如何从流api的列表中获取随机元素?

时间:2017-08-31 12:54:36

标签: java java-8

使用Java8 stream api从列表中获取随机元素的最有效方法是什么?

Arrays.asList(new Obj1(), new Obj2(), new Obj3());

感谢。

9 个答案:

答案 0 :(得分:14)

为什么要使用流?您只需要从0到列表大小的随机数,然后在此索引上调用get

Random r = new Random();
ElementType e = list.get(r.nextInt(list.size()));

Stream在这里没有任何有趣的东西,但您可以尝试:

Random r = new Random();
ElementType e = list.stream().skip(r.nextInt(list.size()-1)).findFirst().get();

想法是跳过任意数量的元素(但不是最后一个元素!),然后获取第一个元素(如果存在)。因此,您将拥有一个非空的Optional<ElementType>,然后使用get提取其值。跳过后,你有很多选择。

在这里使用流是非常低效的......

注意:这些解决方案都没有考虑空列表,但问题是在非空列表中定义的。

答案 1 :(得分:3)

如果你来使用流,我写了一个优雅的,虽然非常低效的收集器来完成这项工作:

/**
 * Returns a random item from the stream (or null in case of an empty stream).
 * This operation can't be lazy and is inefficient, and therefore shouldn't
 * be used on streams with a large number or items or in performance critical sections.
 * @return a random item from the stream or null if the stream is empty.
 */
public static <T> Collector<T, List<T>, T> randomItem() {
    final Random RANDOM = new Random();
    return Collector.of(() -> (List<T>) new ArrayList<T>(), 
                              (acc, elem) -> acc.add(elem),
                              (list1, list2) -> ListUtils.union(list1, list2), // Using a 3rd party for list union, could be done "purely"
                              list -> list.isEmpty() ? null : list.get(RANDOM.nextInt(list.size())));
}

用法:

@Test
public void standardRandomTest() {
    assertThat(Stream.of(1, 2, 3, 4).collect(randomItem())).isBetween(1, 4);
}

答案 2 :(得分:2)

有更有效的方法可以做到这一点,但如果必须是Stream,最简单的方法是创建自己的Comparator,它返回随机结果(-1,0,1)并对你的流进行排序:

 List<String> strings = Arrays.asList("a", "b", "c", "d", "e", "f");
    String randomString = strings
            .stream()
            .sorted((o1, o2) -> ThreadLocalRandom.current().nextInt(-1, 2))
            .findAny()
            .get();

ThreadLocalRandom已准备好“开箱即用”方法,以便在比较器所需的范围内获得随机数。

答案 3 :(得分:1)

您可以执行以下操作:

 yourStream.collect(new RandomListCollector<>(randomSetSize));

我想您必须像这样编写自己的Collector实现,以实现同类随机化:

 public class RandomListCollector<T> implements Collector<T, RandomListCollector.ListAccumulator<T>, List<T>> {

private final Random rand;
private final int size;

public RandomListCollector(Random random , int size) {
    super();
    this.rand = random;
    this.size = size;
}

public RandomListCollector(int size) {
    this(new Random(System.nanoTime()), size);
}

@Override
public Supplier<ListAccumulator<T>> supplier() {
    return () -> new ListAccumulator<T>();
}

@Override
public BiConsumer<ListAccumulator<T>, T> accumulator() {
    return (l, t) -> {
        if (l.size() < size) {
            l.add(t);
        } else if (rand.nextDouble() <= ((double) size) / (l.gSize() + 1)) {
            l.add(t);
            l.remove(rand.nextInt(size));
        } else {
            // in any case gSize needs to be incremented
            l.gSizeInc();
        }
    };

}

@Override
public BinaryOperator<ListAccumulator<T>> combiner() {
    return (l1, l2) -> {
        int lgSize = l1.gSize() + l2.gSize();
        ListAccumulator<T> l = new ListAccumulator<>();
        if (l1.size() + l2.size()<size) {
            l.addAll(l1);
            l.addAll(l2);
        } else {
            while (l.size() < size) {
                if (l1.size()==0 || l2.size()>0 && rand.nextDouble() < (double) l2.gSize() / (l1.gSize() + l2.gSize())) {
                    l.add(l2.remove(rand.nextInt(l2.size()), true));
                } else {
                    l.add(l1.remove(rand.nextInt(l1.size()), true));
                }
            }
        }
        // set the gSize of l :
        l.gSize(lgSize);
        return l;

    };
}

@Override
public Function<ListAccumulator<T>, List<T>> finisher() {

    return (la) -> la.list;
}

@Override
public Set<Characteristics> characteristics() {
    return Collections.singleton(Characteristics.CONCURRENT);
}

static class ListAccumulator<T> implements Iterable<T> {
    List<T> list;
    volatile int gSize;

    public ListAccumulator() {
        list = new ArrayList<>();
        gSize = 0;
    }

    public void addAll(ListAccumulator<T> l) {
        list.addAll(l.list);
        gSize += l.gSize;

