ios - ReactiveCocoa - 发出数组中最新 N 值的 SignalProducer

Question

我有一个 SignalProducer，ProducerA，它以不同的间隔发出值。我正在尝试收集 SignalProducer 发出的最新 N 个值，并创建一个新的生产者 ProducerB，它发出一个包含最新 N 个值的数组。

ProducerB 应该在 ProducerA 发出前 N 个值时开始发出值，然后在每次 ProducerA 发出一个新值时发出一个新数组。

有人能帮我吗？

Answer 1

let (producerA, observerA) = SignalProducer<Int, NoError>.buffer(5)
let n = 3

producerA.take(n).collect()
        .takeUntilReplacement(producerA.skip(n).map { [$0] })
        .scan([], { $0.suffix(n - 1) + $1 })
        .startWithNext {
                print($0)
}

observerA.sendNext(1) // nothing printed
observerA.sendNext(2) // nothing printed
observerA.sendNext(3) // prints [1, 2, 3]
observerA.sendNext(4) // prints [2, 3, 4]
observerA.sendNext(5) // prints [3, 4, 5]

Answer 2

我想出了这段代码

extension SignalProducer {
    /// Creates a new producer that emits an array that contains the latest N values that were emitted 
    /// by the original producer as specified in 'capacity'.
    @warn_unused_result(message="Did you forget to call `start` on the producer?")
    public func latestValues(n:Int) -> SignalProducer<[Value], Error> {
        var array: [Value] = []
        return self.map {
            value in

            array.append(value)

            if array.count >= n {
                array.removeFirst(array.count - n)
            }

            return array
        }
            .filter {
                $0.count == n
        }
    }
}