swift - 在看到特定数量的事件后处理下一个事件

Question

我正在尝试做的是订阅一个发出一系列 Enums 的可观察对象。目标是每当我看到 3 个特定类型的 Enums 时，我的 onNext 就会被调用。以下是我尝试过的但是，它只能使用一次。它不会继续下去。我想知道处理此问题的最佳方法是什么。

enum Baseball {
 case strike, ball, hit
}

let bag = DisposeBag()
let subject = PublishSubject<Baseball>()

subject.filter { $0 == .strike }
  .elementAt(2)
  .subscribe(onNext: { _ in print("3 Strikes you're out") 
}).addDisposableTo(bag)

// First batter
subject.onNext(.strike)
subject.onNext(.ball)
subject.onNext(.ball)
subject.onNext(.ball)
subject.onNext(.strike)
subject.onNext(.strike) // 3 Strikes you're out is printed

// Second batter
subject.onNext(.ball)
subject.onNext(.ball)
subject.onNext(.hit)

// Third batter
subject.onNext(.strike)
subject.onNext(.strike)
subject.onNext(.strike) // Would like this to fire as well

Answer 1

使用缓冲运算符:

subject.filter { $0 == .strike }
    .buffer(timeSpan: 3e7, count: 3, scheduler: MainScheduler.instance)
    .subscribe(onNext: { print("3 Strikes you're out") })
    .addDisposableTo(bag)

现在每次有 3 次罢工时都会发出信号，或者大约每年一次。

如果您不喜欢它每年超时的事实，您可以编写自己的缓冲区运算符，它只需要一个计数:

extension Observable {
    func buffer(count: Int) -> Observable<[E]> {
        return Observable<[E]>.create { observer in
            var elements: [E] = []
            let lock = NSRecursiveLock()
            return self.subscribe { event in
                switch event {
                case .completed:
                    observer.onCompleted()
                case .error(let error):
                    observer.onError(error)
                case .next(let element):
                    lock.lock(); defer { lock.unlock() }
                    elements.append(element)
                    if elements.count == count {
                        observer.onNext(elements)
                        elements = []
                    }
                }
            }
        }
    }
}