java按属性合并列表中对象的实例

Question

我有一个对象人，它有名字、姓氏和电子邮件

class Person {
   String firstname;
   String lastName;
   String email;
}

我有一个Person列表，其中可能有多个Persons相同的firstName和lastName，我想通过电子邮件地址用分隔符将它们合并

即。 人A=

{
    "firstName": "David",
    "lastName": "Guiney",
    "email": "david.guiney@gmail.com"
}

B人=

{
    "firstName": "David",
    "lastName": "Guiney",
    "email": "guiney.david@hotmail.com"
}

我想把这些合并到

{
    "firstName": "David",
    "lastName": "Guiney",
    "email": "david.guiney@gmail.com;guiney.david@hotmail.com"
}

以便在我的列表中创建一个唯一的实例

Answer 1

首先，如果Person可以有多封电子邮件，我会在Person类中通过一个“emails”属性来存储电子邮件的集合来反映这一点

然后可以使用Java8流

public class Person {

private String firstName;

private String lastName;

private List<String> emails;// if a person is susceptible to have several emails, use a collection and not a single String

public Person(String firstName, String lastName, String email) {
    this.firstName = firstName;
    this.lastName = lastName;
    this.emails = new ArrayList<>();
    this.emails.add(email);
}

public static void main(String[] args) {
    List<Person> persons = Arrays.asList(
            new Person("David", "Guiney", "david.guiney@gmail.com"),
            new Person("Jean", "Patrick", "jp@pluton.com"),
            new Person("David", "Guiney", "david.guiney@hotmail.com"),
            new Person("Bruce", "Lee", "bl@kungfu.com")
    );

    // Group together the Person instances having the same firstname and same lastname
    Map<List<String>, Person> keyToPerson = persons.stream().collect(Collectors.toMap(
            p -> Arrays.asList(p.firstName, p.lastName),// define the key allowing to group person as the combination of firstname and lastname
            Function.identity(),// what is associated to the key. Here we want the Person instance itself
            (p1, p2) -> {// the logic to solve collision, i.e. instances having the same key
                p1.emails.addAll(p2.emails);// add the email(s) of the second instances in the email of the first instance
                return p1;
            }
    ));
    keyToPerson.values().forEach(p -> {
        System.out.println(p.firstName + " " + p.lastName + " (" + String.join("; ", p.emails) + ")");
    });
}

}

它应该印一些像那样的东西

David Guiney (david.guiney@gmail.com; david.guiney@hotmail.com)
Bruce Lee (bl@kungfu.com)
Jean Patrick (jp@pluton.com)

您还可以重新定义Person类的equals和hashcode方法，以实现如何将2个实例分组在一起

Answer 2

1. 这取决于您定义的唯一性或相等性。这可以用equals和hashCode方法表示

2. 您可以使用java。util。流动收集器#toMap方法提供合并功能并将列表映射到映射。在merge函数中，您可以实现如何处理具有相同“键”的两个对象的逻辑

   公共阶层人士 { 公众人物（stringfirstname、stringlastname、stringemail） { 这名字=名字； 这lastName=lastName； 这电子邮件=电子邮件； }

    String firstname;
    String lastName;
    String email;
   
    @Override
    public boolean equals(Object o)
    {
        if (this == o)
        {
            return true;
        }
        if (o == null || getClass() != o.getClass())
        {
            return false;
        }
        Person person = (Person) o;
        return Objects.equals(firstname, person.firstname) && Objects.equals(lastName, person.lastName);
    }
   
    @Override
    public int hashCode()
    {
        return Objects.hash(firstname, lastName);
    }
   
    @Override
    public String toString()
    {
        return "Person{" +
            "firstname='" + firstname + '\'' +
            ", lastName='" + lastName + '\'' +
            ", email='" + email + '\'' +
            '}';
    }
   
    public static void main(String[] args)
    {
        List<Person> persons = Arrays.asList(new Person("David", "Guiney", "david.guiney@gmail.com"),
            new Person("David", "Guiney", "david.guiney@gmail.com"),
            new Person("Andreas", "Radauer", "please_no@spam.com")
        );
   
        Map<Integer, Person> uniquePersons =
            persons.stream()
                .collect(Collectors.toMap(
                    Person::hashCode,
                    Function.identity(),
                    (person1, person2) -> {
                        person1.email = person1.email + ";" + person2.email; // this could be improved
                        return person1;
                    }
                ));
   
        System.out.println(uniquePersons.values());
    }
   

   }

如果您不想在这个用例中使用equals和hashCode，那么您当然可以提供自己的getKey逻辑

Answer 3

您可以逐个检查列表中的所有对象，但问题在于时间复杂性。它将采用O（n^2）格式

for(int i=0;i<list.size();i++){
        for(int j=i+1;j<list.size();j++){
            if(list.get(i).firstName.equals(list.get(j).firstName) && list.get(i).lastName.equals(list.get(j).lastName)){
                list.get(i).email += ";"+list.get(j).email;
                list.remove(j);
            }
        }
    }