java - 在 Java 8 中是否有一种简洁的方法来迭代带有索引的流？

Question

是否有一种简洁的方法可以在访问流中的索引的同时迭代流？

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};

List<String> nameList;
Stream<Integer> indices = intRange(1, names.length).boxed();
nameList = zip(indices, stream(names), SimpleEntry::new)
        .filter(e -> e.getValue().length() <= e.getKey())
        .map(Entry::getValue)
        .collect(toList());

与那里给出的 LINQ 示例相比，这似乎相当令人失望

string[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
var nameList = names.Where((c, index) => c.Length <= index + 1).ToList();

有没有更简洁的方法？

此外，似乎拉链已经移动或被移除......

score 524 · Accepted Answer

最干净的方法是从索引流开始：

String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};
IntStream.range(0, names.length)
         .filter(i -> names[i].length() <= i)
         .mapToObj(i -> names[i])
         .collect(Collectors.toList());

结果列表仅包含“Erik”。

当您习惯于 for 循环时，一种看起来更熟悉的替代方法是使用可变对象维护一个临时计数器，例如AtomicInteger：

String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};
AtomicInteger index = new AtomicInteger();
List<String> list = Arrays.stream(names)
                          .filter(n -> n.length() <= index.incrementAndGet())
                          .collect(Collectors.toList());

请注意，在并行流上使用后一种方法可能会中断，因为项目不一定会“按顺序”处理。

score 89 · Accepted Answer

Java 8 流 API 缺乏获取流元素索引的功能以及将流压缩在一起的能力。这是不幸的，因为它使某些应用程序（如 LINQ 挑战）比其他应用程序更加困难。

但是，通常有解决方法。通常这可以通过使用整数范围“驱动”流来完成，并利用原始元素通常位于数组或可通过索引访问的集合中这一事实。例如，挑战 2 问题可以这样解决：

String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};

List<String> nameList =
    IntStream.range(0, names.length)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(toList());

正如我上面提到的，这利用了数据源（名称数组）可直接索引的事实。如果不是这样，这种技术就行不通。

我承认这并不能满足挑战 2 的意图。尽管如此，它确实合理有效地解决了问题。

编辑

我之前的代码示例用于flatMap融合过滤器和映射操作，但这很麻烦并且没有提供任何优势。我已经根据 Holger 的评论更新了示例。

score 56 · Accepted Answer

从番石榴 21 开始，您可以使用

Streams.mapWithIndex()

示例（来自官方文档）：

Streams.mapWithIndex(
    Stream.of("a", "b", "c"),
    (str, index) -> str + ":" + index)
) // will return Stream.of("a:0", "b:1", "c:2")

score 27 · Accepted Answer

我在我的项目中使用了以下解决方案。我认为这比使用可变对象或整数范围要好。

import java.util.*;
import java.util.function.*;
import java.util.stream.Collector;
import java.util.stream.Collector.Characteristics;
import java.util.stream.Stream;
import java.util.stream.StreamSupport;
import static java.util.Objects.requireNonNull;


public class CollectionUtils {
    private CollectionUtils() { }

    /**
     * Converts an {@link java.util.Iterator} to {@link java.util.stream.Stream}.
     */
    public static <T> Stream<T> iterate(Iterator<? extends T> iterator) {
        int characteristics = Spliterator.ORDERED | Spliterator.IMMUTABLE;
        return StreamSupport.stream(Spliterators.spliteratorUnknownSize(iterator, characteristics), false);
    }

    /**
     * Zips the specified stream with its indices.
     */
    public static <T> Stream<Map.Entry<Integer, T>> zipWithIndex(Stream<? extends T> stream) {
        return iterate(new Iterator<Map.Entry<Integer, T>>() {
            private final Iterator<? extends T> streamIterator = stream.iterator();
            private int index = 0;

            @Override
            public boolean hasNext() {
                return streamIterator.hasNext();
            }

            @Override
            public Map.Entry<Integer, T> next() {
                return new AbstractMap.SimpleImmutableEntry<>(index++, streamIterator.next());
            }
        });
    }

    /**
     * Returns a stream consisting of the results of applying the given two-arguments function to the elements of this stream.
     * The first argument of the function is the element index and the second one - the element value. 
     */
    public static <T, R> Stream<R> mapWithIndex(Stream<? extends T> stream, BiFunction<Integer, ? super T, ? extends R> mapper) {
        return zipWithIndex(stream).map(entry -> mapper.apply(entry.getKey(), entry.getValue()));
    }

    public static void main(String[] args) {
        String[] names = {"Sam", "Pamela", "Dave", "Pascal", "Erik"};

