对于求和和求平均的具体问题，请使用 collectAndThen 与 summarizingDouble ：

For the concrete problem of summing and averaging, use collectingAndThen along with summarizingDouble:

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

对于更通用的问题（收集关于你人员的各种事情），你可以创建一个像这个：

For the more generic problem (collect various things about your Persons), you can create a complex collector like this:

// 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)); 

地图值是列表，其中第一个元素是应用第一个收集器的结果，依此类推。您可以使用 Collectors.collectingAndThen（complexCollector，list - > ...）添加自定义修整器步骤，以将此列表转换为更合适的名称。

Map values are lists where first element is the result of applying the first collector and so on. You can add a custom finisher step using Collectors.collectingAndThen(complexCollector, list -> ...) to convert this list to something more appropriate.