Concatenate list of Objects arrays in Java8

Question

I have list of 1D arrays wrapped in an object, I want to collect them in 1 single array using java-8 stream and collect.

Given class Item with an array of

class A{
 int x;
}


class Item{
    A[] parts=new A[];
    public A[] getParts(){
       return parts;
    }
}

So if I have list l

List<Item> l=new ArrayList<>();
l.add(new Item());
l.add(new Item());

I need to collect the content in these two object in one single array of type A.

So if first Item has parts={1,2} and 2nd Item parts={3,4} the output should be somehow like

A[] out={1,2,3,4}

I tried to do

l.stream().map(Item::getParts).collect(Collectors.toList()).toArray(new A[n]);

but apperantly it has meany problems.

I deleted my answer, you will learn more by reading the Stream tutorial docs.oracle.com/javase/tutorial/collections/streams. Your code doesn't compile. — Tunaki
– Tunaki, Commented Apr 17, 2016 at 14:34
Your method is named getParts(), not getparts(). And Stream as a toArray() method, as well as a flatMap() method.. — JB Nizet
– JB Nizet, Commented Apr 17, 2016 at 14:37
In order to stream the elements of an array, you must use Arrays.stream(myArray). Currently, your list is a list of arrays, not a list of elements from the arrays. — Andreas
– Andreas, Commented Apr 17, 2016 at 14:37

Yassin Hajaj · Accepted Answer · 2016-04-18 16:55:13Z

3

Map each Item to its A[] array. Then when having a stream of A[] just flat map it to get each A of each array and make it one array at the end.

A[] array = l.stream().map(x -> x.getParts())
                      .flatMap(x -> Arrays.stream(x))
                      .toArray(A[]::new); // [1, 2, 3, 4]

However, if you only interested in the int value, there is an extra step

int[] array = l.stream().map(x -> x.getParts())
                      .flatMap(x -> Arrays.stream(x))
                      .mapToInt(x -> x.getX())
                      .toArray();

edited Apr 18, 2016 at 16:55

answered Apr 18, 2016 at 5:35

Yassin Hajaj

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1 Comment

Holger Over a year ago

I recommend using Arrays.stream instead of Stream.of in this situation.

Tassos Bassoukos · Accepted Answer · 2016-04-17 14:59:28Z

1

You want to use .flatMap() instead of a .map(); in the lambda just generate a new stream of the items in the array.

answered Apr 17, 2016 at 14:59

Tassos Bassoukos

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Comments

Chota Bheem · Accepted Answer · 2016-04-17 16:31:37Z

Try following:

static void test() {
        List<Item> l = new ArrayList<>();
        l.add(new Item(1, 2,5));
        l.add(new Item(3, 4,10));

        System.out.println(l);

        int totalNoOfParts = l.stream().mapToInt(item->item.getParts().length).reduce(0, Integer::sum);

        A[] allParts = l.stream().map(item -> item.getParts()).collect(() -> new A[totalNoOfParts],
                (c, e) -> System.arraycopy(e, 0, c, getNextPos(c), e.length),
                (c1, c2) -> System.arraycopy(c2, 0, c1, getNextPos(c1), c2.length));

        System.out.println(Arrays.asList(allParts));

    }

    static int getNextPos(A[] a) {
        for (int i = 0; i < a.length; i++) {
            if (a[i] == null)
                return i;
        }
        throw new InvalidParameterException("Invalid index value");
    }

I modified your Item class to have fixed length of array.

class Item {

    A[] parts = new A[3];

    Item(int... values) {

        Objects.requireNonNull(values, "Argument Not Valid");

        for (int i = 0; i < values.length; i++) {
            parts[i] = new A();
            parts[i].x = values[i];
        }
    }

    public A[] getParts() {
        return parts;
    }
}

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