使用 Java Stream 从 ArrayList 获得最高分

Question

我想通过 Id 获得最高分组。如果两个最高分相同，那么我想根据最低可选 ID 获得最高分。我想在 Java Stream 中获得它。到目前为止，我正在尝试以下代码这不起作用示例：

数组列表：

ID：1 Score：80 OptionalId：1

ID：1 Score：90 OptionalId：2

ID：1 Score：90 OptionalId：3

ID：2 Score：80 OptionalId：1

ID：2 Score：100 OptionalId：3

ID：2 Score： 100 可选 ID：5

结果应该是

ID：1 得分 90 OptionalId：2

ID 2 得分 100 OptionalId：3

Map<Long, Optional<Person>> result1 = records.stream()

                  .collect(Collectors.groupingBy(Person::getId,

                          Collectors.maxBy(Comparator.comparing(Person::getScore)),

                          Collector.minBy(Comparator.comparing(Person::getOptionalId))));

        for(Person ns: result1) {

            sb.append(ns.getBatchNumber());

            sb.append(',');

Answer 1

我建议你从一个Comparator<Person>优先考虑最大值 score然后是最小值的自定义开始optionalId。为简洁起见，这是将其传递给变量的好方法：final Comparator<Person> comparator = Comparator&nbsp; &nbsp; .comparing(Person::getScore)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;// descending score first&nbsp; &nbsp; .thenComparing(Comparator&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // then ..&nbsp; &nbsp; &nbsp; &nbsp; .comparing(Person::getOptionalId)&nbsp; &nbsp; &nbsp; &nbsp; // .. optionalId&nbsp; &nbsp; &nbsp; &nbsp; .reversed());&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // .. but ascending现在使用一些收集器java流.Collectors::groupingBy按值和下游对所有Persons进行分组idCollectors::reducing将所有Person相同的 s减少id为一个，使用Comparator<Person>得到最高score和最低的一个optionalId。Collectors::collectingAndThenCollection<Optional<Person>>将结构从to展平，Collection<Person>因为任何减少操作都会导致Optional- 此步骤是可选的，可能会根据您的示例跳过。这是代码：Collection<Person> filtered = records.stream()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;// Stream<Person>&nbsp; &nbsp; .collect(Collectors.groupingBy(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // from Map<Long, List<Person>>&nbsp; &nbsp; &nbsp; &nbsp; Person::getId,&nbsp; &nbsp; &nbsp; &nbsp; Collectors.collectingAndThen(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // .. downstream to ..&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Collectors.reducing((a, b) ->&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // .. Map<Long, Optional<Person>>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; comparator.compare(a, b) > 0 ? a : b),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Optional::get))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // .. Map<Long, Person>&nbsp; &nbsp; .values();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;// .. Collection<Person>[人 [id=1, score=90, optionalId=2], 人 [id=2, score=100, optionalId=3]]

Answer 2

对于给定的 Id 值，必须有一个 Person。Id 值的存在完全取决于 Person。因此，如果存在 Id，则也必须存在 Person。因此，将 aOptional<Person>作为地图的值有什么意义。相比之下，仅将一个Person实例作为map. 在这里，我将toMap收集器与 一起使用BinaryOperator.maxBy来完成工作。这是它的外观。请注意如何将BinaryOperator.maxBy用作mergeFunction。Map<Integer, Person> maxPersonById = records.stream()&nbsp; &nbsp; .collect(Collectors.toMap(Person::getId, Function.identity(),&nbsp; &nbsp; &nbsp; &nbsp; BinaryOperator.maxBy(Comparator.comparing(Person::getScore)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .thenComparing(Comparator.comparing(Person::getOptionalId).reversed()))));这是上述给定输入的输出。{1=Person [id=1, score=90, optionalId=2], 2=Person [id=2, score=100, optionalId=3]}

Answer 3

您可以尝试使用以下聚合的流代码ID，然后使用两级排序找到最大分数，首先是分数，然后是可选的 ID，以防分数平局：import static java.util.Collections.reverseOrder;import static java.util.Comparator.comparing;Map<Long, Optional<Person>> result1 = records.stream()&nbsp; &nbsp; .collect(Collectors.groupingBy(Person::getId,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Collectors.maxBy(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Comparator.comparing(Person::getScore)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .thenComparing(reverseOrder(comparing(Person::getOptionalId))))));Optional[ID: 1 Score: 90 OptionalId: 2]Optional[ID: 2 Score: 100 OptionalId: 3]这里的技巧是仅反转可选 ID 的排序顺序，我们希望它是升序的，而不是降序的。默认情况下排序顺序为降序，因为我们正在调用Collections.maxBy.