首页 > 解决方案 > Java Streams – 如何按值分组并找到每个组的最小值和最大值?

问题描述

对于我的示例,拥有汽车对象并发现基于模型(分组依据)的最小和最大价格值。

List<Car> carsDetails = UserDB.getCarsDetails();
Map<String, DoubleSummaryStatistics> collect4 = carsDetails.stream()
                .collect(Collectors.groupingBy(Car::getMake, Collectors.summarizingDouble(Car::getPrice)));
collect4.entrySet().forEach(e->System.out.println(e.getKey()+" "+e.getValue().getMax()+" "+e.getValue().getMin()));

output :
Lexus 94837.79 17569.59
Subaru 96583.25 8498.41
Chevrolet 99892.59 6861.85

但我找不到哪些汽车对象具有最高和最低价格。我怎样才能做到这一点?

标签: javajava-8java-stream

解决方案


如果您只对Car每组一个感兴趣,您可以使用,例如

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));

但既然你想要最贵和最便宜的,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)));

由于没有与DoubleSummaryStatistics. 如果这种情况不止一次发生,那么值得用这样的类来填补空白:

/**
 * 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);
    }
}

将此添加到您的代码库后,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getPrice返回double,使用Comparator.comparingDouble(Car::getPrice)而不是Comparator.comparing(Car::getPrice).


推荐阅读