Question

我想以小批量大小迭代一个ArrayList。

例如，如果ArrayList大小为75且批量大小为10，我希望它处理0-10，然后是10-20，然后是20-30等记录。

我尝试了这个，但它不起作用：

int batchSize = 10;
int start = 0;
int end = batchSize;

for(int counter = start ; counter < end ; counter ++)
{
    if (start > list.size())
    {
        System.out.println("breaking");
        break;
    }

    System.out.println("counter   " + counter);
    start = start + batchSize;
    end = end + batchSize;
}

Answer 1

您需要的是：Lists.partition(java.util.List, int)来自Google Guava

示例：

final List<String> listToBatch = new ArrayList<>();
final List<List<String>> batch = Lists.partition(listToBatch, 10);
for (List<String> list : batch) {
  // Add your code here
}

Answer 2

您可以像批量大小和列表大小那样使用余数来查找计数。

int batchSize = 10;
int start = 0;
int end = batchSize;

int count = list.size() / batchSize;
int remainder = list.size() % batchSize;
int counter = 0;
for(int i = 0 ; i < count ; i ++)
{
    System.out.println("counter   " + counter);
    for(int counter = start ; counter < end ; counter ++)
    {
        //access array as a[counter]
    }
    start = start + batchSize;
    end = end + batchSize;
}

if(remainder != 0)
{
    end = end - batchSize + remainder;
    for(int counter = start ; counter < end ; counter ++)
    {
       //access array as a[counter]
    }
}

Answer 3

int start = 0;
int end=updateBatchSize;
List finalList = null;
 try {
        while(end < sampleList.size()){
            if(end==sampleList.size()){
                break;
            }
            finalList = sampleList.subList(Math.max(0,start),Math.min(sampleList.size(),end));
            start=Math.max(0,start+updateBatchSize);
            end=Math.min(sampleList.size(),end+updateBatchSize);
        }

循环arraylist批量

3 个答案: