Java 8 流：按多个收集器分组

java java-8 java-stream

2022-09-03 18:26:04

我想使用Java 8 Stream和Group by一个分类器，但有多个收集器函数。因此，在分组时，例如计算一个字段（或另一个字段）的平均值和总和。

我尝试用一个例子来简化这一点：

public void test() {
    List<Person> persons = new ArrayList<>();
    persons.add(new Person("Person One", 1, 18));
    persons.add(new Person("Person Two", 1, 20));
    persons.add(new Person("Person Three", 1, 30));
    persons.add(new Person("Person Four", 2, 30));
    persons.add(new Person("Person Five", 2, 29));
    persons.add(new Person("Person Six", 3, 18));

    Map<Integer, Data> result = persons.stream().collect(
            groupingBy(person -> person.group, multiCollector)
    );
}

class Person {
    String name;
    int group;
    int age;

    // Contructor, getter and setter
}

class Data {
    long average;
    long sum;

    public Data(long average, long sum) {
        this.average = average;
        this.sum = sum;
    }

    // Getter and setter
}

结果应该是将分组结果相关联的地图，例如

1 => Data(average(18, 20, 30), sum(18, 20, 30))
2 => Data(average(30, 29), sum(30, 29))
3 => ....

这对于像“Collectors.counting（）”这样的函数来说完全可以正常工作，但我喜欢链接多个函数（理想情况下从列表中无限）。

List<Collector<Person, ?, ?>>

有可能做这样的事情吗？

答案 1

对于求和求和求和的具体问题，请使用 collectAndThen 以及 summarizingDouble：

Map<Integer, Data> result = persons.stream().collect(
        groupingBy(Person::getGroup, 
                collectingAndThen(summarizingDouble(Person::getAge), 
                        dss -> new Data((long)dss.getAverage(), (long)dss.getSum()))));

对于更一般的问题（收集有关您的Personing的各种内容），您可以创建一个复杂的收集器，如下所示：

// Individual collectors are defined here
List<Collector<Person, ?, ?>> collectors = Arrays.asList(
        Collectors.averagingInt(Person::getAge),
        Collectors.summingInt(Person::getAge));

@SuppressWarnings("unchecked")
Collector<Person, List<Object>, List<Object>> complexCollector = Collector.of(
    () -> collectors.stream().map(Collector::supplier)
        .map(Supplier::get).collect(toList()),
    (list, e) -> IntStream.range(0, collectors.size()).forEach(
        i -> ((BiConsumer<Object, Person>) collectors.get(i).accumulator()).accept(list.get(i), e)),
    (l1, l2) -> {
        IntStream.range(0, collectors.size()).forEach(
            i -> l1.set(i, ((BinaryOperator<Object>) collectors.get(i).combiner()).apply(l1.get(i), l2.get(i))));
        return l1;
    },
    list -> {
        IntStream.range(0, collectors.size()).forEach(
            i -> list.set(i, ((Function<Object, Object>)collectors.get(i).finisher()).apply(list.get(i))));
        return list;
    });

Map<Integer, List<Object>> result = persons.stream().collect(
        groupingBy(Person::getGroup, complexCollector));

映射值是其中第一个元素是应用第一个收集器的结果的列表，依此类推。您可以添加自定义整理器步骤，用于将此列表转换为更合适的内容。Collectors.collectingAndThen(complexCollector, list -> ...)

答案 2

通过使用映射作为输出类型，可以有一个潜在的无限的化约器列表，每个化约器产生自己的统计数据并将其添加到地图中。

public static <K, V> Map<K, V> addMap(Map<K, V> map, K k, V v) {
    Map<K, V> mapout = new HashMap<K, V>();
    mapout.putAll(map);
    mapout.put(k, v);
    return mapout;
}

...

    List<Person> persons = new ArrayList<>();
    persons.add(new Person("Person One", 1, 18));
    persons.add(new Person("Person Two", 1, 20));
    persons.add(new Person("Person Three", 1, 30));
    persons.add(new Person("Person Four", 2, 30));
    persons.add(new Person("Person Five", 2, 29));
    persons.add(new Person("Person Six", 3, 18));

    List<BiFunction<Map<String, Integer>, Person, Map<String, Integer>>> listOfReducers = new ArrayList<>();

    listOfReducers.add((m, p) -> addMap(m, "Count", Optional.ofNullable(m.get("Count")).orElse(0) + 1));
    listOfReducers.add((m, p) -> addMap(m, "Sum", Optional.ofNullable(m.get("Sum")).orElse(0) + p.i1));

    BiFunction<Map<String, Integer>, Person, Map<String, Integer>> applyList
            = (mapin, p) -> {
                Map<String, Integer> mapout = mapin;
                for (BiFunction<Map<String, Integer>, Person, Map<String, Integer>> f : listOfReducers) {
                    mapout = f.apply(mapout, p);
                }
                return mapout;
            };
    BinaryOperator<Map<String, Integer>> combineMaps
            = (map1, map2) -> {
                Map<String, Integer> mapout = new HashMap<>();
                mapout.putAll(map1);
                mapout.putAll(map2);
                return mapout;
            };
    Map<String, Integer> map
            = persons
            .stream()
            .reduce(new HashMap<String, Integer>(),
                    applyList, combineMaps);
    System.out.println("map = " + map);

生产：

map = {Sum=10, Count=6}