是否有Java 8流操作限制流(可能是无限的),直到第一个元素无法匹配谓词?
在Java 9中,我们可以像下面的例子一样使用takeWhile来打印所有小于10的数字。
IntStream
.iterate(1, n -> n + 1)
.takeWhile(n -> n < 10)
.forEach(System.out::println);
因为在Java 8中没有这样的操作,那么以通用的方式实现它的最佳方法是什么呢?
是否有Java 8流操作限制流(可能是无限的),直到第一个元素无法匹配谓词?
在Java 9中,我们可以像下面的例子一样使用takeWhile来打印所有小于10的数字。
IntStream
.iterate(1, n -> n + 1)
.takeWhile(n -> n < 10)
.forEach(System.out::println);
因为在Java 8中没有这样的操作,那么以通用的方式实现它的最佳方法是什么呢?
当前回答
这是从JDK 9 java.util.stream.Stream.takeWhile(Predicate)中复制的源代码。为了使用JDK 8,有一点不同。
static <T> Stream<T> takeWhile(Stream<T> stream, Predicate<? super T> p) {
class Taking extends Spliterators.AbstractSpliterator<T> implements Consumer<T> {
private static final int CANCEL_CHECK_COUNT = 63;
private final Spliterator<T> s;
private int count;
private T t;
private final AtomicBoolean cancel = new AtomicBoolean();
private boolean takeOrDrop = true;
Taking(Spliterator<T> s) {
super(s.estimateSize(), s.characteristics() & ~(Spliterator.SIZED | Spliterator.SUBSIZED));
this.s = s;
}
@Override
public boolean tryAdvance(Consumer<? super T> action) {
boolean test = true;
if (takeOrDrop && // If can take
(count != 0 || !cancel.get()) && // and if not cancelled
s.tryAdvance(this) && // and if advanced one element
(test = p.test(t))) { // and test on element passes
action.accept(t); // then accept element
return true;
} else {
// Taking is finished
takeOrDrop = false;
// Cancel all further traversal and splitting operations
// only if test of element failed (short-circuited)
if (!test)
cancel.set(true);
return false;
}
}
@Override
public Comparator<? super T> getComparator() {
return s.getComparator();
}
@Override
public void accept(T t) {
count = (count + 1) & CANCEL_CHECK_COUNT;
this.t = t;
}
@Override
public Spliterator<T> trySplit() {
return null;
}
}
return StreamSupport.stream(new Taking(stream.spliterator()), stream.isParallel()).onClose(stream::close);
}
其他回答
takeWhile是质子包库提供的函数之一。
Stream<Integer> infiniteInts = Stream.iterate(0, i -> i + 1);
Stream<Integer> finiteInts = StreamUtils.takeWhile(infiniteInts, i -> i < 10);
assertThat(finiteInts.collect(Collectors.toList()),
hasSize(10));
这是在int上做的一个版本-正如问题中所问的那样。
用法:
StreamUtil.takeWhile(IntStream.iterate(1, n -> n + 1), n -> n < 10);
下面是StreamUtil的代码:
import java.util.PrimitiveIterator;
import java.util.Spliterators;
import java.util.function.IntConsumer;
import java.util.function.IntPredicate;
import java.util.stream.IntStream;
import java.util.stream.StreamSupport;
public class StreamUtil
{
public static IntStream takeWhile(IntStream stream, IntPredicate predicate)
{
return StreamSupport.intStream(new PredicateIntSpliterator(stream, predicate), false);
}
private static class PredicateIntSpliterator extends Spliterators.AbstractIntSpliterator
{
private final PrimitiveIterator.OfInt iterator;
private final IntPredicate predicate;
public PredicateIntSpliterator(IntStream stream, IntPredicate predicate)
{
super(Long.MAX_VALUE, IMMUTABLE);
this.iterator = stream.iterator();
this.predicate = predicate;
}
@Override
public boolean tryAdvance(IntConsumer action)
{
if (iterator.hasNext()) {
int value = iterator.nextInt();
if (predicate.test(value)) {
action.accept(value);
return true;
}
}
return false;
}
}
}
去图书馆拿算盘——普通的算盘。它提供了你想要的API和更多:
IntStream.iterate(1, n -> n + 1).takeWhile(n -> n < 10).forEach(System.out::println);
声明:我是AbacusUtil的开发者。
甚至我也有类似的需求——调用web服务,如果失败,重试3次。如果在多次尝试后仍然失败,请发送电子邮件通知。在谷歌搜索了很多之后,anyMatch()成为了救星。我的示例代码如下。在下面的例子中,如果webServiceCall方法在第一次迭代本身中返回true,则stream不会在我们调用anyMatch()时进一步迭代。我相信,这就是你想要的。
import java.util.stream.IntStream;
import io.netty.util.internal.ThreadLocalRandom;
class TrialStreamMatch {
public static void main(String[] args) {
if(!IntStream.range(1,3).anyMatch(integ -> webServiceCall(integ))){
//Code for sending email notifications
}
}
public static boolean webServiceCall(int i){
//For time being, I have written a code for generating boolean randomly
//This whole piece needs to be replaced by actual web-service client code
boolean bool = ThreadLocalRandom.current().nextBoolean();
System.out.println("Iteration index :: "+i+" bool :: "+bool);
//Return success status -- true or false
return bool;
}
您可以使用java8 + rxjava。
import java.util.stream.IntStream;
import rx.Observable;
// Example 1)
IntStream intStream = IntStream.iterate(1, n -> n + 1);
Observable.from(() -> intStream.iterator())
.takeWhile(n ->
{
System.out.println(n);
return n < 10;
}
).subscribe() ;
// Example 2
IntStream intStream = IntStream.iterate(1, n -> n + 1);
Observable.from(() -> intStream.iterator())
.takeWhile(n -> n < 10)
.forEach( n -> System.out.println(n));