        this.type = type;
        this.time1 = Duration.parse(time1);
        this.time2 = Duration.parse(time2);
    }

    public Person(LocalDate startDate, String type, Duration time1, Duration time2) {
        this.startDate = startDate;
        this.type = type;
        this.time1 = time1;
        this.time2 = time2;
    }

    // getters and setters...

    @Override
    public String toString() {
        return "Person(" + startDate + "," + type + "," + time1 + "," + time2 + ")";
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Person person = (Person) o;
        return Objects.equals(startDate, person.startDate) && Objects.equals(type, person.type) && Objects.equals(time1, person.time1) && Objects.equals(time2, person.time2);
    }

    @Override
    public int hashCode() {
        return Objects.hash(startDate, type, time1, time2);
    }
}

并将列表初始化为：

List<Person> personList = Arrays.asList(
        new Person("2023-02-02","member1","PT8H00M","PT4H00M"),
        new Person("2023-02-02","member1","PT50H00M","PT0H45M"),
        new Person("2023-02-02","member2","PT10H00M","PT0H40M"),
        new Person("2023-02-02","member2","PT1H00M","PT1H20M"),
        new Person("2023-02-03","member1","PT8H00M","PT2H00M"),
        new Person("2023-02-03","member1","PT10H00M","PT0H45M"),
        new Person("2023-02-03","member2","PT10H00M","PT1H40M"),
        new Person("2023-02-03","member2","PT2H00M","PT1H20M")
);

所有这些代码都为我提供了问题所需的结果。