How to interleave (merge) two Java 8 Streams?

Question

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

What do I need to do for the output to be the below?

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

I looked into concat but as the javadoc explains, it just appends one after the other, it does not interleave / intersperse.

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

Creates a lazily concatenated stream whose elements are all the elements of the first stream followed by all the elements of the second stream.

Wrongly gives

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

Could do it if I collected them and iterated, but was hoping for something more Java8-y, Streamy :-)

Note

I don't want to zip the streams

“zip” operation will take an element from each collection and combine them.

the result of a zip operation would be something like this: (unwanted)

 // onetwo
 // threefour
 // fivesix

Answer 1

I’d use something like this:

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);
}

It retains the characteristics of the input streams as far as possible, which allows certain optimizations (e.g. for count()and toArray()). Further, it adds the ORDERED even when the input streams might be unordered, to reflect the interleaving.

When one stream has more elements than the other, the remaining elements will appear at the end.

Answer 2

A much dumber solution than Holger did, but may be it would fit your requirements:

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();
}

Answer 3

As you can see from the question comments, I gave this a go using 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);
    }

Answer 4

This may not be a good answer because
(1) it collects to map, which you don't want to do I guess and
(2) it is not completely stateless as it uses AtomicIntegers.

Still adding it because
(1) it is readable and
(2) community can get an idea from this and try to improve it.

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

Output

one
two
three
four
five
six

@Holger's edit

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

Answer 5

One solution with 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);

Or simply by the solution in abacus-common provided by me:

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

Answer 6

without any external lib (using jdk11)

import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class MergeUtil {

    private static <T> Stream<T> zipped(List<T> lista, List<T> listb) {
        int maxSize = Math.max(lista.size(), listb.size());
        final var listStream = IntStream
                .range(0, maxSize)
                .mapToObj(i -> {
                    List<T> result = new ArrayList<>(2);
                    if (i < lista.size()) result.add(lista.get(i));
                    if (i < listb.size()) result.add(listb.get(i));
                    return result;
                });
        return listStream.flatMap(List::stream);
    }

    public static void main(String[] args) {
        var l1 = List.of(1, 2, 3);
        var l2 = List.of(4, 5, 6, 7, 8, 9);
        final var zip = zipped(l1, l2);
        System.out.println(zip.collect(Collectors.toList()));
    }

}

listStream is a Stream<List<A>> that flatted in return.

The result is: [1, 4, 2, 5, 3, 6, 7, 8, 9]

Answer 7

Using Guava's Streams.zip and Stream.flatMap:

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

---

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

Prints:

one
two
three
four
five
six