根据属性在 java 中的多个属性上从对象列表中分离重复项和非重复项 8

Question

我想根据属性 customerId 和 customerRegistration 使用 java 8

从对象列表中分离重复项和非重复项

segregate meaning

如果我有 3 条记录具有相同的 customerId 和 customerRegistration，则第一个记录应添加到非重复列表中，第二条记录应添加到重复列表中。

我在下面的 pojo class 中创建了基于 customerId 和 customerRegistration 的 hashcode 和 equals 方法。

public class Asset {

private String customerName;    
private int customerId;   
private String Type;   
private String customerRegistration;   
private int assetTagList;   
private String rso;    
private String active;   
private int billableUser;    
private int billableAsset;

@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;
    Asset asset = (Asset) o;
    return customerId == asset.customerId && Objects.equals(customerRegistration, asset.customerRegistration);
}

@Override
public int hashCode() {
    return Objects.hash(customerId, customerRegistration);
}

}

我在 java7 中写了下面的逻辑，它可以正常工作，有什么方法可以使用流来编写（将下面代码中的 forloop 逻辑转换为流）？

public  Map<String ,List<Asset>> execute1() throws IOException {

    List<Asset> assetList=new ArrayList<>();
    Set<Asset> set=new HashSet<>();
    List<Asset> duplicateList=new ArrayList<>();
    List<Asset> nonDuplicateList=new ArrayList<>();
    Map<String ,List<Asset>> map = new HashMap<>();
    Asset asset1 = new Asset("French Customer1", 673, "Vehicle", "KNH 9009", 175, "FR", "Yes", 0, 0);
    Asset asset2 = new Asset("French Customer2", 673, "Vehicle", "KNH 9009", 175, "FR", "Yes", 0, 0);
    
    Asset asset3 = new Asset("French Customer3", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
    Asset asset4= new Asset("French Customer4", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
    Asset asset5= new Asset("French Customer4", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
    
    Asset asset6 = new Asset("French Customer5", 674, "Vehicle", "KNH 9008", 175, "FR", "Yes", 0, 0);
    
    assetList.add(asset1); 
    assetList.add(asset2);
    assetList.add(asset3);
    assetList.add(asset4);
    assetList.add(asset5);
    assetList.add(asset6);


    for (Asset assetTemp:assetList){
        if(set.add(assetTemp)){
            nonDuplicateList.add(assetTemp);
        }
        else {
            duplicateList.add(assetTemp);
        }
    }

    map.put("duplicate",duplicateList);
    map.put("nonDuplicateList",nonDuplicateList);
    
    
   return map
}

期望输出如下（来自上面 execute1 函数中定义的输入资产列表）

nonDuplicateList should have (asset1,asset3,asset6):

Asset asset1 = new Asset("French Customer1", 673, "Vehicle", "KNH 9009", 175, "FR", "Yes", 0, 0);
Asset asset3 = new Asset("French Customer3", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
Asset asset6 = new Asset("French Customer5", 674, "Vehicle", "KNH 9008", 175, "FR", "Yes", 0, 0);

duplicateList should have:(asset2,asset4,asset5)

Asset asset2 = new Asset("French Customer2", 673, "Vehicle", "KNH 9009", 175, "FR", "Yes", 0, 0);
Asset asset4= new Asset("French Customer4", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
Asset asset5= new Asset("French Customer4", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);

Answer 1

您可以根据添加到集合中的元素对元素进行分组 returns 对或错：

public  static Map<String ,List<Asset>> execute2() throws IOException {

    List<Asset> assetList=new ArrayList<>();
    Asset asset1 = new Asset("French Customer1", 673, "Vehicle", "KNH 9009", 175, "FR", "Yes", 0, 0);
    Asset asset2 = new Asset("French Customer2", 673, "Vehicle", "KNH 9009", 175, "FR", "Yes", 0, 0);

    Asset asset3 = new Asset("French Customer3", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
    Asset asset4= new Asset("French Customer4", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);
    Asset asset5= new Asset("French Customer4", 673, "Vehicle", null, 175, "FR", "Yes", 0, 0);

    Asset asset6 = new Asset("French Customer5", 674, "Vehicle", "KNH 9008", 175, "FR", "Yes", 0, 0);

    assetList.add(asset1);
    assetList.add(asset2);
    assetList.add(asset3);
    assetList.add(asset4);
    assetList.add(asset5);
    assetList.add(asset6);

    Set<Asset> set=new HashSet<>();
    return assetList.stream().collect(Collectors.groupingBy(x -> set.add(x) ? "nonDuplicateList" : "duplicate"));
}

Answer 2

您可以使用与 Set<Asset> set 相同的方法，并使用 partitioningBy 收集器将其收集到地图中。即使在给定的示例中它不会造成太大问题，请阅读 this 答案以了解 为什么您不应该使用收集器来改变现有集合。

assetList.stream().collect(Collectors.partitioningBy(set::add));

这将是 return 具有 Boolean 键的映射，其中 true 将是 non-duplicates。

根据属性在 java 中的多个属性上从对象列表中分离重复项和非重复项 8

segregate duplicates and non duplicates from a list of objects based on property on multiple property in java 8

java

spring

java-8

根据 属性 在 java 中的多个 属性 上从对象列表中分离重复项和非重复项 8

segregate duplicates and non duplicates from a list of objects based on property on multiple property in java 8

java

spring

java-8

根据属性在 java 中的多个属性上从对象列表中分离重复项和非重复项 8