通过JavaStream实现多个集合的笛卡尔产品

Question

现在我只能实现两个集合的笛卡尔积，下面是代码：

public static <T1, T2, R extends Collection<Pair<T1, T2>>>
R getCartesianProduct(
        Collection<T1> c1, Collection<T2> c2,
        Collector<Pair<T1, T2>, ?, R> collector) {
    return c1.stream()
            .flatMap(e1 -> c2.stream().map(e2 -> new Pair<>(e1, e2)))
            .collect(collector);
}

这段代码在IntelliJ中运行良好，但在Eclipse中不起作用。编译器符合性级别均为1.8：

The method collect(Collector<? super Object,A,R>)
in the type Stream<Object> is not applicable for
the arguments (Collector<Pair<T1,T2>,capture#5-of ?,R>)

这里是Pair.java：

public class Pair<T1, T2> implements Serializable {
    protected T1 first;
    protected T2 second;
    private static final long serialVersionUID = 1360822168806852921L;

    public Pair(T1 first, T2 second) {
        this.first = first;
        this.second = second;
    }

    public Pair(Pair<T1, T2> pair) {
        this(pair.getFirst(), pair.getSecond());
    }

    public T1 getFirst() {
        return this.first;
    }

    public T2 getSecond() {
        return this.second;
    }

    public void setFirst(T1 o) {
        this.first = o;
    }

    public void setSecond(T2 o) {
        this.second = o;
    }

    public String toString() {
        return "(" + this.first + ", " + this.second + ")";
    }

    @Override
    public boolean equals(Object o) {
        if(!(o instanceof Pair))
            return false;
        Pair p = (Pair) o;
        if(!this.first.equals(p.first))
            return false;
        if(!this.second.equals(p.second))
            return false;
        return true;
    }

    @Override
    public int hashCode() {
        int hash = 1;
        hash = hash * 31 + this.first.hashCode();
        hash = hash * 31 + this.second.hashCode();
        return hash;
    }
}

如何修复此错误？

有没有一种优雅的方法来实现几个集合的笛卡尔产品？假设我们有类tuple。

Answer 1

您可以使用Supplier作为收集器的参数。toCollection方法。这简化了代码并使其更通用。不同类型和数量的集合及其元素的笛卡尔积。

在线试试吧！

public static void main(String[] args) {
    Set<Integer> a = Set.of(1, 2);
    Set<Character> b = Set.of('*', '+');
    List<String> c = List.of("A", "B");

    Set<Set<Object>> cpSet = cartesianProduct(HashSet::new, a, b, c);
    List<List<Object>> cpList = cartesianProduct(ArrayList::new, a, b, c);

    // output, order may vary
    System.out.println(cpSet);
    System.out.println(cpList);
}

java prettyprint-override">/**
 * @param nCol the supplier of the output collection
 * @param cols the input array of collections
 * @param <R>  the type of the return collection
 * @return the cartesian product of the multiple collections
 */
@SuppressWarnings("unchecked")
public static <R extends Collection<?>> R cartesianProduct(
        Supplier nCol, Collection<?>... cols) {
    // check if supplier is not null
    if (nCol == null) return null;
    return (R) Arrays.stream(cols)
        // non-null and non-empty collections
        .filter(col -> col != null && col.size() > 0)
        // represent each element of a collection as a singleton collection
        .map(col -> (Collection<Collection<?>>) col.stream()
            .map(e -> Stream.of(e).collect(Collectors.toCollection(nCol)))
            .collect(Collectors.toCollection(nCol)))
        // summation of pairs of inner collections
        .reduce((col1, col2) -> (Collection<Collection<?>>) col1.stream()
            // combinations of inner collections
            .flatMap(inner1 -> col2.stream()
                // concatenate into a single collection
                .map(inner2 -> Stream.of(inner1, inner2)
                    .flatMap(Collection::stream)
                    .collect(Collectors.toCollection(nCol))))
            // collection of combinations
            .collect(Collectors.toCollection(nCol)))
        // otherwise an empty collection
        .orElse((Collection<Collection<?>>) nCol.get());
}

输出（顺序可能有所不同）：

[[1,A,*],[1,B,*],[1,A,+],[A,2,*],[1,B,+],[2,B,*],[A,2,+],[2,B,+]]
[[2,+,A],[2,+,B],[2,*,A],[2,*,B],[1,+,A],[1,+,B],[1,*,A],[1,*,B]]

^{另请参见不太通用的版本：在Java中查找笛卡尔积}

Answer 2

这里有一个解决方案，它概括了编译时不知道必需的平面地图应用程序的数量（即产品的顺序）的情况。

BinaryOperator<Function<String,Stream<String>>> kleisli = (f,g) -> s -> f.apply(s).flatMap(g);

List<String> cartesian(int n, Collection<String> coll) {
    return coll.stream()
           .flatMap( IntStream.range(1, n).boxed()
                     .map(_any -> crossWith(coll::stream, String::concat)) // create (n-1) appropriate crossWith instances
                     .reduce(s -> Stream.of(s), kleisli)                   // compose them sequentially
                    )                                                      // flatMap the stream with the entire function chain
           .collect(toList());
}

你会在我自己的博客上找到这是如何工作的细节。

Answer 3

Eclipse在类型推断方面存在问题。如果添加类型提示。

如果我可以建议一个不同的方法，考虑让你的cartesianProducts不做整个流和集合，而只是一个帮手的平面地图：

static <T1, T2, R> Function<T1, Stream<R>> crossWith(
         Supplier<? extends Stream<T2>> otherSup, 
         BiFunction<? super T1, ? super T2, ? extends R> combiner
) {
    return t1 -> otherSup.get().map(t2 -> combiner.apply(t1, t2));
}

现在，如果希望结果包含Pairs，则只需创建Pair，并且可以通过多次应用flatMap来进行高阶笛卡尔积：

List<String> letters = Arrays.asList("A", "B", "C");
List<Integer> numbers = Arrays.asList(1, 2, 3);

List<Pair<String, Integer>> board = letters.stream()
                .flatMap(crossWith(numbers::stream, Pair::new))
                .collect(toList());


List<String> ops = Arrays.asList("+", "-", "*", "/");

List<String> combinations = letters.stream()
                .flatMap(crossWith(ops::stream, String::concat))
                .flatMap(crossWith(letters::stream, String::concat))
                .collect(toList());   // triple cartesian product