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如何交错(合并)两个Java8流?

java-stream functional-programming java-8

Blundell · 技术社区 · 6 年前

 Stream<String> a = Stream.of("one", "three", "five");
 Stream<String> b = Stream.of("two", "four", "six");

我需要做什么才能使输出如下?

// one
// two
// three
// four
// five
// six

我调查 concat 但正如javadoc所解释的,它只是一个接一个地追加,而不是交错/穿插。

Stream<String> out = Stream.concat(a, b);
out.forEach(System.out::println);

创建一个延迟连接的流,其元素是所有第二流。

错误地给予

 // one
 // three
 // five
 // two
 // four
 // six

如果我收集它们并进行迭代,就可以做到这一点,但我希望得到更多的Java8-y,Streamy:-)

我不想让溪流断流

zip操作将从每个集合中获取一个元素并将它们组合起来。

zip操作的结果如下:(不需要)

 // onetwo
 // threefour
 // fivesix

5 回复 | 直到 5 年前

Holger 5 年前

public static <T> Stream<T> interleave(Stream<? extends T> a, Stream<? extends T> b) {
    Spliterator<? extends T> spA = a.spliterator(), spB = b.spliterator();
    long s = spA.estimateSize() + spB.estimateSize();
    if(s < 0) s = Long.MAX_VALUE;
    int ch = spA.characteristics() & spB.characteristics()
           & (Spliterator.NONNULL|Spliterator.SIZED);
    ch |= Spliterator.ORDERED;

    return StreamSupport.stream(new Spliterators.AbstractSpliterator<T>(s, ch) {
        Spliterator<? extends T> sp1 = spA, sp2 = spB;

        @Override
        public boolean tryAdvance(Consumer<? super T> action) {
            Spliterator<? extends T> sp = sp1;
            if(sp.tryAdvance(action)) {
                sp1 = sp2;
                sp2 = sp;
                return true;
            }
            return sp2.tryAdvance(action);
        }
    }, false);
}

它尽可能保留输入流的特征,从而允许某些优化(例如 count() 和 toArray() ). 此外,它还增加了 ORDERED 即使输入流可能是无序的,也要反映交错。

Eugene 6 年前

这是一个比霍尔格更愚蠢的解决方案,但可能符合您的要求:

private static <T> Stream<T> interleave(Stream<T> left, Stream<T> right) {
    Spliterator<T> splLeft = left.spliterator();
    Spliterator<T> splRight = right.spliterator();

    T[] single = (T[]) new Object[1];

    Stream.Builder<T> builder = Stream.builder();

    while (splRight.tryAdvance(x -> single[0] = x) && splLeft.tryAdvance(builder)) {
        builder.add(single[0]);
    }

    return builder.build();
}

Blundell 6 年前

正如您从问题评论中所看到的,我尝试使用zip:

Stream<String> a = Stream.of("one", "three", "five");
Stream<String> b = Stream.of("two", "four", "six");

Stream<String> out = interleave(a, b);


    public static <T> Stream<T> interleave(Stream<T> streamA, Stream<T> streamB) {
        return zip(streamA, streamB, (o1, o2) -> Stream.of(o1, o2)).flatMap(s -> s);
    }

    /**
    * https://stackoverflow.com/questions/17640754/zipping-streams-using-jdk8-with-lambda-java-util-stream-streams-zip
    **/
    private static <A, B, C> Stream<C> zip(Stream<A> streamA, Stream<B> streamB, BiFunction<A, B, C> zipper) {
        final Iterator<A> iteratorA = streamA.iterator();
        final Iterator<B> iteratorB = streamB.iterator();
        final Iterator<C> iteratorC = new Iterator<C>() {
            @Override
            public boolean hasNext() {
                return iteratorA.hasNext() && iteratorB.hasNext();
            }

            @Override
            public C next() {
                return zipper.apply(iteratorA.next(), iteratorB.next());
            }
        };
        final boolean parallel = streamA.isParallel() || streamB.isParallel();
        return iteratorToFiniteStream(iteratorC, parallel);
    }

    private static <T> Stream<T> iteratorToFiniteStream(Iterator<T> iterator, boolean parallel) {
        final Iterable<T> iterable = () -> iterator;
        return StreamSupport.stream(iterable.spliterator(), parallel);
    }

Kartik 6 年前

今年五月这是一个好答案,因为

(2) 它不是完全无状态的,因为它使用原子整数。

仍然添加它,因为
(1) 它可读性强,而且
(2) 社区可以从中获得想法,并尝试改进它。

Stream<String> a = Stream.of("one", "three", "five");
Stream<String> b = Stream.of("two", "four", "six");

AtomicInteger i = new AtomicInteger(0);
AtomicInteger j = new AtomicInteger(1);

Stream.of(a.collect(Collectors.toMap(o -> i.addAndGet(2), Function.identity())),
        b.collect(Collectors.toMap(o -> j.addAndGet(2), Function.identity())))
        .flatMap(m -> m.entrySet().stream())
        .sorted(Comparator.comparing(Map.Entry::getKey))
        .forEach(e -> System.out.println(e.getValue())); // or collect

输出

one
two
three
four
five
six

Stream.concat(a.map(o -> new AbstractMap.SimpleEntry<>(i.addAndGet(2), o)),
        b.map(o -> new AbstractMap.SimpleEntry<>(j.addAndGet(2), o)))
        .sorted(Map.Entry.comparingByKey())
        .forEach(e -> System.out.println(e.getValue())); // or collect

user_3380739 6 年前

一个解决方案 Iterator

final Iterator<String> iterA = a.iterator();
final Iterator<String> iterB = b.iterator();

final Iterator<String> iter = new Iterator<String>() {
  private final AtomicInteger idx = new AtomicInteger();
  @Override
  public boolean hasNext() { 
    return iterA.hasNext() || iterB.hasNext();
  }
  @Override
  public String next() {
    return idx.getAndIncrement() % 2 == 0 && iterA.hasNext() ? iterA.next() : iterB.next();
  }
};

 // Create target Stream with StreamEx from: https://github.com/amaembo/streamex    
 StreamEx.of(iter).forEach(System.out::println);

 // Or Streams from Google Guava
 Streams.stream(iter).forEach(System.out::println);

abacus-util

 AtomicInteger idx = new AtomicInteger();
 StreamEx.merge(a, b, (s1, s2) -> idx.getAndIncrement() % 2 == 0 ? Nth.FIRST : Nth.SECOND).forEach(Fn.println());

ZhekaKozlov 5 年前

Streams.zip 和 Stream.flatMap :

Stream<String> interleaved = Streams
        .zip(a, b, (x, y) -> Stream.of(x, y))
        .flatMap(Function.identity());

interleaved.forEach(System.out::println);

印刷品:

one
two
three
four
five
six