How to compose two collections item with item?

Question

Is it possible to aggregate two collections one item at a time? For example:

L1: {1,2,3,4}
L2: {1,2,3,4}

I want to aggregate the two, L1 and L2 so that I have a returning

L3: {2, 4, 6, 8}

Hence L3[i] = L2[i] + L3[i]

How can I accomplish this using lambdas?

This looks like the zip function. A search on google wielded [this question](http://stackoverflow.com/questions/17640754/zipping-streams-using-jdk8-with-lambda-java-util-stream-streams-zip). — Mephy, Aug 31 '15 at 12:11
@Mephy seems that zip, didn't make it in the jdk 8 official release — Olimpiu POP, Aug 31 '15 at 12:24

Konstantin Yovkov · Accepted Answer · 2015-08-31T12:43:14.030

4

You could play with IntStream:

List<Integer> result = IntStream.range(0, Math.min(l1.size(), l2.size()))
                                .map(i -> l1.get(i) + l2.get(i))
                                .boxed()
                                .collect(Collectors.toList());

edited Aug 31 '15 at 12:43

answered Aug 31 '15 at 12:14

Konstantin Yovkov

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@kocko - I've tried it, but it doesn't compile...I'm I missing something? – Olimpiu POP Aug 31 '15 at 12:34
@kocko - it works, thank you. You'd be my hero if you adapt your example with an example to collect in a list. – Olimpiu POP Aug 31 '15 at 12:39
8

You can almost always replace `.map(…).boxed()` with `.mapToObj(…)` – Holger Aug 31 '15 at 12:50

Karol Król · Answer 2 · 2015-09-02T07:49:18.433

1

These code snippets do the job:

Solution 1 (simplest approach):

public static Collection<Integer> sumCollectionElements(Collection<Integer> L1, Collection<Integer> L2) {
    final Iterator<Integer> iterator = L2.iterator();
    return L1.stream()
            .limit(Math.min(L1.size(), L2.size()))
            .map(integer -> integer + iterator.next())
            .collect(Collectors.toList());
}

Solution 2 (more general approach -> gist):

public 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> mergedIterator = new Iterator<C>() {
        @Override
        public boolean hasNext() {
            return iteratorA.hasNext() && iteratorB.hasNext();
        }

        @Override
        public C next() {
            return zipper.apply(iteratorA.next(), iteratorB.next());
        }
    };
    return iteratorToFiniteStream(mergedIterator);
}

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

edited Sep 02 '15 at 07:49

answered Aug 31 '15 at 20:46

Karol Król

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any idea how to get rid of the iterator? In order to resume everything in using streams? – Olimpiu POP Sep 01 '15 at 04:27
@OlimpiuPOP there is a solution for you. Please review this GitHub project: https://github.com/poetix/protonpack and check method StreamUtils#zip – Karol Król Sep 01 '15 at 08:34
yeah I know this one, I was thinking about something that exists in the code base of jdk 8. Thank you! – Olimpiu POP Sep 01 '15 at 08:40
@OlimpiuPOP Maybe my gist will satisfy you? – Karol Król Sep 01 '15 at 23:01
With third-party code it could be even shorter: `List l3 = StreamEx.zip(l1, l2, Integer::sum).toList();` (using my StreamEx library, I'm just trying to limit myself and not to advertise it too often, so I didn't post an answer). Sometimes having additional dependencies not an option. – Tagir Valeev Sep 02 '15 at 01:23
@TagirValeev thank you! Leaving your modesty aside, could you please share the link the official repository of your library? Thank you :). – Olimpiu POP Sep 02 '15 at 04:41
@OlimpiuPOP, ok if OP asks, I wrote an answer :-) – Tagir Valeev Sep 02 '15 at 05:20
@TagirValeev I've seen your lib in GitHub its great! – Karol Król Sep 02 '15 at 10:50

score 1 · Answer 3 · answered Sep 02 '15 at 05:20

If you don't mind using third-party libraries, consider my StreamEx library which enhances standard Java Stream API. Your problem can be solved like this:

List<Integer> l3 = StreamEx.zip(l1, l2, Integer::sum).toList();

At a first glance the StreamEx.zip method is similar to protonpack zip method. However it zips not the streams, but random access lists (like ArrayList or Arrays.asList) which allows proper parallelization. Internally it's close to the @kocko solution with minor differences:

If input lists have different size, an IllegalArgumentException will be thrown.
Internally the separate spliterator is used which might work somewhat faster in some scenarios.

Also note the usage of toList() method which is a shortcut for collect(Collectors.toList()).

score -1 · Answer 4 · answered Aug 31 '15 at 12:28

-1

L1 = {1,2,3,4}  
L2 = {1,2,3,4}

L3 = map(lambda L1,L2:L1+L2, L1,L2)

Output: [2, 4, 6, 8]

answered Aug 31 '15 at 12:28

Mayur Koshti

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How to compose two collections item with item?

4 Answers4