在带有lambda b93的JDK 8中,有一个类java.util.stream.Streams.zip in b93可用于压缩流(这在教程Exploring Java8 Lambdas. Part 1 by Dhananjay Nene中有说明)。这个功能:
创建一个惰性和顺序组合的Stream,其元素是 结合两个流的元素的结果。
然而在b98中,这已经消失了。事实上,java.util.stream in b98中甚至无法访问Streams
类。
是否已移动此功能,如果是,我如何使用b98简洁地压缩流?
我想到的应用程序是in this java implementation of Shen,在那里我替换了
中的zip功能static <T> boolean every(Collection<T> c1, Collection<T> c2, BiPredicate<T, T> pred)
static <T> T find(Collection<T> c1, Collection<T> c2, BiPredicate<T, T> pred)
具有相当冗长的代码(不使用b98中的功能)。
答案 0 :(得分:70)
我也需要这个,所以我只是从b93中获取源代码并将其放在“util”类中。我不得不稍微修改它以使用当前的API。
这里的参考是工作代码(风险自负......):
public static<A, B, C> Stream<C> zip(Stream<? extends A> a,
Stream<? extends B> b,
BiFunction<? super A, ? super B, ? extends C> zipper) {
Objects.requireNonNull(zipper);
Spliterator<? extends A> aSpliterator = Objects.requireNonNull(a).spliterator();
Spliterator<? extends B> bSpliterator = Objects.requireNonNull(b).spliterator();
// Zipping looses DISTINCT and SORTED characteristics
int characteristics = aSpliterator.characteristics() & bSpliterator.characteristics() &
~(Spliterator.DISTINCT | Spliterator.SORTED);
long zipSize = ((characteristics & Spliterator.SIZED) != 0)
? Math.min(aSpliterator.getExactSizeIfKnown(), bSpliterator.getExactSizeIfKnown())
: -1;
Iterator<A> aIterator = Spliterators.iterator(aSpliterator);
Iterator<B> bIterator = Spliterators.iterator(bSpliterator);
Iterator<C> cIterator = new Iterator<C>() {
@Override
public boolean hasNext() {
return aIterator.hasNext() && bIterator.hasNext();
}
@Override
public C next() {
return zipper.apply(aIterator.next(), bIterator.next());
}
};
Spliterator<C> split = Spliterators.spliterator(cIterator, zipSize, characteristics);
return (a.isParallel() || b.isParallel())
? StreamSupport.stream(split, true)
: StreamSupport.stream(split, false);
}
答案 1 :(得分:40)
zip是protonpack library提供的功能之一。
Stream<String> streamA = Stream.of("A", "B", "C");
Stream<String> streamB = Stream.of("Apple", "Banana", "Carrot", "Doughnut");
List<String> zipped = StreamUtils.zip(streamA,
streamB,
(a, b) -> a + " is for " + b)
.collect(Collectors.toList());
assertThat(zipped,
contains("A is for Apple", "B is for Banana", "C is for Carrot"));
答案 2 :(得分:26)
如果你的项目中有Guava,你可以使用Streams.zip方法(在Guava 21中添加):
返回一个流,其中每个元素是将streamA和streamB中的每个元素的相应元素传递给function的结果。结果流只会与两个输入流中较短的一个一样长;如果一个流更长,其额外元素将被忽略。得到的流不能有效地分裂。这可能会损害并行性能。
public class Streams {
...
public static <A, B, R> Stream<R> zip(Stream<A> streamA,
Stream<B> streamB, BiFunction<? super A, ? super B, R> function) {
...
}
}
答案 3 :(得分:23)
Zipping two streams using JDK8 with lambda (gist).
public static <A, B, C> Stream<C> zip(Stream<A> streamA, Stream<B> streamB, BiFunction<A, B, C> zipper) {
final Iterator<A> iteratorA = streamA.iterator();
final Iterator<B> iteratorB = streamB.iterator();
final Iterator<C> iteratorC = new Iterator<C>() {
@Override
public boolean hasNext() {
return iteratorA.hasNext() && iteratorB.hasNext();
}
@Override
public C next() {
return zipper.apply(iteratorA.next(), iteratorB.next());
}
};
final boolean parallel = streamA.isParallel() || streamB.isParallel();
return iteratorToFiniteStream(iteratorC, parallel);
}
public static <T> Stream<T> iteratorToFiniteStream(Iterator<T> iterator, boolean parallel) {
final Iterable<T> iterable = () -> iterator;
return StreamSupport.stream(iterable.spliterator(), parallel);
}
答案 4 :(得分:14)
由于我不能设想在索引(Lists)以外的集合上使用压缩,而且我是简单的忠实粉丝,这将是我的解决方案:
<A,B,C> Stream<C> zipped(List<A> lista, List<B> listb, BiFunction<A,B,C> zipper){
int shortestLength = Math.min(lista.size(),listb.size());
return IntStream.range(0,shortestLength).mapToObj( i -> {
return zipper.apply(lista.get(i), listb.get(i));
});
}
答案 5 :(得分:10)
您提到的类的方法已移至Stream
接口本身,而不是默认方法。但似乎zip
方法已被删除。也许是因为不清楚不同大小的流的默认行为应该是什么。但实现所需的行为是直截了当的:
static <T> boolean every(
Collection<T> c1, Collection<T> c2, BiPredicate<T, T> pred) {
Iterator<T> it=c2.iterator();
return c1.stream().allMatch(x->!it.hasNext()||pred.test(x, it.next()));
}
static <T> T find(Collection<T> c1, Collection<T> c2, BiPredicate<T, T> pred) {
Iterator<T> it=c2.iterator();
return c1.stream().filter(x->it.hasNext()&&pred.test(x, it.next()))
.findFirst().orElse(null);
}
答案 6 :(得分:6)
Lazy-Seq库提供了zip功能。
https://github.com/nurkiewicz/LazySeq
这个库受scala.collection.immutable.Stream
的启发,旨在提供不可变,线程安全且易于使用的延迟序列实现,可能是无限的。
答案 7 :(得分:5)
我谦卑地建议这个实施。生成的流被截断为两个输入流中较短的一个。
public static <L, R, T> Stream<T> zip(Stream<L> leftStream, Stream<R> rightStream, BiFunction<L, R, T> combiner) {
Spliterator<L> lefts = leftStream.spliterator();
Spliterator<R> rights = rightStream.spliterator();
return StreamSupport.stream(new AbstractSpliterator<T>(Long.min(lefts.estimateSize(), rights.estimateSize()), lefts.characteristics() & rights.characteristics()) {
@Override
public boolean tryAdvance(Consumer<? super T> action) {
return lefts.tryAdvance(left->rights.tryAdvance(right->action.accept(combiner.apply(left, right))));
}
}, leftStream.isParallel() || rightStream.isParallel());
}
答案 8 :(得分:1)
public class Tuple<S,T> {
private final S object1;
private final T object2;
public Tuple(S object1, T object2) {
this.object1 = object1;
this.object2 = object2;
}
public S getObject1() {
return object1;
}
public T getObject2() {
return object2;
}
}
public class StreamUtils {
private StreamUtils() {
}
public static <T> Stream<Tuple<Integer,T>> zipWithIndex(Stream<T> stream) {
Stream<Integer> integerStream = IntStream.range(0, Integer.MAX_VALUE).boxed();
Iterator<Integer> integerIterator = integerStream.iterator();
return stream.map(x -> new Tuple<>(integerIterator.next(), x));
}
}
答案 9 :(得分:1)
我提供的AOL&#39; cyclops-react也提供了压缩功能,通过extended Stream implementation,也实现了反应流接口ReactiveSeq,并通过StreamUtils提供了大量的通过静态方法向标准Java Streams提供相同的功能。
List<Tuple2<Integer,Integer>> list = ReactiveSeq.of(1,2,3,4,5,6)
.zip(Stream.of(100,200,300,400));
List<Tuple2<Integer,Integer>> list = StreamUtils.zip(Stream.of(1,2,3,4,5,6),
Stream.of(100,200,300,400));
它还提供更广泛的基于Applicative的压缩。例如。
ReactiveSeq.of("a","b","c")
.ap3(this::concat)
.ap(of("1","2","3"))
.ap(of(".","?","!"))
.toList();
//List("a1.","b2?","c3!");
private String concat(String a, String b, String c){
return a+b+c;
}
甚至能够将一个流中的每个项目与另一个项目中的每个项目配对
ReactiveSeq.of("a","b","c")
.forEach2(str->Stream.of(str+"!","2"), a->b->a+"_"+b);
//ReactiveSeq("a_a!","a_2","b_b!","b_2","c_c!","c2")
答案 10 :(得分:1)
使用最新的Guava库(针对Streams
类),您应该能够做到
final Map<String, String> result =
Streams.zip(
collection1.stream(),
collection2.stream(),
AbstractMap.SimpleEntry::new)
.collect(Collectors.toMap(e -> e.getKey(), e -> e.getValue()));
答案 11 :(得分:1)
这对您有用吗?这是一个简短的函数,它懒惰地评估正在压缩的流,因此您可以为它提供无限的流(它不需要占用被压缩的流的大小)。
如果流是有限的,则流之一会在元素用完时停止。
import java.util.Objects;
import java.util.function.BiFunction;
import java.util.stream.Stream;
class StreamUtils {
static <ARG1, ARG2, RESULT> Stream<RESULT> zip(
Stream<ARG1> s1,
Stream<ARG2> s2,
BiFunction<ARG1, ARG2, RESULT> combiner) {
final var i2 = s2.iterator();
return s1.map(x1 -> i2.hasNext() ? combiner.apply(x1, i2.next()) : null)
.takeWhile(Objects::nonNull);
}
}
这里有一些单元测试代码(比代码本身长得多!)
