如何将迭代器转换为流?

我正在寻找一种简洁的方法将迭代器转换为流或更具体地说，将迭代器“查看”为流。

出于性能考虑，我希望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);
}

然而，我得到一个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)

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

一种方法是从迭代器中创建一个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);

另一种可能更具可读性的方法是使用Iterable——使用lambdas从Iterator创建Iterable非常容易，因为Iterable是一个函数式接口:

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

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

2014-07-01 13:29:33

好建议! 以下是我对可重复使用的看法:

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());

2015-01-23 21:16:53

Spliterators类包含多个函数来创建Spliterator，例如这里我使用spliteratorUnknownSize来获取Iterator作为参数，然后使用StreamSupport创建Stream

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

2017-02-11 12:08:32

从版本21开始，Guava库提供了Streams.stream(迭代器)

它做了@assylias的答案所显示的。

2017-03-02 08:53:10

使用Collections.list (iterator) .stream()……

2018-04-30 15:24:45

这在Java 9中是可能的。

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

2018-05-07 08:49:14

import com.google.common.collect.Streams;

并使用Streams.stream(迭代器):

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

2020-03-09 04:49:36

在Java 9+上使用Stream::iterate(T, Predicate, UnaryOperator)的另一种方法:

Stream.iterate(iterator, Iterator::hasNext, UnaryOperator.identity())
        .map(Iterator::next)
        .forEach(System.out::println);

2020-06-11 19:28:36

1 .小梁的溶液包法:

public static <T> Stream<T> toStream(Iterator<T> iterator) {
    return StreamSupport.stream(((Iterable<T>)() -> iterator).spliterator(), false);
}

2番石榴流实现(用@Beta标记):

public static <T> Stream<T> stream(Iterator<T> iterator) {
    return StreamSupport.stream(Spliterators.spliteratorUnknownSize(iterator, 0), false);
}

2021-04-08 08:31:10

如果迭代规模已知，这是可能的:

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.**limit(3)**.forEach(System.out::println);
}

2023-01-31 16:44:58

如何将迭代器转换为流?

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