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Java 8 Streams 不允许重用。这产生了一个难题，即在创建滑动窗口通量以计算像 x(i)*x(i-1) 这样的关系时如何重用流。

以下代码基于移位运算符的思想。我用 skip(1) 移动第一个流以创建第二个流。

Flux<Integer> primary = Flux.fromStream(IntStream.range(1, 10).boxed());
Flux<Integer> secondary = primary.skip(1);
primary.zipWith(secondary)
        .map(t -> t.getT1() * t.getT2())
        .subscribe(System.out::println);

这是上述代码的可视化表示：

1 2 3 4 5 6 7 8 9 10
v v v v v v v v v v  skip(1)
2 3 4 5 6 7 8 9 10
v v v v v v v v v v  zipWith
1 2, 2 3, 3 4, 4 5, 5 6, 6 7, 7 8, 8 9, 9 10 <- sliding window of length 2
v v v v v v v v v v  multiples
2 6 12 20 30 42 56 72 90

不幸的是，此代码错误为：

java.lang.IllegalStateException: stream has already been operated upon or closed

显而易见的解决方法是缓存元素并确保缓存大小大于或等于流大小：

Flux<Integer> primary = Flux.fromStream(IntStream.range(1, 10).boxed()).cache(10);

或使用流替换：

Flux<Integer> primary = Flux.range(0, 10);

第二种解决方案将重新执行 skip(1) 序列的原始序列。

然而，一个有效的解决方案只需要一个大小为 2 的缓冲区。如果流恰好是一个大文件，这很重要：

Files.lines(Paths.get(megaFile));

如何有效地缓冲流，以便对主 Flux 的多个订阅不会导致所有内容都被读入内存或导致重新执行？

4

3 回答 3

5

我终于找到了一个解决方案，尽管它不是面向缓冲区的。灵感是首先解决滑动窗口为 2 的问题：

Flux<Integer> primary = Flux.fromStream(IntStream.range(0, 10).boxed());
primary.flatMap(num -> Flux.just(num, num))
    .skip(1)
    .buffer(2)
    .filter(list -> list.size() == 2)
    .map(list -> Arrays.toString(list.toArray()))
    .subscribe(System.out::println);

该过程的可视化表示如下：

1 2 3 4 5 6 7 8 9 
V V V V V V V V V    Flux.just(num, num)
1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9
V V V V V V V V V    skip(1)
1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9
V V V V V V V V V    bufffer(2)
1 2, 2 3, 3 4, 4 5, 5 6, 6 7, 7 8, 8 9, 9
V V V V V V V V V    filter
1 2, 2 3, 3 4, 4 5, 5 6, 6 7, 7 8, 8 9

这是输出：

[0, 1]
[1, 2]
[2, 3]
[3, 4]
[4, 5]
[5, 6]
[6, 7]
[7, 8]
[8, 9]

然后我将上面的想法概括为一个任意滑动窗口大小的解决方案：

public class SlidingWindow {

    public static void main(String[] args) {
        System.out.println("Different sliding windows for sequence 0 to 9:");
        SlidingWindow flux = new SlidingWindow();
        for (int windowSize = 1; windowSize < 5; windowSize++) {
            flux.slidingWindow(windowSize, IntStream.range(0, 10).boxed())
                .map(SlidingWindow::listToString)
                .subscribe(System.out::print);
            System.out.println();
        }

        //show stream difference: x(i)-x(i-1)
        List<Integer> sequence = Arrays.asList(new Integer[]{10, 12, 11, 9, 13, 17, 21});
        System.out.println("Show difference 'x(i)-x(i-1)' for " + listToString(sequence));
        flux.slidingWindow(2, sequence.stream())
            .doOnNext(SlidingWindow::printlist)
            .map(list -> list.get(1) - list.get(0))
            .subscribe(System.out::println);
        System.out.println();
    }

    public <T> Flux<List<T>> slidingWindow(int windowSize, Stream<T> stream) {
        if (windowSize > 0) {
            Flux<List<T>> flux = Flux.fromStream(stream).map(ele -> Arrays.asList(ele));
            for (int i = 1; i < windowSize; i++) {
                flux = addDepth(flux);
            }
            return flux;
        } else {
            return Flux.empty();
        }
    }

    protected <T> Flux<List<T>> addDepth(Flux<List<T>> flux) {
        return flux.flatMap(list -> Flux.just(list, list))
            .skip(1)
            .buffer(2)
            .filter(list -> list.size() == 2)
            .map(list -> flatten(list));
    }

    protected <T> List<T> flatten(List<List<T>> list) {
        LinkedList<T> newl = new LinkedList<>(list.get(1));
        newl.addFirst(list.get(0).get(0));
        return newl;
    }

    static String listToString(List list) {
        return list.stream()
            .map(i -> i.toString())
            .collect(Collectors.joining(", ", "[ ", " ], "))
            .toString();
    }

    static void printlist(List list) {
        System.out.print(listToString(list));
    }

