如何交错(合并)两个 Java 8 流?-Java问题

How to interleave (merge) two Java 8 Streams?(如何交错(合并)两个 Java 8 流?)

本文介绍了如何交错(合并)两个 Java 8 流?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

 流<字符串>a = Stream.of("一", "三", "五");流<字符串>b = Stream.of("二", "四", "六");

我需要做什么才能使输出如下?

//一个//二//三//四个//五//六

我查看了 concat 但正如 javadoc 解释的那样，它只是一个接一个地附加，它不会交错/穿插.

流<字符串>out = Stream.concat(a, b);out.forEach(System.out::println);

<块引用>

创建一个惰性连接流，其元素都是第一个流的元素，然后是第二个流.

错误地给予

//一个//三//五//二//四个//六

如果我收集它们并进行迭代可以做到，但希望有更多的 Java8-y、Streamy :-)

注意

我不想压缩流

<块引用>

zip"操作将从每个集合中取出一个元素并将它们组合起来.

zip 操作的结果是这样的:(不需要的)

//onetwo//三四//五六

解决方案

我会使用这样的:

公共静态<T>流<T>interleave(Streama，Streamb) {分离器动作){分离器

它尽可能地保留了输入流的特征，这允许进行某些优化(例如对于count()和toArray()).此外，即使输入流可能是无序的，它也会添加 ORDERED 以反映交错.

当一个流的元素多于另一个时，剩余的元素将出现在最后.

 Stream<String> a = Stream.of("one", "three", "five");
 Stream<String> b = Stream.of("two", "four", "six");

What do I need to do for the output to be the below?

// one
// two
// three
// four
// five
// six

I looked into concat but as the javadoc explains, it just appends one after the other, it does not interleave / intersperse.

Stream<String> out = Stream.concat(a, b);
out.forEach(System.out::println);

Creates a lazily concatenated stream whose elements are all the elements of the first stream followed by all the elements of the second stream.

Wrongly gives

 // one
 // three
 // five
 // two
 // four
 // six

Could do it if I collected them and iterated, but was hoping for something more Java8-y, Streamy :-)



Note

I don't want to zip the streams 

  "zip" operation will take an element from each collection and combine them.
the result of a zip operation would be something like this: (unwanted)
 // onetwo
 // threefour
 // fivesix

 解决方案 
I’d use something like this:
public static <T> Stream<T> interleave(Stream<? extends T> a, Stream<? extends T> b) {
    Spliterator<? extends T> spA = a.spliterator(), spB = b.spliterator();
    long s = spA.estimateSize() + spB.estimateSize();
    if(s < 0) s = Long.MAX_VALUE;
    int ch = spA.characteristics() & spB.characteristics()
           & (Spliterator.NONNULL|Spliterator.SIZED);
    ch |= Spliterator.ORDERED;

    return StreamSupport.stream(new Spliterators.AbstractSpliterator<T>(s, ch) {
        Spliterator<? extends T> sp1 = spA, sp2 = spB;

        @Override
        public boolean tryAdvance(Consumer<? super T> action) {
            Spliterator<? extends T> sp = sp1;
            if(sp.tryAdvance(action)) {
                sp1 = sp2;
                sp2 = sp;
                return true;
            }
            return sp2.tryAdvance(action);
        }
    }, false);
}
It retains the characteristics of the input streams as far as possible, which allows certain optimizations (e.g. for count()and toArray()). Further, it adds the ORDERED even when the input streams might be unordered, to reflect the interleaving.

When one stream has more elements than the other, the remaining elements will appear at the end.

                        这篇关于如何交错(合并)两个 Java 8 流?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！