如何将迭代器转换为流？

Question

我正在寻找一种简洁的方法来将Iterator一个Stream或更具体的转换为"视图"迭代器作为流.

出于性能原因,我想在新列表中避免使用迭代器的副本:

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Spliterator并将其用作流的基础:

Iterator<String> sourceIterator = Arrays.asList("A", "B", "C").iterator();
Stream<String> targetStream = StreamSupport.stream(
          Spliterators.spliteratorUnknownSize(sourceIterator, Spliterator.ORDERED),
          false);

一个可能更具可读性的替代方法是使用Iterable - 并且使用lambdas从Iterator创建Iterable非常容易,因为Iterable是一个功能接口:

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

Iterable<String> iterable = () -> sourceIterator;
Stream<String> targetStream = StreamSupport.stream(iterable.spliterator(), false);

Answer 2

从版本21开始,Guava库提供 Streams.stream(iterator)

它确实是@ assylias的答案.

Answer 3

很棒的建议!这是我的可重复使用:

public class StreamUtils {

    public static <T> Stream<T> asStream(Iterator<T> sourceIterator) {
        return asStream(sourceIterator, false);
    }

    public static <T> Stream<T> asStream(Iterator<T> sourceIterator, boolean parallel) {
        Iterable<T> iterable = () -> sourceIterator;
        return StreamSupport.stream(iterable.spliterator(), parallel);
    }
}

和用法(确保静态导入asStream):

List<String> aPrefixedStrings = asStream(sourceIterator)
                .filter(t -> t.startsWith("A"))
                .collect(toList());

Answer 4

这在java 9中可以使用除了流之外的任何东西,但是需要注意一点需要注意.以下天真的示例将无法打印迭代器的最后一个元素.

Stream.generate(iterator::next)
    .takeWhile(i -> iterator.hasNext()) 
    .forEach(System.out::println);

但是,这可以正常工作:

Stream.generate(() -> null)
    .takeWhile(x -> iterator.hasNext())
    .map(n -> iterator.next())
    .forEach(System.out::println);

Answer 5

Spliterator从Iterator使用Spliterators类创建包含了多个用于创建分隔符的函数，例如，此处使用的spliteratorUnknownSize是将迭代器作为参数，然后使用StreamSupport

Spliterator<Model> spliterator = Spliterators.spliteratorUnknownSize(
        iterator, Spliterator.NONNULL);
Stream<Model> stream = StreamSupport.stream(spliterator, false);

Answer 6

import com.google.common.collect.Streams;

并使用Streams.stream(iterator)：

Streams.stream(iterator)
       .map(v-> function(v))
       .collect(Collectors.toList());