Question

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

我已查看this question - 该问题使用的聚合器为BinaryOperator（采用类似类型的两个项目并生成相同类型的项目）。我希望最终结果中的项目为List，而不是输入流中元素的类型。

具体而言，假设我所需的函数名为product，请执行以下操作：

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地懒惰。例如，如果输入流是由Stream.generate()生成的，那么如果这些流的供应商在绝对需要之前没有被执行，那么它就会很棒。

Answer 1

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]

Answer 2

你可以实现如下：

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}

其他流的笛卡尔积的流，每个元素作为List？

2 个答案: