java-如何将CompletableFuture>转换为Stream>？

而已.真的找不到更接近的东西.我要实施以下任务链

List<Item> items = Collections.singletonList(new Item("John Smith", "1997-2014"));

Stream<CompletableFuture<List<ScrappingResult>>> scrappingFutures =
    items.stream().map(item ->
        CompletableFuture.supplyAsync(() -> new ScrappingTask(item).call()));

Stream<CompletableFuture<ScrappingResult>> scrappingFuturesUnwrapped =
    scrappingFutures.map(resultsFuture -> ???);

Stream<CompletableFuture<TrimmingResult>> trimmingResults = scrappingFuturesUnwrapped.map(resultFuture ->
    // thenCompose?
    resultFuture.thenCompose(result -> {
        Path clipsDir = Paths.get("./"
            + result.getItem().getName()
            + "/" + result.getItem().getTimespan());

        AtomicInteger clipIdx = new AtomicInteger();

        return result.getVideo().getClips().stream()
            .map(clip -> CompletableFuture.supplyAsync(() ->
                new TrimmingTask(
                    ffmpegPath,
                    result.getVideo().getVideoUrl(),
                    clip,
                    clipsDir.resolve("clip_" + clipIdx.incrementAndGet() + ".mp3")).call())
            );
    });
);

最后一行在语法上不正确,但是我希望传达出这个想法.所以,我想做两次类似flatMap的操作并得到
Stream< CompletableFuture< TrimmingResult>>在末尾.

我怎么做？

谢谢.

解决方法:

如果我不误解您的意图,那么您只想弄平结果.您可以使用Spliterator懒惰地接收结果&使用flatMap将Streams合并为扁平化流,例如：

Stream<CompletableFuture<ScrappingResult>> scrappingFuturesUnwrapped =
               scrappingFutures.flatMap(each -> unwrap(each));

static <T> Stream<? extends CompletableFuture<T>> 
  unwrap(CompletableFuture<? extends List<? extends T>> master) {

    return generate(new Predicate<Consumer<? super CompletableFuture<T>>>() {

        private Iterator<? extends T> cursor;

        @Override
        public boolean test(Consumer<? super CompletableFuture<T>> consumer) {
            cursor = cursor == null ? await().iterator() : cursor;
            if (cursor.hasNext()) {
                consumer.accept(completedFuture(cursor.next()));
                return true;
            }
            return false;
        }

        //                        v--- blocked at the first time
        private List<? extends T> await() {
            try {
                return master.get();
            } catch (InterruptedException | ExecutionException e) {
                throw new CompletionException(e);
            }
        }
    });
}

static <T> Stream<? extends CompletableFuture<T>> 
  generate(Predicate<Consumer<? super CompletableFuture<T>>> generator) {

    long unknownSize = Long.MAX_VALUE;
    return stream(new AbstractSpliterator<CompletableFuture<T>>(unknownSize, 0) {
        public boolean tryAdvance(Consumer<? super CompletableFuture<T>> action) {
            return generator.test(action);
        }
    }, false);
}

摘要

上面的解决方案只是我的第一个想法,在这种情况下,它不是最好的方法,您可以将其与big design first相对.但是,即使这是一个较差的解决方案,但我还是会保留在这里,因为它可能会让某些人思考以其他方式.有关更多详细信息,您可以从此处和there查看@Holger的评论.

我承认最好的方法是@Holger在下面说的话,由于没有人写下来,请记录下来以服务更多人.

Stream<CompletableFuture<ScrappingResult>> scrappingFuturesUnwrapped =
               scrappingFutures.flatMap(each -> each.join().stream())
                               .map(CompletableFuture::completedFuture);

标签：completable-future,java-8,java
来源： https://codeday.me/bug/20191110/2013379.html