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

您可以将Guava库添加到项目中并使用列表。划分方法,例如:

List<Integer> bigList = ...
List<List<Integer>> smallerLists = Lists.partition(bigList, 10);

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));

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]]

使用StreamEx库,您可以使用StreamEx。ofSubLists(List<T> source, int length)方法:

返回一个新的StreamEx,它由给定源列表的不重叠子列表组成,具有指定的长度(最后一个子列表可能更短)。

// Assuming you don't actually care that the lists are of type ArrayList
List<List<Integer>> sublists = StreamEx.ofSubLists(result, 10).toList();

// If you actually want them to be of type ArrayList, per your question
List<List<Integer>> sublists = StreamEx.ofSubLists(result, 10).toCollection(ArrayList::new);

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());
        }
    }