我为自己编写了一个实用程序,将列表分解为给定大小的批次。我只是想知道是否已经有任何apache commons util用于此。

public static <T> List<List<T>> getBatches(List<T> collection,int batchSize){
    int i = 0;
    List<List<T>> batches = new ArrayList<List<T>>();
    while(i<collection.size()){
        int nextInc = Math.min(collection.size()-i,batchSize);
        List<T> batch = collection.subList(i,i+nextInc);
        batches.add(batch);
        i = i + nextInc;
    }

    return batches;
}

请让我知道是否有任何现有的公用事业已经相同。


当前回答

这里有一个例子:

final AtomicInteger counter = new AtomicInteger();
final int partitionSize=3;
final List<Object> list=new ArrayList<>();
            list.add("A");
            list.add("B");
            list.add("C");
            list.add("D");
            list.add("E");
       
        
final Collection<List<Object>> subLists=list.stream().collect(Collectors.groupingBy
                (it->counter.getAndIncrement() / partitionSize))
                .values();
        System.out.println(subLists);

输入: [a, b, c, d, e]

输出: [a, b, c], [d, e]]

你可以在这里找到例子: https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/

其他回答

我想到了这个:

private static <T> List<List<T>> partition(Collection<T> members, int maxSize)
{
    List<List<T>> res = new ArrayList<>();

    List<T> internal = new ArrayList<>();

    for (T member : members)
    {
        internal.add(member);

        if (internal.size() == maxSize)
        {
            res.add(internal);
            internal = new ArrayList<>();
        }
    }
    if (internal.isEmpty() == false)
    {
        res.add(internal);
    }
    return res;
}
List<T> batch = collection.subList(i,i+nextInc);
->
List<T> batch = collection.subList(i, i = i + nextInc);

您可以使用下面的代码来获得该批列表。

Iterable<List<T>> batchIds = Iterables.partition(list, batchSize);

你需要导入谷歌番石榴库来使用上面的代码。

如果你想要生成一个Java-8的批处理流,你可以尝试下面的代码:

public static <T> Stream<List<T>> batches(List<T> source, int length) {
    if (length <= 0)
        throw new IllegalArgumentException("length = " + length);
    int size = source.size();
    if (size <= 0)
        return Stream.empty();
    int fullChunks = (size - 1) / length;
    return IntStream.range(0, fullChunks + 1).mapToObj(
        n -> source.subList(n * length, n == fullChunks ? size : (n + 1) * length));
}

public static void main(String[] args) {
    List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14);

    System.out.println("By 3:");
    batches(list, 3).forEach(System.out::println);
    
    System.out.println("By 4:");
    batches(list, 4).forEach(System.out::println);
}

输出:

By 3:
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
[10, 11, 12]
[13, 14]
By 4:
[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10, 11, 12]
[13, 14]

在Java 9中,你可以使用带有hasNext条件的IntStream.iterate()。所以你可以把方法的代码简化成这样:

public static <T> List<List<T>> getBatches(List<T> collection, int batchSize) {
    return IntStream.iterate(0, i -> i < collection.size(), i -> i + batchSize)
            .mapToObj(i -> collection.subList(i, Math.min(i + batchSize, collection.size())))
            .collect(Collectors.toList());
}

使用{0,1,2,3,4,5,6,7,8,9},getbatch (numbers, 4)的结果将是:

[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9]]