Applying the `collec()` of stream for counting. Exercise - java-stream

Question

I am trying to create a custom collector in order to count valid elements of a list. I have done it using one of the already provide collectors:
arr.stream()
.filter(e -> e.matches("[:;][~-]?[)D]"))
.map(e -> 1)
.reduce(0, Integer::sum);
but as a challenge for myself, I wanted to create my own custom collector in order to understand it better. And this is where I got stuck.
It is probably something trivial but I am learning this and can't figure a supplier, an accumulator, and a combiner. I guess I still don't understand something about them. For instance, I have a similar stream:
arr1.stream()
.filter(e -> e.matches("[:;][~-]?[)D]"))
.map(e -> 1)
.collect(temporary array, adding to array, reduce);
AFAIK supplier is a function without arguments, which returns something. I studied standard examples and it is usually a method reference for a new collection, e.g. ArrayList::new. I tried to use constant 0, e -> 0 because I want to return a scalar. I think it is wrong because it makes the stream returning 0. If using method reference for a temporary collection, Java complains about a mismatch of types of a supplier and returning object. I am also confused about using an accumulator if the final result is a number as a combiner would reduce all elements to a number, e.g. (a,b) -> a + b`.
I'm completely stumped.

Answer 1

Probably part of your problem is that you cannot obviously create an accumulator for an Integer type since it is immutable.
You start with this:
System.out.println(IntStream.of(1,2,3).reduce(0, Integer::sum));
You can extend to this:
System.out.println(IntStream.of(1,2,3).boxed()
.collect(Collectors.reducing(0, (i1,i2)->i1+i2)));
Or even this, which has an intermediate mapping function
System.out.println(IntStream.of(1,2,3).boxed()
.collect(Collectors.reducing(0, i->i*2, (i1,i2)->i1+i2)));
You can get this far with your own Collector
Collector<Integer, Integer, Integer> myctry = Collector.of(
()->0,
(i1,i2)->{
// what to do here?
},
(i1,i2)->{
return i1+i2;
}
);
The accumulator is A function that folds a value into a mutable result container with mutable being the keyword here.
So, make a mutable integer
public class MutableInteger {
private int value;
public MutableInteger(int value) {
this.value = value;
}
public void set(int value) {
this.value = value;
}
public int intValue() {
return value;
}
}
And now:
Collector<MutableInteger, MutableInteger, MutableInteger> myc = Collector.of(
()->new MutableInteger(0),
(i1,i2)->{
i1.set(i1.intValue()+i2.intValue());
},
(i1,i2)->{
i1.set(i1.intValue()+i2.intValue());
return i1;
}
);
And then:
System.out.println(IntStream.of(1,2,3)
.mapToObj(MutableInteger::new)
.collect(myc).intValue());
Reference:
Example of stream reduction with distinct combiner and accumulator
EDIT: The finisher just does whatever with the final result. If you don't set it on purpose then it is set by default to IDENTITY_FINISH which is Function.identity() which says just to return the final result as is.
EDIT: If you're really desperate:
Collector<int[], int[], int[]> mycai = Collector.of(
()->new int[1],
(i1,i2)->i1[0] += i2[0],
(i1,i2)->{i1[0] += i2[0]; return i1;}
);
System.out.println(IntStream.of(1,2,3)
.mapToObj(v->{
int[] i = new int[1];
i[0] = v;
return i;
})
.collect(mycai)[0]);