Reputation: 3781
Example;
IntStream a = create(3, 1); // => [0,0,1]
IntStream b = create(5, 2); // => [0,0,0,0,2]
The first stream gives an infinite stream of [0,0,1,0,0,1...]
and the second an infinite stream of [0,0,0,0,2,0,0,0,0,2,...]
.
The result stream is ri = ai + bi meaning that I just want to take the sum of the elements at the same position from each stream.
Is this possible in Java ?
Upvotes: 3
Views: 335
Reputation: 834
You can define your own Spliterator to create a stream from it later.
import java.util.Comparator;
import java.util.Spliterators;
import java.util.function.IntConsumer;
public class SumSpliterator extends Spliterators.AbstractIntSpliterator {
private OfInt aSplit;
private OfInt bSplit;
SumSpliterator(OfInt a, OfInt b) {
super(Math.min(a.estimateSize(), b.estimateSize()), Spliterator.ORDERED);
aSplit = a;
bSplit = b;
}
@Override
public boolean tryAdvance(IntConsumer action) {
SummingConsumer consumer = new SummingConsumer();
if (aSplit.tryAdvance(consumer) && bSplit.tryAdvance(consumer)) {
action.accept(consumer.result);
return true;
}
return false;
}
static class SummingConsumer implements IntConsumer {
int result;
@Override
public void accept(int value) {
result += value;
}
}
}
Then create a stream and check the results
IntStream a = //create stream a
IntStream b = //create stream b
SumSpliterator spliterator = new SumSpliterator(a.spliterator(), b.spliterator());
Stream<Integer> stream = StreamSupport.stream(spliterator, false);
stream.limit(20).forEach(System.out::println);
Upvotes: 1
Reputation: 50726
You can use Guava's Streams.zip()
helper:
IntStream sum(IntStream a, IntStream b) {
return Streams.zip(a.boxed(), b.boxed(), Integer::sum)
.map(Integer::intValue);
}
Upvotes: 2