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：
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());
};
这里的累加器是：
// 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());
};
和组合器
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]]
这会产生以下输出：
{{1}}
我可能已完全扼杀了流的概念，但是有没有办法简化，优化或重写它？
这是一个简单的完整程序（遗憾的是，stackoverflow不允许在没有足够描述的情况下发布代码）
答案 0 :(得分：0)
以下是我提出的 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，以后无法修改传递给此函数的流。
用于演示其功能的测试：
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()));
}
/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 我最初会让它返回一个Stream流，但我意识到使用流迭代器会使这个任务比它需要的更困难对于溪流的懒惰性质。