如何将迭代器转换为流？

Question

我正在寻找一种Iterator将a 转换为Stream或更具体地以将“迭代器”作为流“查看”的简洁方法。

出于性能原因，我想避免在新列表中复制迭代器：

Iterator<String> sourceIterator = Arrays.asList("A", "B", "C").iterator();

Collection<String> copyList = new ArrayList<String>();

sourceIterator.forEachRemaining(copyList::add);

Stream<String> targetStream = copyList.stream();

根据评论中的一些建议，我也尝试使用Stream.generate：

public static void main(String[] args) throws Exception {

    Iterator<String> sourceIterator = Arrays.asList("A", "B", "C").iterator();

    Stream<String> targetStream = Stream.generate(sourceIterator::next);

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

}

但是，我得到了NoSuchElementException（因为没有调用hasNext）

Exception in thread "main" java.util.NoSuchElementException

    at java.util.AbstractList$Itr.next(AbstractList.java:364)

    at Main$$Lambda$1/1175962212.get(Unknown Source)

    at java.util.stream.StreamSpliterators$InfiniteSupplyingSpliterator$OfRef.tryAdvance(StreamSpliterators.java:1351)

    at java.util.Spliterator.forEachRemaining(Spliterator.java:326)

    at java.util.stream.ReferencePipeline$Head.forEach(ReferencePipeline.java:580)

    at Main.main(Main.java:20)

我已经看过StreamSupport和Collections，但我没有发现任何东西。

Answer 1

一种方法是从迭代器创建一个拆分器，并将其用作流的基础：Iterator<String> sourceIterator = Arrays.asList("A", "B", "C").iterator();Stream<String> targetStream = StreamSupport.stream(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Spliterators.spliteratorUnknownSize(sourceIterator, Spliterator.ORDERED),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; false);一种可能更易读的替代方法是使用Iterable-使用lambda从Iterator创建Iterable非常容易，因为Iterable是一个功能性接口：Iterator<String> sourceIterator = Arrays.asList("A", "B", "C").iterator();Iterable<String> iterable = () -> sourceIterator;Stream<String> targetStream = StreamSupport.stream(iterable.spliterator(), false);

Answer 2

很棒的建议！这是我可重复使用的方法：public class StreamUtils {&nbsp; &nbsp; public static <T> Stream<T> asStream(Iterator<T> sourceIterator) {&nbsp; &nbsp; &nbsp; &nbsp; return asStream(sourceIterator, false);&nbsp; &nbsp; }&nbsp; &nbsp; public static <T> Stream<T> asStream(Iterator<T> sourceIterator, boolean parallel) {&nbsp; &nbsp; &nbsp; &nbsp; Iterable<T> iterable = () -> sourceIterator;&nbsp; &nbsp; &nbsp; &nbsp; return StreamSupport.stream(iterable.spliterator(), parallel);&nbsp; &nbsp; }}和用法（确保静态导入asStream）：List<String> aPrefixedStrings = asStream(sourceIterator)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .filter(t -> t.startsWith("A"))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(toList());

Answer 3

在Java 9中，仅使用流就可以做到这一点，但是需要注意。以下朴素的示例将无法显示迭代器的最后一个元素。Stream.generate(iterator::next)&nbsp; &nbsp; .takeWhile(i -> iterator.hasNext())&nbsp;&nbsp; &nbsp; .forEach(System.out::println);但是，这确实可以正常工作：Stream.generate(() -> null)&nbsp; &nbsp; .takeWhile(x -> iterator.hasNext())&nbsp; &nbsp; .map(n -> iterator.next())&nbsp; &nbsp; .forEach(System.out::println);