        System.out.println("Test zipWithIndex");
        zipWithIndex(Arrays.stream(names)).forEach(entry -> System.out.println(entry));

        System.out.println();
        System.out.println("Test mapWithIndex");
        mapWithIndex(Arrays.stream(names), (Integer index, String name) -> index+"="+name).forEach((String s) -> System.out.println(s));
    }
}

score 13 · Accepted Answer

除了 protonpack，jOOλ 的 Seq还提供了这个功能（并且通过像cyclops-react这样的扩展库，我是这个库的作者）。

Seq.seq(Stream.of(names)).zipWithIndex()
                         .filter( namesWithIndex -> namesWithIndex.v1.length() <= namesWithIndex.v2 + 1)
                         .toList();

Seq 还仅支持 Seq.of(names) 并将在幕后构建一个 JDK Stream。

简单反应等价物看起来类似

 LazyFutureStream.of(names)
                 .zipWithIndex()
                 .filter( namesWithIndex -> namesWithIndex.v1.length() <= namesWithIndex.v2 + 1)
                 .toList();

simple-react 版本更适合异步/并发处理。

score 12 · Accepted Answer

为了完整起见，这里是涉及我的StreamEx库的解决方案：

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
EntryStream.of(names)
    .filterKeyValue((idx, str) -> str.length() <= idx+1)
    .values().toList();

在这里，我们创建一个EntryStream<Integer, String>扩展Stream<Entry<Integer, String>>并添加一些特定操作，如filterKeyValueor values。还使用了toList()快捷方式。

score 9 · Accepted Answer

当 Stream 由列表或数组创建时（并且您知道大小），我在这里找到了解决方案。但是如果 Stream 的大小未知怎么办？在这种情况下，试试这个变体：

public class WithIndex<T> {
    private int index;
    private T value;

    WithIndex(int index, T value) {
        this.index = index;
        this.value = value;
    }

    public int index() {
        return index;
    }

    public T value() {
        return value;
    }

    @Override
    public String toString() {
        return value + "(" + index + ")";
    }

    public static <T> Function<T, WithIndex<T>> indexed() {
        return new Function<T, WithIndex<T>>() {
            int index = 0;
            @Override
            public WithIndex<T> apply(T t) {
                return new WithIndex<>(index++, t);
            }
        };
    }
}

用法：

public static void main(String[] args) {
    Stream<String> stream = Stream.of("a", "b", "c", "d", "e");
    stream.map(WithIndex.indexed()).forEachOrdered(e -> {
        System.out.println(e.index() + " -> " + e.value());
    });
}

score 6 · Accepted Answer

使用列表，您可以尝试

List<String> strings = new ArrayList<>(Arrays.asList("First", "Second", "Third", "Fourth", "Fifth")); // An example list of Strings
strings.stream() // Turn the list into a Stream
    .collect(HashMap::new, (h, o) -> h.put(h.size(), o), (h, o) -> {}) // Create a map of the index to the object
        .forEach((i, o) -> { // Now we can use a BiConsumer forEach!
            System.out.println(String.format("%d => %s", i, o));
        });

输出：

0 => First
1 => Second
2 => Third
3 => Fourth
4 => Fifth

score 5 · Accepted Answer

如果您碰巧使用 Vavr（以前称为 Javaslang），则可以利用专用方法：

Stream.of("A", "B", "C")
  .zipWithIndex();

如果我们打印出内容，我们会看到一些有趣的东西：

Stream((A, 0), ?)

这是因为Streams他们很懒，我们不知道流中的下一个项目。

score 4 · Accepted Answer

这是AbacusUtil的代码

Stream.of(names).indexed()
      .filter(e -> e.value().length() <= e.index())
      .map(Indexed::value).toList();

披露：我是 AbacusUtil 的开发者。

score 3 · Accepted Answer

There isn't a way to iterate over a Stream whilst having access to the index because a Stream is unlike any Collection. A Stream is merely a pipeline for carrying data from one place to another, as stated in the documentation:

No storage. A stream is not a data structure that stores elements; instead, they carry values from a source (which could be a data structure, a generator, an IO channel, etc) through a pipeline of computational operations.

