我想获取字符串流并将其转换为单词对流。例如:

我有:{ "A", "Apple", "B", "Banana", "C", "Carrot"}

我想要:{ ("A", "Apple"), ("Apple", "B"), ("B", "Banana"), ("Banana", "C") } 。

这与 Zipping 几乎相同，如 Zipping streams using JDK8 with lambda (java.util.stream.Streams.zip) 中所述

但是，这会产生: {(A，苹果)，(B，香蕉)，(C，胡萝卜)}

下面的代码有效，但显然是错误的方法(不是线程安全的等等):

static String buffered = null;

static void output(String s) {
    String result = null;
    if (buffered != null) {
        result = buffered + "," + s;
    } else {
        result = null;
    }

    buffered = s;
    System.out.println(result);
}

// ***** 

Stream<String> testing = Stream.of("A", "Apple", "B", "Banana", "C", "Carrot");
testing.forEach(s -> {output(s);});

最佳答案

如果你:

1. 不喜欢创建包含流中所有字符串的列表的想法
2. 不想使用外部库
3. 喜欢亲自动手

然后您可以创建一个方法来使用 Java 8 低级流构建器对流中的元素进行分组 StreamSupport和 Spliterator :

class StreamUtils {
    public static<T> Stream<List<T>> sliding(int size, Stream<T> stream) {
        return sliding(size, 1, stream);
    }

    public static<T> Stream<List<T>> sliding(int size, int step, Stream<T> stream) {
        Spliterator<T> spliterator = stream.spliterator();
        long estimateSize;

        if (!spliterator.hasCharacteristics(Spliterator.SIZED)) {
            estimateSize = Long.MAX_VALUE;
        } else if (size > spliterator.estimateSize()) {
            estimateSize = 0;
        } else {
            estimateSize = (spliterator.estimateSize() - size) / step + 1;
        }

        return StreamSupport.stream(
                new Spliterators.AbstractSpliterator<List<T>>(estimateSize, spliterator.characteristics()) {
                    List<T> buffer = new ArrayList<>(size);

                    @Override
                    public boolean tryAdvance(Consumer<? super List<T>> consumer) {
                        while (buffer.size() < size && spliterator.tryAdvance(buffer::add)) {
                            // Nothing to do
                        }

                        if (buffer.size() == size) {
                            List<T> keep = new ArrayList<>(buffer.subList(step, size));
                            consumer.accept(buffer);
                            buffer = keep;
                            return true;
                        }
                        return false;
                    }
                }, stream.isParallel());
    }
}

方法和参数命名的灵感来自于它们在 Scala 中的对应项。

让我们测试一下:

Stream<String> testing = Stream.of("A", "Apple", "B", "Banana", "C", "Carrot");
System.out.println(StreamUtils.sliding(2, testing).collect(Collectors.toList()));

[[A, Apple], [Apple, B], [B, Banana], [Banana, C], [C, Carrot]]

关于不重复元素呢:

Stream<String> testing = Stream.of("A", "Apple", "B", "Banana", "C", "Carrot");
System.out.println(StreamUtils.sliding(2, 2, testing).collect(Collectors.toList()));

[[A, Apple], [B, Banana], [C, Carrot]]

现在有了无限的 Stream:

StreamUtils.sliding(5, Stream.iterate(0, n -> n + 1))
        .limit(5)
        .forEach(System.out::println);

[0, 1, 2, 3, 4]
[1, 2, 3, 4, 5]
[2, 3, 4, 5, 6]
[3, 4, 5, 6, 7]
[4, 5, 6, 7, 8]

关于java - 如何将字符串流转换为字符串流对？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32162478/