RxJava：Return 包含过去和当前结果的列表

Question

我想链接一个连续的数据流并创建一个包含过去结果的结果列表。我可以用下面的代码来做。有没有办法在 rx 链之外没有变量？谢谢！

- [0]
- [0, 1]
- [0, 1, 2]
- [0, 1, 2, 3]
- [0, 1, 2, 3, 4]
- [0, 1, 2, 3, 4, 5]

final List<Long> list = new ArrayList<>();

Observable
        .interval(1, TimeUnit.SECONDS)
        .subscribe(new Action1<Long>() {
            @Override
            public void call(Long number) {
                list.add(number);
                System.out.println("- " + list);
            }
        });

Thread.sleep(100000000L);

↓↓↓↓↓↓↓↓↓↓

Observable
        .interval(1, TimeUnit.SECONDS)
        .addToPastResultList()      // <--- something like this?
        .subscribe(new Action1<List<Long>>() {
            @Override
            public void call(List<Long> list) {
                System.out.println("- " + list);
            }
        });

Thread.sleep(100000000L);

Answer 1

我正在回答我的问题...看起来我可以创建一个带有私有变量的地图来处理它。

Observable
        .interval(1, TimeUnit.SECONDS)
        .map(new Func1<Long, List<Long>>() {
            private List<Long> list = new ArrayList<>();
            @Override
            public List<Long> call(Long number) {
                list.add(number);
                return list;
            }
        })
        .subscribe(new Action1<List<Long>>() {
            @Override
            public void call(List<Long> list) {
                System.out.println("- " + list);
            }
        });

Thread.sleep(100000000L);

Answer 2

您正在寻找 scan 操作员

Observable.interval(1, TimeUnit.SECONDS)
        .scan(new ArrayList<>(), (list, integer) -> {
            list.add(integer);
            return list;
        })
        .subscribe(list -> System.out.println(list));

RxJava：Return 包含过去和当前结果的列表

RxJava: Return List Containing Past And Current Result

java

reactive-programming

observable

rx-java

rx-android