Java stream reduce

我与影子孤独终老i 提交于 2019-12-04 02:25:49

I think your solution is pretty nice already, especially as using a reduction enables parallelism easily compared to collecting into a shared outside container. But it's easier to use collect instead of reduce as Holger pointed out. Furthermore, the conditions in the accumulator can be simplified a bit, and you forgot to merge the last and first elements in the combiner:

List<Tuple> reduce = tupleStream.collect(ArrayList::new, WDParser::add, WDParser::combine);

private static List<Tuple> combine(List<Tuple> list1, List<Tuple> list2)
{
    if (!list2.isEmpty())
    {
        add(list1, list2.remove(0)); // merge lists in the middle if necessary
        list1.addAll(list2);         // add all the rest
    }
    return list1;
}

private static List<Tuple> add(List<Tuple> list, Tuple t)
{
    int lastIndex = list.size() - 1;
    if (list.isEmpty() || list.get(lastIndex).getDirection() != t.getDirection())
    {
        list.add(t);
    }
    else
    {
        list.set(lastIndex, list.get(lastIndex).merge(t));
    }
    return list;
}

Instead of using indexes to access the first/last element you could even use LinkedList and the methods add/removeFirst/Last().

How about this. First define a small helper method:

private static Tuple mergeTwo(Tuple left, Tuple right) {
    int[] leftArray = left.getData();
    int[] rightArray = right.getData();
    int[] result = new int[leftArray.length + rightArray.length];
    System.arraycopy(leftArray, 0, result, 0, leftArray.length);
    System.arraycopy(rightArray, 0, result, leftArray.length, rightArray.length);
    return new Tuple(left.getDirection(), result);
}

This is close to your concat/merge I guess, but a single one. Basically a way to merge two Tuple(s) together.

And a helper method to produce the needed Collector, you can put this into a utility so that it can be re-used:

private static Collector<Tuple, ?, List<Tuple>> mergedTuplesCollector() {
    class Acc {

        ArrayDeque<Tuple> deque = new ArrayDeque<>();

        void add(Tuple elem) {
            Tuple head = deque.peek();
            if (head == null || head.getDirection() != elem.getDirection()) {
                deque.offerFirst(elem);
            } else {
                deque.offerFirst(mergeTwo(deque.poll(), elem));
            }
        }

        Acc merge(Acc right) {

            Tuple lastLeft = deque.peekLast();
            Tuple firstRight = right.deque.peekFirst();

            if (lastLeft.getDirection() == firstRight.getDirection()) {
                deque.offerLast(mergeTwo(deque.pollLast(), right.deque.pollFirst()));
            } else {
                deque.addAll(right.deque);
            }

            return this;
        }

        public List<Tuple> finisher() {
            return new ArrayList<>(deque);
        }

    }
    return Collector.of(Acc::new, Acc::add, Acc::merge, Acc::finisher);
}

And usage would be, for example:

List<Tuple> merged = tuples.stream()
            .parallel()
            .collect(mergedTuplesCollector());

This is an alternative approach that uses slightly different data structures.

If this is an option, changing from int[] to List<Integer> allows for more flexibility (not to mention avoiding creating/copying arrays multiple times):

class Tuple {
    Direction direction;
    List<Integer> data;
}

And the following function does the merging on a Deque collection:

private static List<Integer> next(Deque<Tuple> t, Direction d) {
    if (!t.isEmpty() && t.peekLast().getDirection() == d) {
        return t.peekLast().getData();
    } else {
        Tuple next = new Tuple();
        next.direction = d;
        next.data = new ArrayList<>();
        t.addLast(next);
        return next.data;
    }
}

And with that, the stream can look as simple as:

Deque<Tuple> deq = new LinkedList<>(); //the final collection of tuples

tuples.stream()
.flatMap(tp -> tp.getData().stream()
                 .map(d -> Pair.of(tp.getDirection(), Integer.valueOf(d))))
.forEach(el -> next(deq, el.getLeft()).add(el.getRight()));

I've got two ideas on this topic. First one is getting the indices like in this answer and group it accordingly.

The second idea - if you already got a Stream a custom Collector should be used (similar to the other solutions, though using Deque):

private Collector<Tuple, ?, List<Tuple>> squashTuples() {
  return new Collector<Tuple, Deque<Tuple>, List<Tuple>>() {
    @Override
    public Supplier<Deque<Tuple>> supplier() {
      return ArrayDeque::new;
    }

    @Override
    public BiConsumer<Deque<Tuple>, Tuple> accumulator() {
      return (acc, e) -> {
        Objects.requireNonNull(e);
        if (!acc.isEmpty() && acc.peekLast().getDirection() == e.getDirection()) {
          acc.offerLast(acc.pollLast().merge(e));
        } else {
          acc.offerLast(e);
        }
      };
    }

    @Override
    public BinaryOperator<Deque<Tuple>> combiner() {
      return (left, right) -> {
        if (!left.isEmpty() && !right.isEmpty() && left.peekLast().getDirection() == right.peekFirst().getDirection()) {
          left.offerLast(left.pollLast().merge(right.pollFirst()));
        }
        left.addAll(right);
        return left;
      };
    }

    @Override
    public Function<Deque<Tuple>, List<Tuple>> finisher() {
      return ArrayList::new;
    }

    @Override
    public Set<Characteristics> characteristics() {
      return EnumSet.noneOf(Characteristics.class);
    }
  };
}
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!