kotlin - 我如何设计一个流，它是另一个流的每最新 5 个数据的平均值？

Question

有一个Flow，每100ms发出一次数据，我希望对Flow的每最新5条数据取一个平均值，并将平均值转换为Double值，转换成另一个Flow。

如何设计流程？

代码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:

致 plplmax:谢谢!

我假设代码 B 会发出 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，然后再计算(6+7+8+9+10)/5 第二，....？这是我的期望。

我担心代码 C 可能会先计算 (1+2+3+4+5)/5，然后计算 (2+3+4+5+6)/5 第二个，...

代码 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

你可以像这样写一个分 block 运算符:

/**
 * 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) }