Question

我有以下用例。我有一个具有以下结构的嵌套地图：

Map<String, Map<WorkType, List<CostLineItem>>>

我必须遍历地图并获取CLObject的列表。如果列表中的单个条目的标识符为null。我必须为每个EnumType生成唯一的标识符。我不确定如何使用流吗？遵循迭代逻辑将明确我要完成的事情

for(Map.Entry<String, Map<WorkType, List<CostLineItem>>> cliByWorkTypeIterator: clisByWorkType.entrySet()) {
       Map<WorkType, List<CostLineItem>> entryValue = cliByWorkTypeIterator.getValue();
       for(Map.Entry<WorkType, List<CostLineItem>>cliListIterator : entryValue.entrySet()) {
           List<CostLineItem> clis = cliListIterator.getValue();
           //if any CLI settlementNumber is zero this means we are in standard upload
           //TODO: Should we use documentType here? Revisit this check while doing dispute file upload
           if(clis.get(0).getSettlementNumber() == null) {
               clis.forEach(f -> f.toBuilder().settlementNumber(UUID.randomUUID().toString()).build());
           }
       }
   }

嵌套的循环使代码位的模板变脏。有人可以在这里帮助我吗？

Answer 1

您可以使用flatMap遍历所有内部List<CostLineItem>的所有Map值。

clisByWorkType.values() // returns Collection<Map<WorkType, List<CostLineItem>>>
              .stream() // returns Stream<Map<WorkType, List<CostLineItem>>>
              .flatMap(v->v.values().stream()) // returns Stream<List<CostLineItem>>
              .filter(clis -> clis.get(0).getSettlementNumber() == null) // filters that Stream
              .forEach(clis -> {do whatever logic you need to perform on the List<CostLineItem>});

Answer 2

clisByWorkType.values()
              .stream()
              .flatMap(e -> e.values().stream())
              .filter(clis -> clis.get(0).getSettlementNumber() == null)
              .flatMap(Collection::stream)
              .forEach(f -> f.toBuilder().settlementNumber(UUID.randomUUID().toString()).build());

Answer 3

以下等效于您的for循环：

clisByWorkType.entrySet()
    .map(Map.Entry::getValue) // cliByWorkTypeIterator.getValue();
    .flatMap(m -> m.entrySet().stream())
    .map(Map.Entry::getValue)
    .map(CostLineItem::getValue)
    .filter(clis.get(0).getSettlementNumber() == null) //filter before flattening
    .flatMap(List::stream)
    .forEach(f -> f.toBuilder().settlementNumber(UUID.randomUUID().toString()).build());

嵌套地图上的流

3 个答案: