执行groupBy操作后如何将多个对象合并为一个? [英] How to merge several objects into a single one after execute the groupBy operation?

问题描述

我创建了下面的类来说明我的疑问.

I have the class below that I've created to illustrate my doubt.

在对 Flowable 进行初始转换后，我有:

After doing the initial transformations on my flowable I have:

UserScoreTO{id=1, name='john', score=4}
UserScoreTO{id=1, name='john', score=5}
UserScoreTO{id=1, name='john', score=1}
UserScoreTO{id=2, name='paul', score=4}
UserScoreTO{id=2, name='paul', score=2}
UserScoreTO{id=3, name='mark', score=1}
UserScoreTO{id=3, name='mark', score=7}

我想将具有相同 id 的 UserScoreTO 对象组合到一个 Flowable 中，该 Flowable 为每个组发出一个对象，其中包含用户信息和分数总和.

I want to combine UserScoreTO objects with the same id into a Flowable that emits one single object for each group, that contains the user information and the sum of the scores.

所以结果将是可流动的:

So the result will be flowable that emits:

User (1, "john", 10);
User (2, "paul", 6);
User (3, "mark", 8);

如果可能的话，我想用 RxJava 来做这件事(我知道我可以用 HashMap 做一些事情来达到同样的结果).

I want to do this with RxJava if possible (I know I could achieve same result doing something with HashMaps).

package examples.rxjava;

import java.util.Arrays;
import java.util.List;

import io.reactivex.Flowable;

import static java.lang.System.out;

public class TestUsers {

    public static void main(String[] args) {
        new TestUsers().execute();
    }


    public void execute() {
        getUsers()
                .flatMap(list -> Flowable.fromIterable(list))
                .groupBy(userScoreTO -> userScoreTO.id).subscribe(group -> group.subscribe(out::println));

    }





    Flowable<List<UserScoreTO>> getUsers() {
        return Flowable.fromCallable(
                () -> Arrays.asList(
                        new UserScoreTO(1, "john", 4),
                        new UserScoreTO(1, "john", 5),
                        new UserScoreTO(1, "john", 1),

                        new UserScoreTO(2, "paul", 4),
                        new UserScoreTO(2, "paul", 2),

                        new UserScoreTO(3, "mark", 1),
                        new UserScoreTO(3, "mark", 7))
        );


    }

    private class User {
        private int id;
        private String name;
        private int totalScore;

        public User(int id, String name, int totalScore) {
            this.id = id;
            this.name = name;
            this.totalScore = totalScore;
        }

        @Override
        public String toString() {
            return "User{" +
                    "id=" + id +
                    ", name='" + name + '\'' +
                    ", totalScore=" + totalScore +
                    '}';
        }
    }

    private class UserScoreTO {
        private int id;
        private String name;
        private int score;


        public UserScoreTO(int id, String name, int score) {
            this.id = id;
            this.name = name;
            this.score = score;
        }

        @Override
        public String toString() {
            return "UserScoreTO{" +
                    "id=" + id +
                    ", name='" + name + '\'' +
                    ", score=" + score +
                    '}';
        }
    }

}

推荐答案

您提供的代码已经完全符合您的要求.groupBy 会将具有相同 id 的 UserScoreTO 对象分组到同一个 GroupedFlowable 对象中.

The code you provided already does exactly what you want. groupBy will group UserScoreTO objects with the same id into the same GroupedFlowable object.

好的，我明白了，也许以下可以更好地满足您的需求

Okay I see, maybe the following does what you want a little bit better

  public void execute() {
getUsers()
    .flatMap(Flowable::fromIterable)
    .groupBy(userScoreTO -> userScoreTO.id)
    .map(group -> group.reduce(new User(group.getKey(), "", 0),
        (user, userScoreTO) -> {
          user.name = userScoreTO.name;
          user.totalScore += userScoreTO.score;
          return user;
        }))
    .subscribe(userSingle -> userSingle.subscribe(System.out::println));

}

您想要的是通过应用 reduce 运算符将 UserScoreTo 聚合到同一个 User 对象中.

What you want is to aggregate the UserScoreTo into the same User object by applying the reduce operator.

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