这与 How to short-circuit reduce on Stream? 本质上是同一个问题。 .但是，由于该问题侧重于 boolean 值流，并且它的答案不能推广到其他类型和减少操作，所以我想问一个更一般的问题。

我们如何对流进行归约，以便在遇到 absorbing element 时短路进行归约操作？

乘法的典型数学情况是 0。这个流:

int product = IntStream.of(2, 3, 4, 5, 0, 7, 8)
        .reduce(1, (a, b) -> a * b);

将消耗最后两个元素(7 和 8)，而不管是否遇到 0 后产品是已知的。

最佳答案

不幸的是，Stream API 创建您自己的短路操作的能力有限。不太干净的解决方案是抛出一个 RuntimeException 并捕获它。这是 IntStream 的实现，但它也可以推广到其他流类型:

public static int reduceWithCancelEx(IntStream stream, int identity, 
                      IntBinaryOperator combiner, IntPredicate cancelCondition) {
    class CancelException extends RuntimeException {
        private final int val;

        CancelException(int val) {
            this.val = val;
        }
    }

    try {
        return stream.reduce(identity, (a, b) -> {
            int res = combiner.applyAsInt(a, b);
            if(cancelCondition.test(res))
                throw new CancelException(res);
            return res;
        });
    } catch (CancelException e) {
        return e.val;
    }
}

使用示例:

int product = reduceWithCancelEx(
        IntStream.of(2, 3, 4, 5, 0, 7, 8).peek(System.out::println), 
        1, (a, b) -> a * b, val -> val == 0);
System.out.println("Result: "+product);

输出:

2
3
4
5
0
Result: 0

请注意，即使它适用于并行流，也不能保证其他并行任务在其中一个引发异常时立即完成。已经开始的子任务可能会一直运行到完成，因此您可能会处理比预期更多的元素。

更新:更长的替代解决方案，但对并行更友好。它基于自定义拆分器，最多返回一个元素，这是所有底层元素的累积结果)。当您在顺序模式下使用它时，它会在单个 tryAdvance 调用中完成所有工作。当您拆分它时，每个部分都会生成相应的单个部分结果，这些结果由 Stream 引擎使用组合器功能进行缩减。这是通用版本，但也可以进行原始特化。

final static class CancellableReduceSpliterator<T, A> implements Spliterator<A>,
        Consumer<T>, Cloneable {
    private Spliterator<T> source;
    private final BiFunction<A, ? super T, A> accumulator;
    private final Predicate<A> cancelPredicate;
    private final AtomicBoolean cancelled = new AtomicBoolean();
    private A acc;

    CancellableReduceSpliterator(Spliterator<T> source, A identity,
            BiFunction<A, ? super T, A> accumulator, Predicate<A> cancelPredicate) {
        this.source = source;
        this.acc = identity;
        this.accumulator = accumulator;
        this.cancelPredicate = cancelPredicate;
    }

    @Override
    public boolean tryAdvance(Consumer<? super A> action) {
        if (source == null || cancelled.get()) {
            source = null;
            return false;
        }
        while (!cancelled.get() && source.tryAdvance(this)) {
            if (cancelPredicate.test(acc)) {
                cancelled.set(true);
                break;
            }
        }
        source = null;
        action.accept(acc);
        return true;
    }

    @Override
    public void forEachRemaining(Consumer<? super A> action) {
        tryAdvance(action);
    }

    @Override
    public Spliterator<A> trySplit() {
        if(source == null || cancelled.get()) {
            source = null;
            return null;
        }
        Spliterator<T> prefix = source.trySplit();
        if (prefix == null)
            return null;
        try {
            @SuppressWarnings("unchecked")
            CancellableReduceSpliterator<T, A> result = 
                (CancellableReduceSpliterator<T, A>) this.clone();
            result.source = prefix;
            return result;
        } catch (CloneNotSupportedException e) {
            throw new InternalError();
        }
    }

    @Override
    public long estimateSize() {
        // let's pretend we have the same number of elements
        // as the source, so the pipeline engine parallelize it in the same way
        return source == null ? 0 : source.estimateSize();
    }

    @Override
    public int characteristics() {
        return source == null ? SIZED : source.characteristics() & ORDERED;
    }

    @Override
    public void accept(T t) {
        this.acc = accumulator.apply(this.acc, t);
    }
}

类似于 Stream.reduce(identity, accumulator, combiner) 的方法和 Stream.reduce(identity, combiner) , 但使用 cancelPredicate:

public static <T, U> U reduceWithCancel(Stream<T> stream, U identity,
        BiFunction<U, ? super T, U> accumulator, BinaryOperator<U> combiner,
        Predicate<U> cancelPredicate) {
    return StreamSupport
            .stream(new CancellableReduceSpliterator<>(stream.spliterator(), identity,
                    accumulator, cancelPredicate), stream.isParallel()).reduce(combiner)
            .orElse(identity);
}

public static <T> T reduceWithCancel(Stream<T> stream, T identity,
        BinaryOperator<T> combiner, Predicate<T> cancelPredicate) {
    return reduceWithCancel(stream, identity, combiner, combiner, cancelPredicate);
}

让我们测试两个版本并计算实际处理了多少元素。让我们把 0 放在最后。异常(exception)版本:

AtomicInteger count = new AtomicInteger();
int product = reduceWithCancelEx(
        IntStream.range(-1000000, 100).filter(x -> x == 0 || x % 2 != 0)
                .parallel().peek(i -> count.incrementAndGet()), 1,
        (a, b) -> a * b, x -> x == 0);
System.out.println("product: " + product + "/count: " + count);
Thread.sleep(1000);
System.out.println("product: " + product + "/count: " + count);

典型输出:

product: 0/count: 281721
product: 0/count: 500001

所以当只处理一些元素时返回结果，任务继续在后台工作并且计数器仍在增加。这是拆分器版本:

AtomicInteger count = new AtomicInteger();
int product = reduceWithCancel(
        IntStream.range(-1000000, 100).filter(x -> x == 0 || x % 2 != 0)
                .parallel().peek(i -> count.incrementAndGet()).boxed(), 
                1, (a, b) -> a * b, x -> x == 0);
System.out.println("product: " + product + "/count: " + count);
Thread.sleep(1000);
System.out.println("product: " + product + "/count: " + count);

典型输出:

product: 0/count: 281353
product: 0/count: 281353

当返回结果时，所有的任务都真正完成了。

关于java - 如何短路 Stream 上的 reduce() 操作？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32495069/