Java 8 groupingBy list list

Question

Say we have a Draw object:

class Draw {

private int num1, num2, num3;

public Draw (int num1, int num2, int num3) {
    this.num1 = num1;
    this.num2 = num2;
    this.num3 = num3;
}

public int getNum1() {
    return num1;
}

public void setNum1(int num1) {
    this.num1 = num1;
}

public int getNum2() {
    return num2;
}

public void setNum2(int num2) {
    this.num2 = num2;
}

public int getNum3() {
    return num3;
}

public void setNum3(int num3) {
    this.num3 = num3;
}

public List<Integer> getResultAsList() {
    return Arrays.asList(num1, num2, num3);
}

}

Now, having a Draws list, I need to get a map, where the key is the number in the draw and the value is the number (how many times this number was drawn)

For example, having

List<Draw> drawList = Arrays.asList(new Draw(1,2,5), new Draw(1,5,6));

I want to get the following card: {1 = 2, 2 = 1, 5 = 2, 6 = 1}

Is it possible to implement this with new groupingBy java 8 operations or another new function?

Thank.

Answer 1

If everything works out correctly:

Map<Integer, Long> map = drawList
      .stream()
      .flatMap(x -> Stream.of(x.getNum1(), x.getNum2(), x.getNum3()))
      .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

System.out.println(map); // {1=2, 2=1, 5=2, 6=1}

Or, obviously, like this:

Map<Integer, Long> map2 = drawList
            .stream()
            .flatMap(x -> x.getResultAsList().stream())
            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));