是否有一种简洁的方法在流上迭代，同时访问流中的索引?
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());
与这里给出的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();
有更简洁的方式吗?
此外，似乎拉链已经移动或被拆除…
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();
这里我们创建了一个EntryStream<Integer, String>，它扩展了Stream<Entry<Integer, String>>，并添加了一些特定的操作，如filterKeyValue或values。也使用了toList()快捷方式。
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));
}
}
最简洁的方法是从一系列指数开始:
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());
结果列表只包含“Erik”。
当你习惯for循环时，另一种看起来更熟悉的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());
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。
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();
这里我们创建了一个EntryStream<Integer, String>，它扩展了Stream<Entry<Integer, String>>，并添加了一些特定的操作，如filterKeyValue或values。也使用了toList()快捷方式。
