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

如果你不想导入apache Commons库,试试下面这段简单的代码:

final static int MAX_ELEMENT = 20;

public static void main(final String[] args) {

    final List<String> list = new ArrayList<String>();

    for (int i = 1; i <= 161; i++) {
        list.add(String.valueOf(i));
        System.out.print("," + String.valueOf(i));
    }
    System.out.println("");
    System.out.println("### >>> ");
    final List<List<String>> result = splitList(list, MAX_ELEMENT);

    for (final List<String> entry : result) {
        System.out.println("------------------------");
        for (final String elm : entry) {
            System.out.println(elm);
        }
        System.out.println("------------------------");
    }

}

private static List<List<String>> splitList(final List<String> list, final int maxElement) {

    final List<List<String>> result = new ArrayList<List<String>>();

    final int div = list.size() / maxElement;

    System.out.println(div);

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

        final int startIndex = i * maxElement;

        if (startIndex >= list.size()) {
            return result;
        }

        final int endIndex = (i + 1) * maxElement;

        if (endIndex < list.size()) {
            result.add(list.subList(startIndex, endIndex));
        } else {
            result.add(list.subList(startIndex, list.size()));
        }

    }

    return result;
}

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

Apache Commons Collections 4在ListUtils类中有一个分区方法。下面是它的工作原理:

import org.apache.commons.collections4.ListUtils;
...

int targetSize = 100;
List<Integer> largeList = ...
List<List<Integer>> output = ListUtils.partition(largeList, targetSize);

你可以使用subList(int fromIndex, int toIndex)来获取原始列表的一部分。

来自API:

返回该列表中指定的fromIndex(包含)和toIndex(不包含)之间部分的视图。(如果fromIndex和toIndex相等,返回的列表为空。)返回的列表由该列表支持,因此返回列表中的非结构性更改会反映在该列表中,反之亦然。返回的列表支持该列表支持的所有可选列表操作。

例子:

List<Integer> numbers = new ArrayList<Integer>(
    Arrays.asList(5,3,1,2,9,5,0,7)
);

List<Integer> head = numbers.subList(0, 4);
List<Integer> tail = numbers.subList(4, 8);
System.out.println(head); // prints "[5, 3, 1, 2]"
System.out.println(tail); // prints "[9, 5, 0, 7]"

Collections.sort(head);
System.out.println(numbers); // prints "[1, 2, 3, 5, 9, 5, 0, 7]"

tail.add(-1);
System.out.println(numbers); // prints "[1, 2, 3, 5, 9, 5, 0, 7, -1]"

如果您需要这些切碎的列表不是一个视图,那么只需从subblist创建一个新的List。下面是一个把这些东西放在一起的例子:

// chops a list into non-view sublists of length L
static <T> List<List<T>> chopped(List<T> list, final int L) {
    List<List<T>> parts = new ArrayList<List<T>>();
    final int N = list.size();
    for (int i = 0; i < N; i += L) {
        parts.add(new ArrayList<T>(
            list.subList(i, Math.min(N, i + L)))
        );
    }
    return parts;
}


List<Integer> numbers = Collections.unmodifiableList(
    Arrays.asList(5,3,1,2,9,5,0,7)
);
List<List<Integer>> parts = chopped(numbers, 3);
System.out.println(parts); // prints "[[5, 3, 1], [2, 9, 5], [0, 7]]"
parts.get(0).add(-1);
System.out.println(parts); // prints "[[5, 3, 1, -1], [2, 9, 5], [0, 7]]"
System.out.println(numbers); // prints "[5, 3, 1, 2, 9, 5, 0, 7]" (unmodified!)

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));
    } 

Java 8

我们可以根据大小或条件拆分列表。

static Collection<List<Integer>> partitionIntegerListBasedOnSize(List<Integer> inputList, int size) {
        return inputList.stream()
                .collect(Collectors.groupingBy(s -> (s-1)/size))
                .values();
}
static <T> Collection<List<T>> partitionBasedOnSize(List<T> inputList, int size) {
        final AtomicInteger counter = new AtomicInteger(0);
        return inputList.stream()
                    .collect(Collectors.groupingBy(s -> counter.getAndIncrement()/size))
                    .values();
}
static <T> Collection<List<T>> partitionBasedOnCondition(List<T> inputList, Predicate<T> condition) {
        return inputList.stream().collect(Collectors.partitioningBy(s-> (condition.test(s)))).values();
}

然后我们可以把它们用作:

final List<Integer> list = Arrays.asList(1,2,3,4,5,6,7,8,9,10);
System.out.println(partitionIntegerListBasedOnSize(list, 4));  // [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
System.out.println(partitionBasedOnSize(list, 4));  // [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
System.out.println(partitionBasedOnSize(list, 3));  // [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
System.out.println(partitionBasedOnCondition(list, i -> i<6));  // [[6, 7, 8, 9, 10], [1, 2, 3, 4, 5]]