如何将一个数组列表(size=1000)拆分为多个相同大小(=10)的数组列表?

ArrayList<Integer> results;

当前回答

你可以使用Eclipse Collections中的chunk方法:

ArrayList<Integer> list = new ArrayList<>(Interval.oneTo(1000));
RichIterable<RichIterable<Integer>> chunks = Iterate.chunk(list, 10);
Verify.assertSize(100, chunks);

这篇DZone文章中还包含了一些块方法的示例。

注意:我是Eclipse Collections的提交者。

其他回答

我猜你遇到的问题是命名100个数组列表并填充它们。您可以创建一个数组列表数组,并使用循环填充每个数组列表。

最简单(也是最愚蠢的)的方法是这样的:

ArrayList results = new ArrayList(1000);
    // populate results here
    for (int i = 0; i < 1000; i++) {
        results.add(i);
    }
    ArrayList[] resultGroups = new ArrayList[100];
    // initialize all your small ArrayList groups
    for (int i = 0; i < 100; i++) {
            resultGroups[i] = new ArrayList();
    }
    // put your results into those arrays
    for (int i = 0; i < 1000; i++) {
       resultGroups[i/10].add(results.get(i));
    } 

Java8流,一个表达式,没有其他库(两个解决方案,无需创建不必要的映射):

List<List<Integer>> partitionedList = IntStream.range(0, (list.size()-1)/targetSize+1)
        .mapToObj(i -> list.subList(i*targetSize, Math.min(i*targetSize+targetSize, list.size())))
        .collect(Collectors.toList());

List<List<Integer>> partitionedList2 = IntStream.iterate(0, i -> i < list.size(), i -> i + targetSize)
        .mapToObj(i -> list.subList(i, Math.min(i + targetSize, list.size())))
        .collect(Collectors.toList());

请记住,这些是子列表,因此对原始列表的更改也会影响这些子列表。

如果你不希望它们是子列表,而是新创建的独立列表,可以这样修改:

List<List<Integer>> partitionedList = IntStream.range(0, (list.size()-1)/targetSize+1)
        .mapToObj(i -> IntStream.range(i*targetSize, Math.min(i*targetSize+targetSize, list.size())).mapToObj(j -> list.get(j)).collect(Collectors.toList()))
        .collect(Collectors.toList());

List<List<Integer>> partitionedList2 = IntStream.iterate(0, i -> i < list.size(), i -> i + targetSize)
        .mapToObj(i -> IntStream.range(i, Math.min(i + targetSize, list.size())).mapToObj(j -> list.get(j)).collect(Collectors.toList()))
        .collect(Collectors.toList());

您需要知道您划分列表的块大小。假设您有一个包含108个条目的列表,您需要25个块大小。因此,你最终会得到5个列表:

4项各有25项; 有8个元素的。

代码:

public static void main(String[] args) {

        List<Integer> list = new ArrayList<Integer>();
        for (int i=0; i<108; i++){
            list.add(i);
        }
        int size= list.size();
        int j=0;
                List< List<Integer> > splittedList = new ArrayList<List<Integer>>()  ;
                List<Integer> tempList = new ArrayList<Integer>();
        for(j=0;j<size;j++){
            tempList.add(list.get(j));
        if((j+1)%25==0){
            // chunk of 25 created and clearing tempList
            splittedList.add(tempList);
            tempList = null;
            //intializing it again for new chunk 
            tempList = new ArrayList<Integer>();
        }
        }
        if(size%25!=0){
            //adding the remaining enteries 
            splittedList.add(tempList);
        }
        for (int k=0;k<splittedList.size(); k++){
            //(k+1) because we started from k=0
            System.out.println("Chunk number: "+(k+1)+" has elements = "+splittedList.get(k).size());
        }
    }
    **Divide a list to lists of n size**

    import java.util.AbstractList;
    import java.util.ArrayList;
    import java.util.List;

    public final class PartitionUtil<T> extends AbstractList<List<T>> {

        private final List<T> list;
        private final int chunkSize;

        private PartitionUtil(List<T> list, int chunkSize) {
            this.list = new ArrayList<>(list);
            this.chunkSize = chunkSize;
        }

        public static <T> PartitionUtil<T> ofSize(List<T> list, int chunkSize) {
            return new PartitionUtil<>(list, chunkSize);
        }

        @Override
        public List<T> get(int index) {
            int start = index * chunkSize;
            int end = Math.min(start + chunkSize, list.size());

            if (start > end) {
                throw new IndexOutOfBoundsException("Index " + index + " is out of the list range <0," + (size() - 1) + ">");
            }

            return new ArrayList<>(list.subList(start, end));
        }

        @Override
        public int size() {
            return (int) Math.ceil((double) list.size() / (double) chunkSize);
        }
    }





Function call : 
              List<List<String>> containerNumChunks = PartitionUtil.ofSize(list, 999)

详情:https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/

创建一个新列表,并使用addAll()方法添加源列表的子列表视图以创建新的子列表

List<T> newList = new ArrayList<T>();
newList.addAll(sourceList.subList(startIndex, endIndex));