Functional implementation of unzip for potentially infinite threads

Question

I implemented the functional operation unzip()as follows:

public static <T, U, V> Tuple2<Stream<U>, Stream<V>> unzip(
        Stream<T> stream, 
        Function<T, Tuple2<U, V>> unzipper) {

    return stream.map(unzipper)
        .reduce(new Tuple2<>(Stream.<U>empty(), Stream.<V>empty()),
            (unzipped, tuple) -> new Tuple2<>(
                Stream.concat(unzipped.$1(), Stream.of(tuple.$1())),
                Stream.concat(unzipped.$2(), Stream.of(tuple.$2()))),
            (unzipped1, unzipped2) -> new Tuple2<>(
                Stream.concat(unzipped1.$1(), unzipped2.$1()),
                Stream.concat(unzipped1.$2(), unzipped2.$2())));
}

This works fine because there are not many elements in the data streams. This is because access to a deeply concatenated stream element can cause StackOverflowException. According to the docsStream.concat() :

Implementation Note:

Use caution when building streams from re-concatenation. Access to a deeply concatenated stream item can lead to deep chains of calls or even to StackOverflowException.

For several elements, my implementation is done unzip. For class Person:

class Person {

    public final String name;

    public final int age;

    Person(String name, int age) {
        this.name = name;
        this.age = age;
    }
}

If I have a stream of people:

Stream<Person> people = Stream.of(
    new Person("Joe", 52), 
    new Person("Alan", 34), 
    new Person("Peter", 42));

I can use the implementation unzip()this way:

Tuple2<Stream<String>, Stream<Integer>> result = StreamUtils.unzip(people, 
        person -> new Tuple2<>(person.name, person.age));

List<String> names = result.$1()
    .collect(Collectors.toList()); // ["Joe", "Alan", "Peter"]
List<Integer> ages = result.$2()
    .collect(Collectors.toList()); // [52, 34, 42]

It is right.

, : unzip() ( )?

: , Tuple2:

public final class Tuple2<A, B> {

    private final A $1;

    private final B $2;

    public Tuple2(A $1, B $2) {
        this.$1 = $1;
        this.$2 = $2;
    }

    public A $1() {
        return $1;
    }

    public B $2() {
        return $2;
    }
}

Answer 1

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public static <T, A1, A2, R1, R2> Collector<T, ?, Tuple2<R1,R2>> both(
    Collector<T, A1, R1> first, Collector<T, A2, R2> second) {

    Supplier<A1> s1=first.supplier();
    Supplier<A2> s2=second.supplier();
    BiConsumer<A1, T> a1=first.accumulator();
    BiConsumer<A2, T> a2=second.accumulator();
    BinaryOperator<A1> c1=first.combiner();
    BinaryOperator<A2> c2=second.combiner();
    Function<A1,R1> f1=first.finisher();
    Function<A2,R2> f2=second.finisher();
    return Collector.of(
        ()->new Tuple2<>(s1.get(), s2.get()),
        (p,t)->{ a1.accept(p.$1(), t); a2.accept(p.$2(), t); },
        (p1,p2)->new Tuple2<>(c1.apply(p1.$1(), p2.$1()), c2.apply(p1.$2(), p2.$2())),
        p -> new Tuple2<>(f1.apply(p.$1()), f2.apply(p.$2())));
}

Tuple2<List<String>, List<Integer>> namesAndAges=
    Stream.of(new Person("Joe", 52), new Person("Alan", 34), new Person("Peter", 42))
        .collect(both(
            Collectors.mapping(p->p.name, Collectors.toList()),
            Collectors.mapping(p->p.age,  Collectors.toList())));
List<String> names = namesAndAges.$1(); // ["Joe", "Alan", "Peter"]
List<Integer> ages = namesAndAges.$2(); // [52, 34, 42]

. , .

, Tuple2, ,

public static <T, T1, T2, A1, A2, R1, R2> Collector<T, ?, Tuple2<R1,R2>> both(
    Function<? super T, ? extends Tuple2<? extends T1, ? extends T2>> f,
    Collector<T1, A1, R1> first, Collector<T2, A2, R2> second) {

    return Collectors.mapping(f, both(
            Collectors.mapping(Tuple2::$1, first),
            Collectors.mapping(Tuple2::$2, second)));
}

Tuple2<List<String>, List<Integer>> namesAndAges=
    Stream.of(new Person("Joe", 52), new Person("Alan", 34), new Person("Peter", 42))
        .collect(both(
            p -> new Tuple2<>(p.name, p.age), Collectors.toList(), Collectors.toList()));

p -> new Tuple2<>(p.name, p.age), , unzip. , "", . Stream , unzip, , , concat, :

public static <T, U, V> Tuple2<Stream<U>, Stream<V>> unzip(
    Stream<T> stream,  Function<T, Tuple2<U, V>> unzipper) {

    return stream.map(unzipper)
        .collect(Collector.of(()->new Tuple2<>(Stream.<U>builder(), Stream.<V>builder()),
            (unzipped, tuple) -> {
                unzipped.$1().accept(tuple.$1()); unzipped.$2().accept(tuple.$2());
            },
            (unzipped1, unzipped2) -> {
                unzipped2.$1().build().forEachOrdered(unzipped1.$1());
                unzipped2.$2().build().forEachOrdered(unzipped1.$2());
                return unzipped1;
            },
            tuple -> new Tuple2<>(tuple.$1().build(), tuple.$2().build())
        ));
}

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