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

import org.apache.commons.collections4.ListUtils;
ArrayList<Integer> mainList = .............;
List<List<Integer>> multipleLists = ListUtils.partition(mainList,100);
int i=1;
for (List<Integer> indexedList : multipleLists){
  System.out.println("Values in List "+i);
  for (Integer value : indexedList)
    System.out.println(value);
i++;
}

其他回答

private ArrayList<List<String>> chunkArrayList(ArrayList<String> arrayToChunk, int chunkSize) {
    ArrayList<List<String>> chunkList = new ArrayList<>();
    int guide = arrayToChunk.size();
    int index = 0;
    int tale = chunkSize;
    while (tale < arrayToChunk.size()){
            chunkList.add(arrayToChunk.subList(index, tale));
            guide = guide - chunkSize;
            index = index + chunkSize;
            tale = tale + chunkSize;
    }
    if (guide >0) {
       chunkList.add(arrayToChunk.subList(index, index + guide));
    }
    Log.i("Chunked Array: " , chunkList.toString());
    return chunkList;
}

例子

    ArrayList<String> test = new ArrayList<>();
    for (int i=1; i<=1000; i++){
        test.add(String.valueOf(i));
    }

    chunkArrayList(test,10);

输出

分块:[[1,2,3,4,5,6,7,8,9,10),(11、12、13、14、15、16、17、18、19、20],[21日,22日,23日,24日,25日,26日,27日,28日,29日,30日],[第三十一条、第三十二条、第三十三,34岁,35岁,36岁,37岁,38岁,39岁,40],[41、42、43、44、45、46岁,47岁,48岁,49岁,50],[51岁,52岁,53岁,54岁,55岁,56岁,57岁的58岁的59岁60],[61,62,63,64,65,66,67,68,69,70],[71,72,73,74,75,76,77,78,79,80],[81,82,83,84,85,86,87,88,89,90],[91,92,93,94,95,96,97,98,99,100 ], .........

你会在日志里看到的

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。

Java8流,一个表达式,没有其他库:

List<String> input = ...
int partitionSize = ...

 Collection<List<String>> partitionedList = IntStream.range(0, input.size())
    .boxed()
        .collect(Collectors.groupingBy(partition -> (partition / partitionSize), Collectors.mapping(elementIndex -> input.get(elementIndex), Collectors.toList())))
            .values();

测试:

List<String> input = Arrays.asList("a", "b", "c", "d", "e", "f", "g", "h" ,"i");

partitionSize = 1 - > [[a], [b], [c], [d], [e], [f], [g], [h],[我]] partitionSize = 2 - > [[a, b], c, d, e, f, g, h,[我]] partitionSize = 3 - > [a, b, c, d, e, f, g, h,我]] partitionSize = 7 - > [[a, b, c, d, e, f, g], [h,我]] partitionSize = 100 -> [[a, b, c, d, e, f, g, h, i]]

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));
    } 
import org.apache.commons.collections4.ListUtils;
ArrayList<Integer> mainList = .............;
List<List<Integer>> multipleLists = ListUtils.partition(mainList,100);
int i=1;
for (List<Integer> indexedList : multipleLists){
  System.out.println("Values in List "+i);
  for (Integer value : indexedList)
    System.out.println(value);
i++;
}