![](/img/trans.png)
[英]Java8 transform a [List<Object>, String] in a Map<Object, String>
[英]Transform List of object to a Map in java
我有一個項目List<Item> resultBl
列表,如下所示:
id = 18003 amount = 128 nameType = SUBMITTED
id = 18189 amount = 95 nameType = SUBMITTED
id = 18192 amount = 160 nameType = POSITIVE
id = 18192 amount = 30 nameType = DRAFT
id = 18192 amount = 873 nameType = SUBMITTED
id = 18237 amount = 390 nameType = POSITIVE
id = 18237 amount = 60 nameType = DRAFT
id = 18237 amount = 2731 nameType = SUBMITTED
我想將此列表轉換為具有這種形式的map
,鍵是id
,值是對象list
:
Map<Integer,List<ItemDetails>> mapTest= new HashMap<Integer,List<ItemDetails>>();
[18237 , [amount = 390 ,nameType = POSITIVE],[amount = 60 nameType = DRAFT], [amount = 2731 nameType = SUBMITTED]], ...
我嘗試了不同的方法,但總是有重復的元素:
List<Integer> ids2 = new ArrayList<Integer>();
List<Integer> ids = new ArrayList<Integer>();
for(Item item: resultBl) {
ids.add(item.getId());
}
ids2 =ids.stream().distinct().collect(Collectors.toList());
Map<Integer,List<ItemDetails>> mapTest= new HashMap<Integer,List<ItemDetails>>();
List<ItemDetails> itemDetailsList = new ArrayList<ItemDetails>();
for(Integer s:ids2) {
for(Item i : resultBl) {
if(s.equals(i.getId())) {
ItemDetails it =new ItemDetails();
it.setAmount(i.getAmount());
it.setNameType(i.getNameType()) ;
itemDetailsList .add(it);
}
}
mapTest.put(s, itemDetailsList);
}
downstream
Collectors.groupingBy
和Collectors.mapping
應該可以工作:
Map<Integer, List<ItemDetails>> result = resultBl.stream().collect(Collectors.groupingBy(Item::getId, Collectors.mapping(ItemDetails::new, Collectors.toList())));
我會使用流和groupingBy
來做到這一點。
當你有你的物品清單resultBl
你所要做的就是
Map<Integer, List<Item>> resultMap = resultBl.stream().collect(groupingBy(Item::getId, toList()));
進口:
import static java.util.stream.Collectors.groupingBy;
import static java.util.stream.Collectors.toList;
這將按id
參數對您的項目進行分組。
您最好的選擇是向ItemDetails
添加一個接受Item
對象的構造函數:
public class ItemDetails {
// properties, getters and setters
public ItemDetails(Item item) {
this.amount = item.getAmount();
this.nameType = item.getNameType();
}
}
然后使用以下 Java 8 功能:
Map<Integer, List<ItemDetails>> itemsPerId =
itemsList.stream().collect(Collectors.groupingBy(Item::getId, Collectors.mapping(ItemDetails::new, Collectors.toList())));
Map<Integer,List<Item>> map = resultBL.stream(Collectors.toMap(Item::getId, Function.identity()));
它是拉姆達。 Java 版本 >= 8
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.