Java 8 partition list

Question

Is it possible to partition a List in pure Jdk8 into equal chunks (sublists).

I know it is possible using Guava Lists class, but can we do it with pure Jdk? I don\'t wan

Answer 1

That can be done easily using the subList() method:

List<String> collection = new ArrayList<>(21);
// fill collection
int chunkSize = 10;
List<List<String>> lists = new ArrayList<>();
for (int i = 0; i < collection.size(); i += chunkSize) {
    int end = Math.min(collection.size(), i + chunkSize);
    lists.add(collection.subList(i, end));
}

Answer 2

private final String dataSheet = "103343262,6478342944, 103426540,84528784843, 103278808,263716791426, 103426733,27736529279, 
103426000,27718159078, 103218982,19855201547, 103427376,27717278645, 
103243034,81667273413";

    final int chunk = 2;
    AtomicInteger counter = new AtomicInteger();
    Collection<List<String>> chuncks= Arrays.stream(dataSheet.split(","))
            .map(String::trim)
            .collect(Collectors.groupingBy(i->counter.getAndIncrement()/chunk))
            .values();

result:

pairs =
 "103218982" -> "19855201547"
 "103278808" -> "263716791426"
 "103243034" -> "81667273413"
 "103426733" -> "27736529279"
 "103426540" -> "84528784843"
 "103427376" -> "27717278645"
 "103426000" -> "27718159078"
 "103343262" -> "6478342944"

We need to group each 2 elements into key, value pairs, so will partion the list into chunks of 2, (counter.getAndIncrement() / 2) will result same number each 2 hits ex:

IntStream.range(0,6).forEach((i)->System.out.println(counter.getAndIncrement()/2));
prints:
0
0
1
1
2
2

You may ajust chunk sizee to partition lists sizes.

Answer 3

I have tried my own solution with a custom made Collector. I hope someone will find it useful, or help me improve it.

class PartitioningCollector<T> implements Collector<T, List<List<T>>, List<List<T>>> {

        private final int batchSize;
        private final List<T> batch;

        public PartitioningCollector(int batchSize) {
            this.batchSize = batchSize;
            this.batch = new ArrayList<>(batchSize);
        }

        @Override
        public Supplier<List<List<T>>> supplier() {
            return LinkedList::new;
        }

        @Override
        public BiConsumer<List<List<T>>, T> accumulator() {
            return (total, element) -> {
                batch.add(element);
                if (batch.size() >= batchSize) {
                    total.add(new ArrayList<>(batch));
                    batch.clear();
                }
            };
        }

        @Override
        public BinaryOperator<List<List<T>>> combiner() {
            return (left, right) -> {
                List<List<T>> result = new ArrayList<>();
                result.addAll(left);
                result.addAll(left);
                return result;
            };
        }

        @Override
        public Function<List<List<T>>, List<List<T>>> finisher() {
            return result -> {
                if (!batch.isEmpty()) {
                    result.add(new ArrayList<>(batch));
                    batch.clear();
                }
                return result;
            };
        }

        @Override
        public Set<Characteristics> characteristics() {
            return emptySet();
        }
    }

Answer 4

Guava Lists class has a partition() method that does exactly this. See https://guava.dev/releases/21.0/api/docs/com/google/common/collect/Lists.html#partition-java.util.List-int-

Answer 5

Try using this code, it uses Java 8:

public static Collection<List<Integer>> splitListBySize(List<Integer> intList, int size) {

    if (!intList.isEmpty() && size > 0) {
        final AtomicInteger counter = new AtomicInteger(0);
        return intList.stream().collect(Collectors.groupingBy(it -> counter.getAndIncrement() / size)).values();
    }
    return null;
}