Java 8 流上的缓冲区运算符答案

【问题标题】：Buffer Operator on Java 8 StreamsJava 8 流上的缓冲区运算符
【发布时间】：2018-05-30 06:50:50
【问题描述】：

我正在尝试编写一个反映 rxjava buffer operator 功能的 java 8 流收集器

我有一个工作代码：

// This will gather numbers 1 to 13 and combine them in groups of
// three while preserving the order even if its a parallel stream.
final List<List<String>> triads = IntStream.range(1, 14)
        .parallel()
        .boxed()
        .map(Object::toString)
        .collect(ArrayList::new, accumulator, combiner);

System.out.println(triads.toString())

这里的累加器是这样的：

final BiConsumer<List<List<String>>, String> accumulator = (acc, a) -> {
      StringBuilder stringBuilder = new StringBuilder();
      stringBuilder.append("Accumulator|");
      stringBuilder.append("Before: ").append(acc.toString());
      int accumulatorSize = acc.size();
      if (accumulatorSize == 0) {
        List<String> newList = new ArrayList<>();
        newList.add(a);
        acc.add(newList);
      } else {
        List<String> lastList = acc.get(accumulatorSize - 1);
        if (lastList.size() != 3) {
          lastList.add(a);
        } else {
          List<String> newList = new ArrayList<>();
          newList.add(a);
          acc.add(newList);
        }
      }
      stringBuilder.append("|After: ").append(acc.toString());
      stringBuilder.append("|a: ").append(a);
      System.out.println(stringBuilder.toString());
    };

还有组合器

// Utility method to make first list of size 3 
// by shifting elements from second to first list
final BiConsumer<List<String>, List<String>> fixSize = (l1, l2) -> {
  while(l1.size() != 3 && l2.size() > 0) {
    l1.add(l2.remove(0));
  }
};

final BiConsumer<List<List<String>>, List<List<String>>> combiner = (l1, l2) -> {
  StringBuilder stringBuilder = new StringBuilder();
  stringBuilder.append("Combiner|");
  stringBuilder.append("Before, l1: ").append(l1).append(", l2: ").append(l2);
  if (l1.isEmpty()) {
    // l1 is empty
    l1.addAll(l2);
  } else {
    // l1 is not empty
    List<String> lastL1List = l1.get(l1.size() - 1);
    if (lastL1List.size() == 3) {
      l1.addAll(l2);
    } else {
      if (l2.isEmpty()) {
        // do nothing
      } else {
        List<List<String>> fixSizeList = new ArrayList<>(1 + l2.size());
        fixSizeList.add(lastL1List);
        fixSizeList.addAll(l2);
        for (int i = 0; i < fixSizeList.size() - 1; i++) {
          List<String> x = fixSizeList.get(i), y = fixSizeList.get(i + 1);
          fixSize.accept(x, y);
        }
        l2.stream().filter(l -> !l.isEmpty()).forEach(l1::add);
        // everything is now of size three except, may be last
      }
    }
  }
  stringBuilder.append("|After, l1: ").append(l1).append(", l2: ").append(l2);
  System.out.println(stringBuilder.toString());
};