Of course, as you appear to be hinting at in your question, you could always convert your Stream<V> to a Collection<V>, such as a List<V>, in which you will have access to the indexes.

score 3 · Accepted Answer

使用https://github.com/poetix/protonpack 你可以做那个 zip：

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};

List<String> nameList;
Stream<Integer> indices = IntStream.range(0, names.length).boxed(); 

nameList = StreamUtils.zip(indices, stream(names),SimpleEntry::new)
        .filter(e -> e.getValue().length() <= e.getKey()).map(Entry::getValue).collect(toList());                   

System.out.println(nameList);

score 3 · Accepted Answer

您可以使用IntStream.iterate()获取索引：

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i < names.length, i -> i + 1)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(Collectors.toList());

这仅适用于 Java 8 中的 Java 9 以上，您可以使用它：

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> nameList = IntStream.iterate(0, i -> i + 1)
        .limit(names.length)
        .filter(i -> names[i].length() <= i)
        .mapToObj(i -> names[i])
        .collect(Collectors.toList());

score 3 · Accepted Answer

如果您不介意使用第三方库，Eclipse Collections可以跨多种类型使用。下面是针对 JDK 类型和 Eclipse Collections 类型的这一挑战的一组解决方案，使用.zipWithIndexforEachWithIndexzipWithIndex

String[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
ImmutableList<String> expected = Lists.immutable.with("Erik");
Predicate<Pair<String, Integer>> predicate =
    pair -> pair.getOne().length() <= pair.getTwo() + 1;

// JDK Types
List<String> strings1 = ArrayIterate.zipWithIndex(names)
    .collectIf(predicate, Pair::getOne);
Assert.assertEquals(expected, strings1);

List<String> list = Arrays.asList(names);
List<String> strings2 = ListAdapter.adapt(list)
    .zipWithIndex()
    .collectIf(predicate, Pair::getOne);
Assert.assertEquals(expected, strings2);

// Eclipse Collections types
MutableList<String> mutableNames = Lists.mutable.with(names);
MutableList<String> strings3 = mutableNames.zipWithIndex()
    .collectIf(predicate, Pair::getOne);
Assert.assertEquals(expected, strings3);

ImmutableList<String> immutableNames = Lists.immutable.with(names);
ImmutableList<String> strings4 = immutableNames.zipWithIndex()
    .collectIf(predicate, Pair::getOne);
Assert.assertEquals(expected, strings4);

MutableList<String> strings5 = mutableNames.asLazy()
    .zipWithIndex()
    .collectIf(predicate, Pair::getOne, Lists.mutable.empty());
Assert.assertEquals(expected, strings5);

这是一个使用forEachWithIndex代替的解决方案。

MutableList<String> mutableNames =
    Lists.mutable.with("Sam", "Pamela", "Dave", "Pascal", "Erik");
ImmutableList<String> expected = Lists.immutable.with("Erik");

List<String> actual = Lists.mutable.empty();
mutableNames.forEachWithIndex((name, index) -> {
        if (name.length() <= index + 1)
            actual.add(name);
    });
Assert.assertEquals(expected, actual);

如果您将 lambda 更改为上面的匿名内部类，那么所有这些代码示例也将在 Java 5 - 7 中运行。

注意：我是 Eclipse Collections 的提交者

score 2 · Accepted Answer

如果您尝试基于谓词获取索引，请尝试以下操作：

如果您只关心第一个索引：

OptionalInt index = IntStream.range(0, list.size())
    .filter(i -> list.get(i) == 3)
    .findFirst();

或者，如果您想查找多个索引：

IntStream.range(0, list.size())
   .filter(i -> list.get(i) == 3)
   .collect(Collectors.toList());

添加.orElse(-1);以防万一你想返回一个值，如果它没有找到它。

score 1 · Accepted Answer

一种可能的方法是索引流中的每个元素：

AtomicInteger index = new AtomicInteger();
Stream.of(names)
  .map(e->new Object() { String n=e; public i=index.getAndIncrement(); })
  .filter(o->o.n.length()<=o.i) // or do whatever you want with pairs...
  .forEach(o->System.out.println("idx:"+o.i+" nam:"+o.n));

