其他流的笛卡尔积流，每个元素作为一个列表？答案

【问题标题】：Stream of cartesian product of other streams, each element as a List?其他流的笛卡尔积流，每个元素作为一个列表？
【发布时间】：2017-05-22 13:24:42
【问题描述】：

如何使用 Java 8 实现一个函数以获取一些流，并生成一个流，其中每个元素都是一个列表，由流的笛卡尔积的一个成员组成？

我查看了this question——该问题使用了一个BinaryOperator 的聚合器（获取两个相同类型的项目并生成一个相同类型的项目）。我希望最终结果中的项目是Lists，而不是输入流中元素的类型。

具体来说，假设我想要的函数叫做product，如下：

Stream<List<String>> result =
    product(
        Stream.of("A", "B", "C", "D"),
        Stream.of("I", "J", "K"),
        Stream.of("Y", "Z")
    );

result.forEach(System.out::println);

应该打印：

[A, I, Y]
[A, I, Z]
[A, J, Y]
[A, J, Z]
[A, K, Y]
[A, K, Z]
[B, I, Y]
...
[D, K, Y]
[D, K, Z]

理想情况下，我希望此操作尽可能懒惰。例如，如果输入流是由Stream.generate() 生成的，那么如果这些流的提供者在绝对需要时才执行，那就太好了。

【问题讨论】：

标签： java java-8 java-stream cartesian-product

【解决方案1】：

一个可能的解决方案如下：

private static <T> Stream<List<T>> product(Stream<T>... streams) {
    if (streams.length == 0) {
        return Stream.empty();
    }
    List<List<T>> cartesian = streams[streams.length - 1]
            .map(x -> Collections.singletonList(x))
            .collect(Collectors.toList());
    for (int i = streams.length - 2; i >= 0; i--) {
        final List<List<T>> previous = cartesian;
        cartesian = streams[i].flatMap(x -> previous.stream().map(p -> {
            final List<T> list = new ArrayList<T>(p.size() + 1);
            list.add(x);
            list.addAll(p);
            return list;
        })).collect(Collectors.toList());
    }
    return cartesian.stream();
}

public static void main(String... args) {
    final Stream<List<String>> result =
            product(
                    Stream.of("A", "B", "C", "D"),
                    Stream.of("I", "J", "K"),
                    Stream.of("Y", "Z")
            );

    result.forEach(System.out::println);
}

产品调用返回一个Stream<List<String>> 结果，打印为

[A, I, Y]
[A, I, Z]
[A, J, Y]
[A, J, Z]
[A, K, Y]
[A, K, Z]
[B, I, Y]
[B, I, Z]
[B, J, Y]
[B, J, Z]
[B, K, Y]
[B, K, Z]
[C, I, Y]
[C, I, Z]
[C, J, Y]
[C, J, Z]
[C, K, Y]
[C, K, Z]
[D, I, Y]
[D, I, Z]
[D, J, Y]
[D, J, Z]
[D, K, Y]
[D, K, Z]

【讨论】：

【解决方案2】：

你可以这样实现：

List<Stream<String>> listStreams = List.of(
        Stream.of("A", "B", "C", "D"),
        Stream.of("I", "J", "K"),
        Stream.of("Y", "Z"));

Stream<List<String>> streamLists = listStreams.stream()
        // represent each list element as SingletonList<Object>
        .map(stream -> stream.map(Collections::singletonList))
        // summation of pairs of inner lists
        .reduce((stream1, stream2) -> {
            // list of lists from second stream
            List<List<String>> list2 = stream2.collect(Collectors.toList());
            // append to the first stream
            return stream1.flatMap(inner1 -> list2.stream()
                    // combinations of inner lists
                    .map(inner2 -> {
                        List<String> list = new ArrayList<>();
                        list.addAll(inner1);
                        list.addAll(inner2);
                        return list;
                    }));
        }).orElse(Stream.empty());

// output
streamLists.forEach(System.out::println);

输出：

[A, I, Y]
[A, I, Z]
[A, J, Y]
[A, J, Z]
[A, K, Y]
[A, K, Z]
[B, I, Y]
[B, I, Z]
[B, J, Y]
[B, J, Z]
[B, K, Y]
[B, K, Z]
[C, I, Y]
[C, I, Z]
[C, J, Y]
[C, J, Z]
[C, K, Y]
[C, K, Z]
[D, I, Y]
[D, I, Z]
[D, J, Y]
[D, J, Z]
[D, K, Y]
[D, K, Z]

^{另见：Find cartesian product of 2 lists}

【讨论】：