根据java 8中多个属性的属性从对象列表中分离重复和非重复

Question

我想使用java 8根据属性customerId和customerRegistration从对象列表中分离重复项和非重复项

分离意义

如果我有3条记录具有相同的customerId和customerRegistration，则第一条记录应添加到非重复列表中，第二条记录应被添加到重复列表中。

我在下面的pojo类中创建了基于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);
}

}

我已经在java7中编写了以下逻辑，这将正常工作，有没有办法使用流编写（将以下代码中的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
}

预期输出如下(来自上述execute1函数中定义的输入资产列表)

非重复列表应具有（资产1，资产3，资产6）：

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);

CopicateList应该有：（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

您可以使用相同的方法使用＜code＞集

java prettyprint-override">assetList.stream().collect(Collectors.partitioningBy(set::add));

这将返回一个带有< code>Boolean键的映射，其中< code>true是非重复的。

Answer 2

您可以根据添加到集合中返回true或false对元素进行分组：

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"));
}