根据源 List 内容将 List 转换为 List<List>

Question

我有以下对象DataPoint：

public class DataPoint {
   private String name:
   private int uploadIndex;
   ...
}

我有清单：

{uploadIndex:1, name:"name_1"}
{uploadIndex:5, name:"name_2"}
{uploadIndex:3, name:"name_3"}
{uploadIndex:3, name:"name_4"}
{uploadIndex:4, name:"name_5"}
{uploadIndex:5, name:"name_6"}
{uploadIndex:2, name:"name_7"}
{uploadIndex:4, name:"name_8"}

我想将其转换为按uploadIndex 排序的List<List<DataPoint>>。
结果应该是（内部列表中的排序并不重要）：

{
   {uploadIndex:1, name:"name_1"}
} 
{
   {uploadIndex:2, name:"name_7"}
} 
{
   {uploadIndex:3, name:"name_3"}
   {uploadIndex:3, name:"name_4"}
} 
{
   {uploadIndex:4, name:"name_5"}
   {uploadIndex:4, name:"name_8"}
} 
{    
   {uploadIndex:5, name:"name_2"}
   {uploadIndex:5, name:"name_6"}
} 

我写了以下代码：

List<DataPoint> dataPoints = getFromDb();
Map<Integer, List<DataPoint>> groupedDataPoints = dataPoints.stream()
                    .collect(Collectors.groupingBy(DataPoint::getUploadIndex, TreeMap::new, Collectors.toList()));
List<List<DataPoint>> groupedList = new ArrayList<>(groupedDataPoints.values());

1.此代码是否保证订购？
2、有没有更简洁的实现方式？

Answer 1

 dataPoints.stream()
           .collect(
                 Collectors.collectingAndThen(
                      Collectors.groupingBy(
                         DataPoint::getUploadIndex, 
                         TreeMap::new, 
                         Collectors.toList()
                      )
                 ),
                      map -> new ArrayList(map.values())
           );

你可以使用Collectors.collectingAndThen;除此之外，您的解决方案对我来说看起来不错。