    }

    public T remove(int index) {
        return remove(index, false);
    }

    public T remove(int index, boolean global) {
        T t = list.remove(index);
        if (t != null && global)
            gSize--;
        return t;
    }

    public void add(T t) {
        list.add(t);
        gSize++;

    }

    public int gSize() {
        return gSize;
    }

    public void gSize(int gSize) {
        this.gSize = gSize;

    }

    public void gSizeInc() {
        gSize++;
    }

    public int size() {
        return list.size();
    }

    @Override
    public Iterator<T> iterator() {
        return list.iterator();
    }
}

}

如果您希望更轻松一些,但又不想将所有列表都加载到内存中,请执行以下操作:

public <T> Stream<T> getRandomStreamSubset(Stream<T> stream, int subsetSize) {
    int cnt = 0;

    Random r = new Random(System.nanoTime());
    Object[] tArr = new Object[subsetSize];
    Iterator<T> iter = stream.iterator();
    while (iter.hasNext() && cnt <subsetSize) {
        tArr[cnt++] = iter.next();          
    }

    while (iter.hasNext()) {
        cnt++;
        T t = iter.next();
        if (r.nextDouble() <= (double) subsetSize / cnt) {
            tArr[r.nextInt(subsetSize)] = t;                

        }

    }

    return Arrays.stream(tArr).map(o -> (T)o );
}

但是您随后将离开流api,并且可以使用基本的迭代器进行相同操作

答案 4 :(得分:1)

另一个想法是实现自己的Spliterator,然后将其用作Stream的来源:

import java.util.List;
import java.util.Random;
import java.util.Spliterator;
import java.util.function.Consumer;
import java.util.function.Supplier;

public class ImprovedRandomSpliterator<T> implements Spliterator<T> {

    private final Random random;
    private final T[] source;
    private int size;

    ImprovedRandomSpliterator(List<T> source, Supplier<? extends Random> random) {
        if (source.isEmpty()) {
            throw new IllegalArgumentException("RandomSpliterator can't be initialized with an empty collection");
        }
        this.source = (T[]) source.toArray();
        this.random = random.get();
        this.size = this.source.length;
    }

    @Override
    public boolean tryAdvance(Consumer<? super T> action) {
        if (size > 0) {
            int nextIdx = random.nextInt(size); 
            int lastIdx = size - 1; 

            action.accept(source[nextIdx]); 
            source[nextIdx] = source[lastIdx]; 
            source[lastIdx] = null; // let object be GCed 
            size--;
            return true;
        } else {
            return false;
        }
    }

    @Override
    public Spliterator<T> trySplit() {
        return null;
    }

    @Override
    public long estimateSize() {
        return source.length;
    }

    @Override
    public int characteristics() {
        return SIZED;
    }
}


public static <T> Collector<T, ?, Stream<T>> toShuffledStream() {
    return Collectors.collectingAndThen(
      toCollection(ArrayList::new),
      list -> !list.isEmpty()
        ? StreamSupport.stream(new ImprovedRandomSpliterator<>(list, Random::new), false)
        : Stream.empty());
}

然后简单地:

list.stream()
  .collect(toShuffledStream())
  .findAny();

Details can be found here.

...但这绝对是一个过大的杀伤力,因此,如果您正在寻找一种务实的方法。肯定要去Jean's solution

答案 5 :(得分:1)

虽然所有给定的答案都有效,但是有一个简单的单行代码可以解决问题,而不必先检查列表是否为空:

List<String> list = List.of("a", "b", "c");
list.stream().skip((int) (list.size() * Math.random())).findAny();

对于空白列表,这将返回Optional.empty

答案 6 :(得分:1)

上次我需要做类似的事情:

List<String> list = Arrays.asList("a", "b", "c");
Collections.shuffle(list);
String letter = list.stream().findAny().orElse(null);
System.out.println(letter);

答案 7 :(得分:0)

所选答案的流解决方案中有错误... 在这种情况下,不能将Random#nextInt与非正长数(0)一起使用。 流解决方案也永远不会选择列表中的最后一个 示例:

List<Integer> intList = Arrays.asList(0, 1, 2, 3, 4);

// #nextInt is exclusive, so here it means a returned value of 0-3
// if you have a list of size = 1, #next Int will throw an IllegalArgumentException (bound must be positive)
int skipIndex = new Random().nextInt(intList.size()-1);

// randomInt will only ever be 0, 1, 2, or 3. Never 4
int randomInt = intList.stream()
                       .skip(skipIndex) // max skip of list#size - 2
                       .findFirst()
                       .get();

我的建议是与让-巴蒂斯特·尤涅斯(Jean-BaptisteYunès)提出的非流式处理方法配合使用,但是如果必须采用流式处理方法,则可以执行以下操作(但这有点难看):

list.stream()
    .skip(list.isEmpty ? 0 : new Random().nextInt(list.size()))
    .findFirst();

答案 8 :(得分:0)

您可以将以下部分添加到Stream中,成为findAny

.sorted((f1, f2) -> (new Random().nextInt(1)) == 0 ? -1 : 1)