java create 2D array from Stream<Integer>

Question

I have Stream Stream<Integer> stream // 1,2,3,4,5,6,7,8,9

I need create int[3][3] from this stream

How can I do it?

I tried

int[][] ints = stream
            .map(i -> new int[]{i})
            .toArray(int[][]::new);

But I get: [[1], [2], [3], [4], [5], [6], [7], [8], [9]]

But I need: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

EDIT: I think it is not dublicate this because I need int[3][3] array and not String[][]

I tried this example

int[][] array =
            IntStream.range(0, 3)
                .mapToObj(x -> IntStream.range(0, 3).boxed()
                    .toArray(Integer[]::new))
                .toArray(int[][]::new);

And I get error

Exception in thread "main" java.lang.ArrayStoreException: [Ljava.lang.Integer;
    at java.util.stream.Nodes$FixedNodeBuilder.accept(Nodes.java:1222)
    at java.util.stream.IntPipeline$4$1.accept(IntPipeline.java:250)
    at java.util.stream.Streams$RangeIntSpliterator.forEachRemaining(Streams.java:110)
    at java.util.Spliterator$OfInt.forEachRemaining(Spliterator.java:693)
    at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:481)
    at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:471)
    at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:545)
    at java.util.stream.AbstractPipeline.evaluateToArrayNode(AbstractPipeline.java:260)
    at java.util.stream.ReferencePipeline.toArray(ReferencePipeline.java:438)
    at SparseMatrixSupportImpl.fromStream(SparseMatrixSupportImpl.java:25)
    at SparseMatrixSupportImpl.fromStream(SparseMatrixSupportImpl.java:4)
    at Main.main(Main.java:12)

EDIT:

int[][] array = stream.collect(() -> new int[3][3],
            (a, i) -> a[(i - 1) / 3][(i - 1) % 3] = i, (a, i) -> {
            });

error

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
    at SparseMatrixSupportImpl.lambda$fromStream$1(SparseMatrixSupportImpl.java:28)
    at java.util.stream.ReduceOps$4ReducingSink.accept(ReduceOps.java:220)
    at java.util.stream.IntPipeline$4$1.accept(IntPipeline.java:250)
    at java.util.Spliterators$IntArraySpliterator.forEachRemaining(Spliterators.java:1032)
    at java.util.Spliterator$OfInt.forEachRemaining(Spliterator.java:693)
    at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:481)
    at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:471)
    at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
    at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
    at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:510)
    at SparseMatrixSupportImpl.fromStream(SparseMatrixSupportImpl.java:27)
    at SparseMatrixSupportImpl.fromStream(SparseMatrixSupportImpl.java:4)
    at Main.main(Main.java:12)



int[] values = new int[]{1, 2, 3,5,6,7,8,9,11};
Stream<Integer> integerStream = Arrays.stream(values).boxed();

Possible duplicate of Filling a Multidimensional Array using a Stream — Anton Balaniuc
– Anton Balaniuc, Commented Nov 2, 2017 at 10:46
@AntonBalaniuc That might not work if the Stream already exists, e.g. comes from some source OP can not change. — tobias_k
– tobias_k, Commented Nov 2, 2017 at 10:48

gil.fernandes · Accepted Answer · 2017-11-03 10:12:17Z

4

You can try this code with Java 9 (this throws a compilation error in Java 8, so beware):

int[][] array = Stream.iterate(1, i -> i < 10, i -> i + 1).collect(() -> new int[3][3],
        (a, i) -> a[(i - 1) / 3][(i - 1) % 3] = i, (a, i) -> {});
Stream.of(array).forEach(a -> System.out.println(Arrays.toString(a)));

Here is the result:

[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

With Java 8:

int[][] array = IntStream.range(1, 10).collect(() -> new int[3][3],
    (a, i) -> a[(i - 1) / 3][(i - 1) % 3] = i, (a, i) -> {});
Stream.of(array).forEach(a -> System.out.println(Arrays.toString(a)));

Here is another solution which supports numbers other than the pre-defined sequence (1,2,3,4,5,6,7,8,9), but is not that clean as a solution as it uses a counter array:

int[] counter = {0};
int[][] array = Stream.of(4, 2, 3, 4, 5, 8, 9, 8, 11).collect(() -> new int[3][3],
        (a, i) -> {
            a[counter[0] / 3][counter[0] % 3] = i;
            counter[0]++;
        }, (a, i) -> {});

Output:

[4, 2, 3]
[4, 5, 8]
[9, 8, 11]

If you want a bit more generic solution based on the solution above you can try this code snippet which allows to set a variable number of columns:

List<Integer> list = Arrays.asList(4, 2, 3, 4, 5, 8, 9, 8, 11, 12, 13, 17, 32, 45, 89, 91, 91, 98, 87);
int[] counter = {0};
int cols = 5;
int rows = (int) Math.ceil(list.size() / cols) + 1;
int[][] array = list.stream().collect(() -> new int[rows][cols],
        (a, i) -> a[counter[0] / cols][counter[0]++ % cols] = i, (a, i) -> {});
Stream.of(array).forEach(a -> System.out.println(Arrays.toString(a)));

The snippet above prints:

[4, 2, 3, 4, 5]
[8, 9, 8, 11, 12]
[13, 17, 32, 45, 89]
[91, 91, 98, 87, 0]

edited Nov 3, 2017 at 10:12

answered Nov 2, 2017 at 11:17

gil.fernandes

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8 Comments

ip696 Over a year ago

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3

gil.fernandes Over a year ago

@ip696: Yes, you can get that, if you use other inputs other than 1,2,3,4,5,6,7,8,9. But the snippets above run fine.

ip696 Over a year ago

I added the original value as I make them

gil.fernandes Over a year ago

@ip696 See updated solution with counter for alternative with other numbers besides 1,2,3,4,5,6,7,8,9.

ip696 Over a year ago

but if I need int[N][M] where N = 1000 000 and M = 1000 000?

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