How to filter and combine objects if they has same fields

Question

I've tried to solve the problem and am stuck. I have class User:

public class User {

   public String name;
   public String email;
   public Integer age;
   public String group;

   public User() {
   }

   public User(String name, String email, Integer age, String group) {
      this.name = name;
      this.email = email;
      this.age = age;
      this.group = group;
   }
}

And list of users looks like:

List<User> users = new ArrayList<>();
users.add(new User("Max" , "test@test", 20 , "n1"));
users.add(new User("John" , "list@test", 21 , "n2"));
users.add(new User("Nancy" , "must@test", 22 , "n3"));
users.add(new User("Nancy" , "must@test", 22 , "n4"));
users.add(new User("Max" , "test@test", 20 , "n5"));

But this list contains duplicate objects with a difference only in the group. So I need to combine duplicate objects to new object looks like :

User: name: "Max", email: "test@test" , age: 20, groups: "n1, n5"

User: name: "John", email: "list@test" , age: 21, groups: "n2"

User: name: "Nancy", email: "must@test" , age: 22, groups: "n3, n4"

I understand that I need to use steams from Java 8, but don't understand exactly how.

Please, help

I don't really think Streams will help you here. What's your level of Java: are you ok with pseudo-code and implement it yourself of do you want a full solution? — Mihai, May 16 '19 at 11:22
The expected out doesn't match the given description. Please, [edit] the post and correct it in order to help you better. Also see coment from @Zabuza. Images are less helpful than actual code. — lealceldeiro, May 16 '19 at 11:23
I would suggest adding an `id` Field for your user object, as identify unique users easily. — BlackPearl, May 16 '19 at 11:34
Possible duplicate of [Java stream merge or reduce duplicate objects](https://stackoverflow.com/questions/41539663/java-stream-merge-or-reduce-duplicate-objects) — Jack Flamp, May 16 '19 at 11:37
@Mihai Im ok with it , even pseodo-code can be helpful for me — Vlad Khurtin, May 16 '19 at 11:50
@JackFlamp I saw this topic, but could not understand the part of the code with the merge function. — Vlad Khurtin, May 16 '19 at 12:05
@VladKhurtin check [this answer](https://stackoverflow.com/questions/56167125/how-to-filter-and-combine-objects-if-they-has-same-fields/56186469#56186469) - you can do this operation in one pass — Adrian, May 18 '19 at 13:12

score 2 · Answer 1 · answered May 17 '19 at 12:29

you can take advantage of toMap collector as it has a merge function which will join your duplicate objects, for example I will create a new object every time a duplicate is found, but you can just modify the existing object

static User join(User a, User b) {
    return new User(a.getName(), a.getEmail(), a.getAge(), a.getGroup() + "," + b.getGroup());
}

and the stream op.

List<User> collect = users.stream()
            .collect(Collectors.collectingAndThen(Collectors.toMap(User::getEmail,
                            Function.identity(), 
                            (a, b) -> join(a, b)),
                    map -> new ArrayList<>(map.values())));

score 1 · Answer 2 · answered May 16 '19 at 12:12

This is a working example of what you need (I hope :)).

It considers the combination of the first 3 fields as a unique key. It then goes through the list and adds Users to a Map based on the key and having the group as a value. I use a Map because it makes retrieval faster. Before inserting a new User I check if it is already in the map. If it is then I append the new group. If it is not I insert it with the current group.

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class User {

    public String name;
    public String email;
    public Integer age;
    public String group;

    public static final void main(String[] args) {
        List<User> users = new ArrayList<>();
        users.add(new User("Max", "test@test", 20, "n1"));
        users.add(new User("John", "list@test", 21, "n2"));
        users.add(new User("Nancy", "must@test", 22, "n3"));
        users.add(new User("Nancy", "must@test", 22, "n4"));
        users.add(new User("Max", "test@test", 20, "n5"));

        List<User> filtered = filter(users);
        filtered.stream().forEach(System.out::println);
    }

    public User() {
    }

    public User(String key, String group) {
        String[] keys = key.split("-");
        this.name = keys[0];
        this.email = keys[1];
        this.age = Integer.parseInt(keys[2]);
        this.group = group;
    }

    public User(String name, String email, Integer age, String group) {
        this.name = name;
        this.email = email;
        this.age = age;
        this.group = group;
    }

    public String toString() {
        return name + " : " + email + " : " + " : " + age + " : " + group;
    }

    public String getUniqueKey() {
        return name + "-" + email + "-" + age;
    }

    public static List<User> filter(List<User> users) {
        Map<String, String> uniqueGroup = new HashMap<>();
        for (User user : users) {
            String found = uniqueGroup.get(user.getUniqueKey());
            if (null == found) {
                uniqueGroup.put(user.getUniqueKey(), user.group);
            } else {
                uniqueGroup.put(user.getUniqueKey(), found + ", " + user.group);
            }
        }

        List<User> newUsers = new ArrayList<>();
        for (String key : uniqueGroup.keySet()) {
            newUsers.add(new User(key, uniqueGroup.get(key)));
        }

        return newUsers;
    }

}

Nicholas K · Accepted Answer · 2019-05-16T13:20:31.310

You may simply do:

List<User> sortedUsers = new ArrayList<>();
// group by email-id
Map<String, List<User>> collectMap = 
                 users.stream().collect(Collectors.groupingBy(User::getEmail));

collectMap.entrySet().forEach(e -> {
    String group = e.getValue().stream()                     // collect group names
                               .map(i -> i.getGroup())
                               .collect(Collectors.joining(","));
    User user = e.getValue().get(0);
    sortedUsers.add(new User(user.getName(), user.getEmail(), user.getAge(), group));
});

which outputs:

[
   User [name=John, email=list@test, age=21, group=n2], 
   User [name=Max, email=test@test, age=20, group=n1,n5], 
   User [name=Nancy, email=must@test, age=22, group=n3,n4]
]

Make sure to add getters and setters, also override the toString() of User.

How to filter and combine objects if they has same fields

3 Answers3