import org.junit.jupiter.api.Test;
import org.junit.jupiter.params.ParameterizedTest;
import org.junit.jupiter.params.provider.Arguments;
import org.junit.jupiter.params.provider.MethodSource;
import java.util.List;
import java.util.concurrent.atomic.AtomicInteger;
import java.util.function.BiFunction;
import java.util.stream.Collectors;
import java.util.stream.Stream;
import static org.junit.jupiter.api.Assertions.assertEquals;
class StreamUtilsTest {
@ParameterizedTest
@MethodSource("shouldZipTestCases")
<ARG1, ARG2, RESULT>
void shouldZip(
String testName,
Stream<ARG1> s1,
Stream<ARG2> s2,
BiFunction<ARG1, ARG2, RESULT> combiner,
Stream<RESULT> expected) {
var actual = StreamUtils.zip(s1, s2, combiner);
assertEquals(
expected.collect(Collectors.toList()),
actual.collect(Collectors.toList()),
testName);
}
private static Stream<Arguments> shouldZipTestCases() {
return Stream.of(
Arguments.of(
"Two empty streams",
Stream.empty(),
Stream.empty(),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.empty()),
Arguments.of(
"One singleton and one empty stream",
Stream.of(1),
Stream.empty(),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.empty()),
Arguments.of(
"One empty and one singleton stream",
Stream.empty(),
Stream.of(1),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.empty()),
Arguments.of(
"Two singleton streams",
Stream.of("blah"),
Stream.of(1),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.of(pair("blah", 1))),
Arguments.of(
"One singleton, one multiple stream",
Stream.of("blob"),
Stream.of(2, 3),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.of(pair("blob", 2))),
Arguments.of(
"One multiple, one singleton stream",
Stream.of("foo", "bar"),
Stream.of(4),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.of(pair("foo", 4))),
Arguments.of(
"Two multiple streams",
Stream.of("nine", "eleven"),
Stream.of(10, 12),
(BiFunction<Object, Object, Object>) StreamUtilsTest::combine,
Stream.of(pair("nine", 10), pair("eleven", 12)))
);
}
private static List<Object> pair(Object o1, Object o2) {
return List.of(o1, o2);
}
static private <T1, T2> List<Object> combine(T1 o1, T2 o2) {
return List.of(o1, o2);
}
@Test
void shouldLazilyEvaluateInZip() {
final var a = new AtomicInteger();
final var b = new AtomicInteger();
final var zipped = StreamUtils.zip(
Stream.generate(a::incrementAndGet),
Stream.generate(b::decrementAndGet),
(xa, xb) -> xb + 3 * xa);
assertEquals(0, a.get(), "Should not have evaluated a at start");
assertEquals(0, b.get(), "Should not have evaluated b at start");
final var takeTwo = zipped.limit(2);
assertEquals(0, a.get(), "Should not have evaluated a at take");
assertEquals(0, b.get(), "Should not have evaluated b at take");
final var list = takeTwo.collect(Collectors.toList());
assertEquals(2, a.get(), "Should have evaluated a after collect");
assertEquals(-2, b.get(), "Should have evaluated b after collect");
assertEquals(List.of(2, 4), list);
}
}
答案 12 :(得分:0)
这很棒。我不得不将两个流压缩成一个Map,其中一个流是关键,另一个是值
Stream<String> streamA = Stream.of("A", "B", "C");
Stream<String> streamB = Stream.of("Apple", "Banana", "Carrot", "Doughnut");
final Stream<Map.Entry<String, String>> s = StreamUtils.zip(streamA,
streamB,
(a, b) -> {
final Map.Entry<String, String> entry = new AbstractMap.SimpleEntry<String, String>(a, b);
return entry;
});
System.out.println(s.collect(Collectors.toMap(e -> e.getKey(), e -> e.getValue())));
输出: {A = Apple,B =香蕉,C =胡萝卜}
答案 13 :(得分:0)
如果有人需要此功能,则streamex库中有StreamEx.zipWith
个函数:
StreamEx<String> givenNames = StreamEx.of("Leo", "Fyodor")
StreamEx<String> familyNames = StreamEx.of("Tolstoy", "Dostoevsky")
StreamEx<String> fullNames = givenNames.zipWith(familyNames, (gn, fn) -> gn + " " + fn);
fullNames.forEach(System.out::println); // prints: "Leo Tolstoy\nFyodor Dostoevsky\n"