Use flatMap or zip in RxJava2?

Question

I have a class called Student and it has two fields grade and school. Both of two fields need to be fetched from remote server. When two result returned, I new a Student object.

With help of RxJava, I have done it in two ways, one in flatMap and another in zip operator.

Observable<String> gradeObservable =
        Observable.create((ObservableOnSubscribe<String>) emitter -> {
            Thread.sleep(1000);
            emitter.onNext("senior");
        }).subscribeOn(Schedulers.io());

Observable<String> schoolObservable =
        Observable.create((ObservableOnSubscribe<String>) emitter -> {
            emitter.onNext("MIT");
        }).subscribeOn(Schedulers.io());

flatMap version

gradeObservable
        .flatMap(grade ->
                schoolObservable.map(school -> {
                    Student student = new Student();
                    student.grade = grade;
                    student.school = school;
                    return student;
                }))
        .subscribe(student -> {
            System.out.println(student.grade);
            System.out.println(student.school);
        });

zip version

 Observable.zip(gradeObservable, schoolObservable, (grade, school) -> {
    Student student = new Student();
    student.grade = grade;
    student.school = school;
    return student;
}).subscribe(student -> {
    System.out.println(student.grade);
    System.out.println(student.school);
});

In my opinion, zip seems more clearly. So in this situation, operator flatMap or zip is better?

Geoffrey Marizy Geoffrey Marizy · Accepted Answer · 2018-03-08T16:29:08

You are clearly composing two observable, which is the purpose of zip(). Not only that but gradeObservable and schoolObservable would be executed in parallel with zip() whereas your flatmap() solution would serialize requests. So, yes, zip() is better.

Use flatMap or zip in RxJava2?

flatMap version

zip version

1 Answers