如何设计一个流，它是另一个流的每一个最新的5个数据的平均值？

Question

有一个流，它每100 as发出一次数据，我希望得到该流的每5个最新数据的平均值，并将平均值作为双值转换为另一个流。

我如何设计流程？

码A

   fun soundDbFlow(period: Long = 100) = flow {
        while (true) {
            var data = getAmplitude()
            emit(data)
            delay(period)
        }
    }
    .get_Average_Per5_LatestData {...} //How can I do? or is there other way?
    .map { soundDb(it) }

    private fun getAmplitude(): Int {
        var result = 0
        mRecorder?.let {
            result = it.maxAmplitude
        }
        return result
    }    
        
    private fun soundDb(input:Int, referenceAmp: Double = 1.0): Double {
        return 20 * Math.log10(input / referenceAmp)
    }

添加的内容：

给plplmax:谢谢！

我假设代码B将发射1，2，3，4，5，6，7，8，9，10.

你保证代码C会先计算(1+2+3+4+5)/5，然后计算(6+7+8+9+10)/5第二，.？这是我的期望。

我担心C代码可能首先计算(1+2+3+4+5)/5，计算(2+3+4+5+6)/5感受器，

码B

suspend fun soundDbFlow(period: Long) = flow {
    while (true) {
        val data = getAmplitude()
        emit(data)
        delay(period)
    }
}

代码C

private fun reduceFlow(period: Long = 100) = flow {
    while (true) {
        val result = soundDbFlow(period)
            .take(5)
            .map { soundDb((it / 5.0).roundToInt()) }
            .reduce { accumulator, value -> accumulator + value }
        emit(result)
    }
}

Answer 1

您可以编写这样的分块操作符：

/**
 * Returns a Flow that emits sequential [size]d chunks of data from the source flow, 
 * after transforming them with [transform].
 * 
 * The list passed to [transform] is transient and must not be cached.
 */
fun <T, R> Flow<T>.chunked(size: Int, transform: suspend (List<T>)-> R): Flow<R> = flow {
    val cache = ArrayList<T>(size)
    collect {
        cache.add(it)
        if (cache.size == size) {
            emit(transform(cache))
            cache.clear()
        }
    }
}

然后像这样使用它：

suspend fun soundDbFlow(period: Long) = flow {
    while (true) {
        val data = getAmplitude()
        emit(data)
        delay(period)
    }
}
    .chunked(5) { (it.sum() / 5.0).roundToInt() }
    .map { soundDb(it) }

Answer 2

这是你想要的吗？

var result = 0

fun soundDbFlow(period: Long) = flow {
    while (true) {
        delay(period)
        val data = getAmplitude()
        emit(data)
    }
}

fun reduceFlow(period: Long = 100) = flow {
    while (true) {
        val sum = soundDbFlow(period)
            .take(5)
            .reduce { accumulator, value -> accumulator + value }

        val average = (sum / 5.0).roundToInt()
        emit(soundDb(average))
    }
}

fun getAmplitude(): Int {
    return ++result
}

fun soundDb(input: Int, referenceAmp: Double = 1.0): Double {
    return 20 * log10(input / referenceAmp)
}

fun main(): Unit = runBlocking {
    reduceFlow().collect { println(it) }
}