How to generate a 2D int array using Java 8 stream?

Question

I am trying to generate a int[][] using Java 8 streams.

This is what I have done so far:

objects.parallelStream()
            .map(o -> o.getPropertyOnes()
                    .parallelStream()
                    .map(t-> t.getIndex())  //<-- getIndex() returns int
                    .mapToInt(i -> i)
                    .toArray())            //<-- here I have a Stream<int[]>
            .toArray();                   //lost here

At the end of outer.map() I have a Stream<int[]>, but not sure how to convert that to int[][]. Please suggest.

@Eugene No, don't want to flatten the collection, rather want to create a 2d int array. — Sayan Pal
– Sayan Pal, Commented Nov 30, 2016 at 13:57
how would you convert a Stream<int[]> to int[][]? Can you provide and example? Even better show us your input and the desired output, it's unclear what you are trying to do — Eugene
– Eugene, Commented Nov 30, 2016 at 14:00
Instead of the two step .map(t -> t.getIndex()) .mapToInt(i -> i), you can use straight-forwardly mapToInt(t -> t.getIndex()). — Holger
– Holger, Commented Nov 30, 2016 at 15:29

Flown · Accepted Answer · 2016-12-01 10:20:30Z

7

First you can simplify the map().mapToInt() to mapToInt(t-> t.getIndex()) (maybe you should use a method reference like <type>::getIndex).

As you said you have Stream<int[]> after the map stage. Then you only need to provide an array generator function like:

int[][] array = Stream.of(1, 2, 3, 4)
                      .map(i -> IntStream.range(0, i).toArray())
                      .toArray(int[][]::new);

Output:

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

edited Dec 1, 2016 at 10:20

answered Nov 30, 2016 at 14:02

Flown

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2 Comments

Sayan Pal Over a year ago

Thank you for the answer. I still need to wait couple of minutes to accept it. One question though, I tried to do toArray(int[]::new) for the inner map, i.e. instead of .map(t-> t.getIndex()).mapToInt(i -> i).toArray(), tried this: .map(t-> t.getIndex()).toArray(int[]::new), but it didn't work. Any idea why?

Flown Over a year ago

@SayanPal Well you have a Stream<Integer> this cannot be stored int an int[]. Stream::mapToInt returns an IntStream which provides a toArray method which returns an int[]. This is exactly what you need.

Andrew · Accepted Answer · 2016-11-30 15:04:51Z

You need a toArray(generator) method that helps us to specify a return type by an IntFunction<T[]> function:

int[][] a = Stream.of(new int[]{1, 2, 3}, new int[]{4, 5, 6}).toArray(int[][]::new);

instead of toArray(), which returns Object[] regardless of an incoming type in a Stream (invokes toArray(Object[]::new) internally):

Object[] a = Stream.of(new int[]{1, 2, 3}, new int[]{4, 5, 6}).toArray();

If you are interested in, behind the scenes, all of this have the following look:

forming a Node [an ArrayNode in our case] (an immutable container for keeping an ordered sequence of elements) from a Pipeline of the previous stage;
array size calculation by using the node size (node.count());
getting a new empty array with a needed length by our IntFunction<T[]> generator (i -> new int[i][] or simply int[][]::new);
copying from the node to this array and returning the latter.

Ritam Chakraborty · Accepted Answer · 2019-07-20 16:14:51Z

1

If you want to generate 2D array of sequential number of n X n where n in the degree.

int n = 4;
System.out.println(Arrays.deepToString(IntStream.rangeClosed(0, n - 1)
        .boxed().map(x -> IntStream.rangeClosed(x * n + 1, (x + 1) * n)
            .boxed().toArray()).toArray()));

Output: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

answered Jul 20, 2019 at 16:14

Ritam Chakraborty

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