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

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]]

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);
List<List<Integer>> allChunkLists = new ArrayList<List<Integer>>();
List<Integer> chunkList = null;
int fromIndex = 0;
int toIndex = CHUNK_SIZE;

while (fromIndex < origList.size()) {
   chunkList = origList.subList(fromIndex, (toIndex > origList.size() ? origList.size() : toIndex));
   allChunkLists.add(chunkList);
   fromIndex = toIndex;
   toIndex += CHUNK_SIZE;
}

没有库,只有Java的subList()。toIndex需要适当地有界,以避免在subList()中出现越界错误。

polygenelubricants提供的答案将基于给定数组的大小。我正在寻找将数组分割成给定数量的部分的代码。以下是我对代码所做的修改:

public static <T>List<List<T>> chopIntoParts( final List<T> ls, final int iParts )
{
    final List<List<T>> lsParts = new ArrayList<List<T>>();
    final int iChunkSize = ls.size() / iParts;
    int iLeftOver = ls.size() % iParts;
    int iTake = iChunkSize;

    for( int i = 0, iT = ls.size(); i < iT; i += iTake )
    {
        if( iLeftOver > 0 )
        {
            iLeftOver--;

            iTake = iChunkSize + 1;
        }
        else
        {
            iTake = iChunkSize;
        }

        lsParts.add( new ArrayList<T>( ls.subList( i, Math.min( iT, i + iTake ) ) ) );
    }

    return lsParts;
}

希望它能帮助到别人。

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());
        }
    }