Collect successive pairs from a stream

Question

Given a stream such as { 0, 1, 2, 3, 4 },

how can I most elegantly transform it into given form:

{ new Pair(0, 1), new Pair(1, 2), new Pair(2, 3), new Pair(3, 4) }

(assuming, of course, I've defined class Pair)?

Edit: This isn't strictly about ints or primitive streams. The answer should be general for a stream of any type.

Answer 1

My StreamEx library which extends standard streams provides a pairMap method for all stream types. For primitive streams it does not change the stream type, but can be used to make some calculations. Most common usage is to calculate differences:

int[] pairwiseDiffs = IntStreamEx.of(input).pairMap((a, b) -> (b-a)).toArray();

For object stream you can create any other object type. My library does not provide any new user-visible data structures like Pair (that's the part of library concept). However if you have your own Pair class and want to use it, you can do the following:

Stream<Pair> pairs = IntStreamEx.of(input).boxed().pairMap(Pair::new);

Or if you already have some Stream:

Stream<Pair> pairs = StreamEx.of(stream).pairMap(Pair::new);

This functionality is implemented using custom spliterator. It has quite low overhead and can parallelize nicely. Of course it works with any stream source, not just random access list/array like many other solutions. In many tests it performs really well. Here's a JMH benchmark where we find all input values preceding a larger value using different approaches (see this question).

Answer 2

We can use RxJava (very powerful reactive extension library)

IntStream intStream  = IntStream.iterate(1, n -> n + 1);

Observable<List<Integer>> pairObservable = Observable.from(intStream::iterator).buffer(2,1);

pairObservable.take(10).forEach(b -> {
            b.forEach(n -> System.out.println(n));
            System.out.println();
        });

The buffer operator transforms an Observable that emits items into an Observable that emits buffered collections of those items..

Answer 3

You can achieve that by using a bounded queue to store elements which flows through the stream (which is basing on the idea which I described in detail here: Is it possible to get next element in the Stream?)

Belows example first defines instance of BoundedQueue class which will store elements going through the stream (if you don't like idea of extending the LinkedList, refer to link mentioned above for alternative and more generic approach). Later you just combine two subsequent elements into instance of Pair:

public class TwoSubsequentElems {
  public static void main(String[] args) {
    List<Integer> input = new ArrayList<Integer>(asList(0, 1, 2, 3, 4));

    class BoundedQueue<T> extends LinkedList<T> {
      public BoundedQueue<T> save(T curElem) {
        if (size() == 2) { // we need to know only two subsequent elements
          pollLast(); // remove last to keep only requested number of elements
        }

        offerFirst(curElem);

        return this;
      }

      public T getPrevious() {
        return (size() < 2) ? null : getLast();
      }

      public T getCurrent() {
        return (size() == 0) ? null : getFirst();
      }
    }

    BoundedQueue<Integer> streamHistory = new BoundedQueue<Integer>();

    final List<Pair<Integer>> answer = input.stream()
      .map(i -> streamHistory.save(i))
      .filter(e -> e.getPrevious() != null)
      .map(e -> new Pair<Integer>(e.getPrevious(), e.getCurrent()))
      .collect(Collectors.toList());

    answer.forEach(System.out::println);
  }
}

Answer 4

I agree with @aepurniet but instead map you have to use mapToObj

range(0, 100).mapToObj((i) -> new Pair(i, i+1)).forEach(System.out::println);

Answer 5

The proton-pack library provides the windowed functionnality. Given a Pair class and a Stream, you can do it like this:

Stream<Integer> st = Stream.iterate(0 , x -> x + 1);
Stream<Pair<Integer, Integer>> pairs = StreamUtils.windowed(st, 2, 1)
                                                  .map(l -> new Pair<>(l.get(0), l.get(1)))
                                                  .moreStreamOps(...);

Now the pairs stream contains:

(0, 1)
(1, 2)
(2, 3)
(3, 4)
(4, ...) and so on

Answer 6

Finding successive pairs

If you're willing to use a third party library and don't need parallelism, then jOOλ offers SQL-style window functions as follows

System.out.println(
Seq.of(0, 1, 2, 3, 4)
   .window()
   .filter(w -> w.lead().isPresent())
   .map(w -> tuple(w.value(), w.lead().get())) // alternatively, use your new Pair() class
   .toList()
);

Yielding

[(0, 1), (1, 2), (2, 3), (3, 4)]

The lead() function accesses the next value in traversal order from the window.

Finding successive triples / quadruples / n-tuples

A question in the comments was asking for a more general solution, where not pairs but n-tuples (or possibly lists) should be collected. Here's thus an alternative approach:

int n = 3;

System.out.println(
Seq.of(0, 1, 2, 3, 4)
   .window(0, n - 1)
   .filter(w -> w.count() == n)
   .map(w -> w.window().toList())
   .toList()
);

Yielding a list of lists

[[0, 1, 2], [1, 2, 3], [2, 3, 4]]

Without the filter(w -> w.count() == n), the result would be

[[0, 1, 2], [1, 2, 3], [2, 3, 4], [3, 4], [4]]

Disclaimer: I work for the company behind jOOλ