这会产生以下输出：

Accumulator|Before: []|After: [[12]]|a: 12
Accumulator|Before: []|After: [[2]]|a: 2
Accumulator|Before: []|After: [[11]]|a: 11
Accumulator|Before: []|After: [[6]]|a: 6
Accumulator|Before: []|After: [[4]]|a: 4
Accumulator|Before: []|After: [[1]]|a: 1
Accumulator|Before: []|After: [[13]]|a: 13
Accumulator|Before: []|After: [[8]]|a: 8
Accumulator|Before: []|After: [[3]]|a: 3
Accumulator|Before: []|After: [[5]]|a: 5
Accumulator|Before: []|After: [[10]]|a: 10
Accumulator|Before: []|After: [[7]]|a: 7
Accumulator|Before: []|After: [[9]]|a: 9
Combiner|Before, l1: [[5]], l2: [[6]]|After, l1: [[5, 6]], l2: [[]]
Combiner|Before, l1: [[12]], l2: [[13]]|After, l1: [[12, 13]], l2: [[]]
Combiner|Before, l1: [[2]], l2: [[3]]|After, l1: [[2, 3]], l2: [[]]
Combiner|Before, l1: [[8]], l2: [[9]]|After, l1: [[8, 9]], l2: [[]]
Combiner|Before, l1: [[10]], l2: [[11]]|After, l1: [[10, 11]], l2: [[]]
Combiner|Before, l1: [[4]], l2: [[5, 6]]|After, l1: [[4, 5, 6]], l2: [[]]
Combiner|Before, l1: [[1]], l2: [[2, 3]]|After, l1: [[1, 2, 3]], l2: [[]]
Combiner|Before, l1: [[7]], l2: [[8, 9]]|After, l1: [[7, 8, 9]], l2: [[]]
Combiner|Before, l1: [[10, 11]], l2: [[12, 13]]|After, l1: [[10, 11, 12], [13]], l2: [[13]]
Combiner|Before, l1: [[1, 2, 3]], l2: [[4, 5, 6]]|After, l1: [[1, 2, 3], [4, 5, 6]], l2: [[4, 5, 6]]
Combiner|Before, l1: [[7, 8, 9]], l2: [[10, 11, 12], [13]]|After, l1: [[7, 8, 9], [10, 11, 12], [13]], l2: [[10, 11, 12], [13]]
Combiner|Before, l1: [[1, 2, 3], [4, 5, 6]], l2: [[7, 8, 9], [10, 11, 12], [13]]|After, l1: [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12], [13]], l2: [[7, 8, 9], [10, 11, 12], [13]]
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12], [13]]

我可能已经完全扼杀了流的概念，但是有没有办法简化、优化或重写它？

为了简单起见，这是完整的程序（遗憾的是，stackoverflow 不允许在没有足够描述的情况下按原样发布代码）

http://rextester.com/TCMG7963

【问题讨论】：

对于初学者，要将其捆绑到单个类中，您可以将其实现为 Collector
所以你想基本上将元素分组？如果是这样，这里已经问了很多次了
@Eugene 你介意分享一个吗？
这里有一个这样的线程：stackoverflow.com/questions/32434592/partition-a-java-8-stream。如果流是有序的，最简单的解决方案可能就是将其转换为 ArrayList，使用 Guava 中的 Lists.partition() 并从中创建另一个流。

标签： java java-8 functional-programming rx-java java-stream

【解决方案1】：

这是我想出的¹：

public static <T> Stream<List<T>> buffer(Stream<T> stream, final long count) {
    final Iterator<T> streamIterator = stream.iterator();

    return StreamSupport.stream(Spliterators.spliteratorUnknownSize(new Iterator<List<T>>() {
        @Override
        public boolean hasNext() {
            return streamIterator.hasNext();
        }

        @Override
        public List<T>next() {
            if (!hasNext()) {
                throw new NoSuchElementException();
            }
            List<T> intermediate = new ArrayList<>();
            for (long v = 0; v < count && hasNext(); v++) {
                intermediate.add(streamIterator.next());
            }
            return intermediate;
        }
    }, 0), false);
}

显然，您传递给此函数的流以后无法修改。

展示其功能的测试：

public class Test {
    public static void main(String[] args) {
        List<List<Integer>> triads = buffer(IntStream.range(1, 14)
                                                    .boxed()
                                                    .parallel(), 3).collect(Collectors.toList());

        System.out.println(triads);
        System.out.println("Empty stream test");
        System.out.println(buffer(Stream.<Integer>empty(), 4).collect(Collectors.toList()));

        System.out.println("Intermediate size greater than stream size");
        System.out.println(buffer(IntStream.range(1, 14)
                                          .boxed()
                                          .parallel(), 15).collect(Collectors.toList()));

        System.out.println("Intermediate size same as stream size");
        System.out.println(buffer(IntStream.range(1, 14)
                                          .boxed()
                                          .parallel(), 14).collect(Collectors.toList()));

        System.out.println("Intermediate size is a multiple of stream size");
        System.out.println(buffer(IntStream.range(0, 14)
                                          .boxed()
                                          .parallel(), 7).collect(Collectors.toList()));
    }

输出

/tmp 
➜ javac Test.java

/tmp 
➜ java Test      
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12], [13]]
Empty stream test
[]
Intermediate size greater than stream size
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]]
Intermediate size same as stream size
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]]
Intermediate size is a multiple of stream size
[[0, 1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12, 13]]

¹_{我最初打算让它返回一个 Stream of 流，但我意识到使用流迭代器会使这项任务变得比预期的要困难得多流的惰性。}

【讨论】：