java - 根据源List内容将List转换为List

Question

我有以下对象 DataPoint :

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

我有 list :

{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 List<List<DataPoint>>按 uploadIndex 排序.
结果应该是(在内部列表中排序并不重要):

{
   {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"}
} 

我写了下面的代码:

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());

<强>1。此代码是否保证顺序？
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())
           );

您可以使用Collectors.collectingAndThen；除此之外，您的解决方案对我来说还不错。