Java Streams – 如何按值分组并找到每个组的最小值和最大值？

如果您只对每个组一个感兴趣，则可以使用，例如Car

Map<String, Car> mostExpensives = carsDetails.stream()
    .collect(Collectors.toMap(Car::getMake, Function.identity(),
        BinaryOperator.maxBy(Comparator.comparing(Car::getPrice))));
mostExpensives.forEach((make,car) -> System.out.println(make+" "+car));

但是，既然你想要最昂贵和最便宜的，你需要这样的东西：

Map<String, List<Car>> mostExpensivesAndCheapest = carsDetails.stream()
    .collect(Collectors.toMap(Car::getMake, car -> Arrays.asList(car, car),
        (l1,l2) -> Arrays.asList(
            (l1.get(0).getPrice()>l2.get(0).getPrice()? l2: l1).get(0),
            (l1.get(1).getPrice()<l2.get(1).getPrice()? l2: l1).get(1))));
mostExpensivesAndCheapest.forEach((make,cars) -> System.out.println(make
        +" cheapest: "+cars.get(0)+" most expensive: "+cars.get(1)));

此解决方案带来了一些不便，因为没有等效于 的通用统计对象。如果这种情况发生不止一次，那么值得用这样的类来填补空白：DoubleSummaryStatistics

/**
 * Like {@code DoubleSummaryStatistics}, {@code IntSummaryStatistics}, and
 * {@code LongSummaryStatistics}, but for an arbitrary type {@code T}.
 */
public class SummaryStatistics<T> implements Consumer<T> {
    /**
     * Collect to a {@code SummaryStatistics} for natural order.
     */
    public static <T extends Comparable<? super T>> Collector<T,?,SummaryStatistics<T>>
                  statistics() {
        return statistics(Comparator.<T>naturalOrder());
    }
    /**
     * Collect to a {@code SummaryStatistics} using the specified comparator.
     */
    public static <T> Collector<T,?,SummaryStatistics<T>>
                  statistics(Comparator<T> comparator) {
        Objects.requireNonNull(comparator);
        return Collector.of(() -> new SummaryStatistics<>(comparator),
            SummaryStatistics::accept, SummaryStatistics::merge);
    }
    private final Comparator<T> c;
    private T min, max;
    private long count;
    public SummaryStatistics(Comparator<T> comparator) {
        c = Objects.requireNonNull(comparator);
    }

    public void accept(T t) {
        if(count == 0) {
            count = 1;
            min = t;
            max = t;
        }
        else {
            if(c.compare(min, t) > 0) min = t;
            if(c.compare(max, t) < 0) max = t;
            count++;
        }
    }
    public SummaryStatistics<T> merge(SummaryStatistics<T> s) {
        if(s.count > 0) {
            if(count == 0) {
                count = s.count;
                min = s.min;
                max = s.max;
            }
            else {
                if(c.compare(min, s.min) > 0) min = s.min;
                if(c.compare(max, s.max) < 0) max = s.max;
                count += s.count;
            }
        }
        return this;
    }

    public long getCount() {
        return count;
    }

    public T getMin() {
        return min;
    }

    public T getMax() {
        return max;
    }

    @Override
    public String toString() {
        return count == 0? "empty": (count+" elements between "+min+" and "+max);
    }
}

将其添加到代码库后，您可以像这样使用它

Map<String, SummaryStatistics<Car>> mostExpensives = carsDetails.stream()
    .collect(Collectors.groupingBy(Car::getMake,
        SummaryStatistics.statistics(Comparator.comparing(Car::getPrice))));
mostExpensives.forEach((make,cars) -> System.out.println(make+": "+cars));

如果返回，则使用可能比 使用 更有效。getPricedoubleComparator.comparingDouble(Car::getPrice)Comparator.comparing(Car::getPrice)

这是一个非常简洁的解决方案。它将所有 s 收集到 a 中，因此无需任何附加类即可工作。CarSortedSet

Map<String, SortedSet<Car>> grouped = carDetails.stream()
        .collect(groupingBy(Car::getMake, toCollection(
                () -> new TreeSet<>(comparingDouble(Car::getPrice)))));

grouped.forEach((make, cars) -> System.out.println(make
        + " cheapest: " + cars.first()
        + " most expensive: " + cars.last()));

一个可能的缺点是性能，因为收集了所有s，而不仅仅是当前的最小值和最大值。但除非数据集非常大，否则我认为它不会引人注目。Car