}

上述代码的输出如下：

Different sliding windows for sequence 0 to 9:
[ 0 ], [ 1 ], [ 2 ], [ 3 ], [ 4 ], [ 5 ], [ 6 ], [ 7 ], [ 8 ], [ 9 ], 
[ 0, 1 ], [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ], [ 5, 6 ], [ 6, 7 ], [ 7, 8 ], [ 8, 9 ], 
[ 0, 1, 2 ], [ 1, 2, 3 ], [ 2, 3, 4 ], [ 3, 4, 5 ], [ 4, 5, 6 ], [ 5, 6, 7 ], [ 6, 7, 8 ], [ 7, 8, 9 ], 
[ 0, 1, 2, 3 ], [ 1, 2, 3, 4 ], [ 2, 3, 4, 5 ], [ 3, 4, 5, 6 ], [ 4, 5, 6, 7 ], [ 5, 6, 7, 8 ], [ 6, 7, 8, 9 ], 

Show difference 'x(i)-x(i-1)' for [ 10, 12, 11, 9, 13, 17, 21 ], 
[ 10, 12 ], 2
[ 12, 11 ], -1
[ 11, 9 ], -2
[ 9, 13 ], 4
[ 13, 17 ], 4
[ 17, 21 ], 4
于 2017-02-02T04:39:45.880 回答
0

我已经实施了以下解决方案：

public <T> Flux<Flux<T>> toSlidingWindow(Flux<T> source, int size) {
    return toSlidingWindow(source, deque -> {
        while (deque.size() > size) {
            deque.poll();
        }
        return Flux.fromIterable(deque);
    });
}

public <T> Flux<Flux<T>> toSlidingWindow(Flux<T> source, Function<Deque<T>, Flux<T>> dequePruneFunction) {
    return source.map(ohlc -> {
        Deque<T> deque = dequeAtomicReference.get();
        deque.offer(ohlc);
        return dequePruneFunction.apply(deque);
    });
}

这可以是固定大小的滑动窗口，也可以使用自定义函数来确定每个窗口的范围。

如果像这样使用它出现任何多线程问题，您可以复制Deque似乎受. 这将确保生成的窗口在其他线程中保持不变。acquirereleaseAtomicReferenceFlux

也许像这样：

public <T> Flux<Flux<T>> toSlidingWindowAsync(Flux<T> source, int size) {
    return toSlidingWindowAsync(source, deque -> {
        while (deque.size() > size) {
            deque.poll();
        }
        return Flux.fromIterable(new LinkedList<>(deque));
    });
}

public <T> Flux<Flux<T>> toSlidingWindowAsync(Flux<T> source, Function<Deque<T>, Flux<T>> dequePruneFunction) {
    AtomicReference<Deque<T>> dequeAtomicReference = new AtomicReference<>(new LinkedList<>());
    return source.map(ohlc -> {
        Deque<T> deque = dequeAtomicReference.getAcquire();
        deque.offer(ohlc);
        Flux<T> windowFlux = dequePruneFunction.apply(deque);
        dequeAtomicReference.setRelease(deque);
        return windowFlux;
    });
}

这将复制Deque用于每个结果滑动窗口的那个。

于 2019-05-31T21:00:01.843 回答
0

如果您使用的是 Reactor Core 3（我不确定此运算符何时发布），您可以简单地使用

    Flux.fromStream(IntStream.rangeClosed(1, 10).boxed())
            .buffer(2, 1)
            .skipLast(1)
            .map(t -> t.stream().reduce((a, b)-> a*b))
            .subscribe(System.out::println);

神奇的是 buffer(2, 1) 部分：这里 maxSize 为 2，skip 为 1。由于 maxSize 大于 skip，这会在通量上创建重叠缓冲区（即滑动窗口），并将每个缓冲区作为列表。需要 skipLast(1)，因为最后一个缓冲区将是单个元素（10 个），因此需要跳过。

于 2019-12-03T00:21:17.637 回答