是否有一个通用的Java实用程序来将列表分成批处理?

时间:2012-08-19 13:34:23

标签: java collections

我自己写了一个实用程序来将列表分成给定大小的批次。我只是想知道是否已经有任何apache commons util。

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;
}

如果已有相同的实用程序,请告诉我。

17 个答案:

答案 0 :(得分:214)

Lists.partition(java.util.List, int)查看Google Guava

  

返回列表的连续子列表,每个列表具有相同的大小(最终列表可能更小)。例如,对包含[a, b, c, d, e]且分区大小为3的列表进行分区会产生[[a, b, c][d, e]] - 包含两个内部列表的外部列表,包含三个和两个元素,所有这些都按原始顺序排列

答案 1 :(得分:42)

如果您想生成Java-8批处理流,可以尝试以下代码:

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]

答案 2 :(得分:12)

另一种方法是使用Collectors.groupingBy索引,然后将分组索引映射到实际元素:

    final List<Integer> numbers = range(1, 12)
            .boxed()
            .collect(toList());
    System.out.println(numbers);

    final List<List<Integer>> groups = range(0, numbers.size())
            .boxed()
            .collect(groupingBy(index -> index / 4))
            .values()
            .stream()
            .map(indices -> indices
                    .stream()
                    .map(numbers::get)
                    .collect(toList()))
            .collect(toList());
    System.out.println(groups);

输出:

  

[1,2,3,4,5,6,7,8,9,10,11]

     

[[1,2,3,4],[5,6,7,8],[9,10,11]]

答案 3 :(得分:6)

我想出了这个:

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;
}

答案 4 :(得分:5)

以下示例演示了List的分块:

package de.thomasdarimont.labs;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class SplitIntoChunks {

    public static void main(String[] args) {

        List<Integer> ints = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11);

        List<List<Integer>> chunks = chunk(ints, 4);

        System.out.printf("Ints:   %s%n", ints);
        System.out.printf("Chunks: %s%n", chunks);
    }

    public static <T> List<List<T>> chunk(List<T> input, int chunkSize) {

        int inputSize = input.size();
        int chunkCount = (int) Math.ceil(inputSize / (double) chunkSize);

        Map<Integer, List<T>> map = new HashMap<>(chunkCount);
        List<List<T>> chunks = new ArrayList<>(chunkCount);

        for (int i = 0; i < inputSize; i++) {

            map.computeIfAbsent(i / chunkSize, (ignore) -> {

                List<T> chunk = new ArrayList<>();
                chunks.add(chunk);
                return chunk;

            }).add(input.get(i));
        }

        return chunks;
    }
}

输出:

Ints:   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
Chunks: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]

答案 5 :(得分:1)

List<T> batch = collection.subList(i,i+nextInc);
->
List<T> batch = collection.subList(i, i = i + nextInc);

答案 6 :(得分:1)

这里有个例子:

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]]

您可以在此处找到示例: https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/

答案 7 :(得分:1)

类似于OP,没有流和库,但简洁:

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

答案 8 :(得分:1)

使用Apache Commons ListUtils.partition

答案 9 :(得分:1)

在Java 9中,可以在hasNext条件下使用IntStream.iterate()。因此,您可以将方法的代码简化为:

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}getBatches(numbers, 4)的结果将是:

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

答案 10 :(得分:1)

曾经有another question被关闭,因为它是与此书的复制品,但是如果您仔细阅读,它的确会有所不同。因此,如果有人(例如我)实际上想要将列表拆分为给定数量的几乎相等大小的子列表,然后继续阅读。

我只是将here中描述的算法移植到Java。

@Test
public void shouldPartitionListIntoAlmostEquallySizedSublists() {

    List<String> list = Arrays.asList("a", "b", "c", "d", "e", "f", "g");
    int numberOfPartitions = 3;

    List<List<String>> split = IntStream.range(0, numberOfPartitions).boxed()
            .map(i -> list.subList(
                    partitionOffset(list.size(), numberOfPartitions, i),
                    partitionOffset(list.size(), numberOfPartitions, i + 1)))
            .collect(toList());

    assertThat(split, hasSize(numberOfPartitions));
    assertEquals(list.size(), split.stream().flatMap(Collection::stream).count());
    assertThat(split, hasItems(Arrays.asList("a", "b", "c"), Arrays.asList("d", "e"), Arrays.asList("f", "g")));
}

private static int partitionOffset(int length, int numberOfPartitions, int partitionIndex) {
    return partitionIndex * (length / numberOfPartitions) + Math.min(partitionIndex, length % numberOfPartitions);
}

答案 11 :(得分:0)

import com.google.common.collect.Lists;

List<List<T>> batches = Lists.partition(List<T>,batchSize)

使用Lists.partition(List,batchSize)。您需要从Google通用软件包(Lists)导入com.google.common.collect.Lists

它将返回List<T>的列表,且每个元素的大小等于您的batchSize

答案 12 :(得分:0)

Java 8的单行代码是:

import static java.util.function.Function.identity;
import static java.util.stream.Collectors.*;

private static <T> Collection<List<T>> partition(List<T> xs, int size) {
    return IntStream.range(0, xs.size())
            .boxed()
            .collect(collectingAndThen(toMap(identity(), xs::get), Map::entrySet))
            .stream()
            .collect(groupingBy(x -> x.getKey() / size, mapping(Map.Entry::getValue, toList())))
            .values();

}

答案 13 :(得分:0)

这是Java 8+的简单解决方案:

public static <T> Collection<List<T>> prepareChunks(List<T> inputList, int chunkSize) {
    AtomicInteger counter = new AtomicInteger();
    return inputList.stream().collect(Collectors.groupingBy(it -> counter.getAndIncrement() / chunkSize)).values();
}

答案 14 :(得分:0)

您可以使用下面的代码来获取列表的批次。

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

您需要导入Google Guava库才能使用上述代码。

答案 15 :(得分:0)

解决此问题的另一种方法:

public class CollectionUtils {

    /**
    * Splits the collection into lists with given batch size
    * @param collection to split in to batches
    * @param batchsize size of the batch
    * @param <T> it maintains the input type to output type
    * @return nested list
    */
    public static <T> List<List<T>> makeBatch(Collection<T> collection, int batchsize) {

        List<List<T>> totalArrayList = new ArrayList<>();
        List<T> tempItems = new ArrayList<>();

        Iterator<T> iterator = collection.iterator();

        for (int i = 0; i < collection.size(); i++) {
            tempItems.add(iterator.next());
            if ((i+1) % batchsize == 0) {
                totalArrayList.add(tempItems);
                tempItems = new ArrayList<>();
            }
        }

        if (tempItems.size() > 0) {
            totalArrayList.add(tempItems);
        }

        return totalArrayList;
    }

}

答案 16 :(得分:0)

请注意,List#subList() 返回基础集合的副本,这在编辑较小的列表时可能会导致意外后果 - 编辑将反映在原始集合中或可能引发 ConcurrentModificationException