在流中使用匿名类并没有很好地使用，但非常有用。

score 1 · Accepted Answer

您不一定需要 map 与
LINQ 示例最接近的 lambda：

int[] idx = new int[] { 0 };
Stream.of(names)
    .filter(name -> name.length() <= idx[0]++)
    .collect(Collectors.toList());

score 1 · Accepted Answer

如果您需要 forEach 中的索引，那么这提供了一种方法。

  public class IndexedValue {

    private final int    index;
    private final Object value;

    public IndexedValue(final int index, final Object value) { 
        this.index = index;
        this.value = value;
    }

    public int getIndex() {
        return index;
    }

    public Object getValue() {
        return value;
    }
}

然后如下使用它。

@Test
public void withIndex() {
    final List<String> list = Arrays.asList("a", "b");
    IntStream.range(0, list.size())
             .mapToObj(index -> new IndexedValue(index, list.get(index)))
             .forEach(indexValue -> {
                 System.out.println(String.format("%d, %s",
                                                  indexValue.getIndex(),
                                                  indexValue.getValue().toString()));
             });
}

score 0 · Accepted Answer

String[] namesArray = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
String completeString
         =  IntStream.range(0,namesArray.length)
           .mapToObj(i -> namesArray[i]) // Converting each array element into Object
           .map(String::valueOf) // Converting object to String again
           .collect(Collectors.joining(",")); // getting a Concat String of all values
        System.out.println(completeString);

输出：山姆、帕梅拉、戴夫、帕斯卡、埃里克

String[] namesArray = {"Sam","Pamela", "Dave", "Pascal", "Erik"};

IntStream.range(0,namesArray.length)
               .mapToObj(i -> namesArray[i]) // Converting each array element into Object
               .map(String::valueOf) // Converting object to String again
               .forEach(s -> {
                //You can do various operation on each element here
                System.out.println(s);
               }); // getting a Concat String of all

在列表中收集：

String[] namesArray = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
 List<String> namesList
                =  IntStream.range(0,namesArray.length)
                .mapToObj(i -> namesArray[i]) // Converting each array element into Object
                .map(String::valueOf) // Converting object to String again
                .collect(Collectors.toList()); // collecting elements in List
        System.out.println(listWithIndex);

score 0 · Accepted Answer

这个问题（Stream Way to get index of first element matching boolean）已将当前问题标记为重复，所以我无法在那里回答；我在这里回答。

这是获取不需要外部库的匹配索引的通用解决方案。

如果你有一个清单。

public static <T> int indexOf(List<T> items, Predicate<T> matches) {
        return IntStream.range(0, items.size())
                .filter(index -> matches.test(items.get(index)))
                .findFirst().orElse(-1);
}

并这样称呼它：

int index = indexOf(myList, item->item.getId()==100);

如果使用一个集合，试试这个。

   public static <T> int indexOf(Collection<T> items, Predicate<T> matches) {
        int index = -1;
        Iterator<T> it = items.iterator();
        while (it.hasNext()) {
            index++;
            if (matches.test(it.next())) {
                return index;
            }
        }
        return -1;
    }

score 0 · Accepted Answer

这是标准Java的解决方案：

在线解决方案：

Arrays.stream("zero,one,two,three,four".split(","))
        .map(new Function<String, Map.Entry<Integer, String>>() {
            int index;

            @Override
            public Map.Entry<Integer, String> apply(String s) {
                return Map.entry(index++, s);
            }
        })
        .forEach(System.out::println);

以及更易读的实用方法解决方案：

static <T> Function<T, Map.Entry<Integer, T>> mapWithIntIndex() {
    return new Function<T, Map.Entry<Integer, T>>() {
        int index;

        @Override
        public Map.Entry<Integer, T> apply(T t) {
            return Map.entry(index++, t);
        }
    };
}

...
Arrays.stream("zero,one,two,three,four".split(","))
        .map(mapWithIntIndex())
        .forEach(System.out::println);

score 0 · Accepted Answer

您可以创建一个静态内部类来封装索引器，就像我在下面的示例中需要做的那样：

static class Indexer {
    int i = 0;
}

public static String getRegex() {
    EnumSet<MeasureUnit> range = EnumSet.allOf(MeasureUnit.class);
    StringBuilder sb = new StringBuilder();
    Indexer indexer = new Indexer();
    range.stream().forEach(
            measureUnit -> {
                sb.append(measureUnit.acronym);
                if (indexer.i < range.size() - 1)
                    sb.append("|");

                indexer.i++;
            }
    );
    return sb.toString();
}

score 0 · Accepted Answer

正如 jean-baptiste-yunès 所说，如果您的流基于 java List，那么使用 AtomicInteger 及其 incrementAndGet 方法是解决该问题的一个很好的方法，只要您返回的整数确实对应于原始 List 中的索引不要使用并行流。

java - 在 Java 8 中是否有一种简洁的方法来迭代带有索引的流？

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