如果边界值或其他标准取决于未来值，那么唯一的解决方案(没有时间机器，或其他关于未来值的知识)是推迟任何决定，直到有足够的未来值。如果你想要一个高于均值的水平，例如，20点，那么你必须等到你至少有19点才能做出任何峰值决策，否则下一个新点可能会完全超过你19点之前的阈值。

Added: If the statistical distribution of the peak heights could be heavy tailed, instead of Uniform or Gaussian, then you may need to wait until you see several thousand peaks before it starts to become unlikely that a hidden Pareto distribution won't produce a peak many times larger than any you currently have seen before or have in your current plot. Unless you somehow know in advance that the very next point can't be 1e20, it could appear, which after rescaling your plot's Y dimension, would be flat up until that point.

c++ (Qt)演示端口，交互式参数

我已经将这个算法的演示应用程序移植到c++ (Qt)上。

代码可以在GitHub上找到这里。带有安装程序的Windows(64位)构建在发布页面上。最后，我将添加一些文档和其他发布版本。

您不能绘制点，但可以从文本文件中导入它们(用空格分隔点——换行也算作空格)。您还可以调整算法参数，实时查看效果。这对于针对特定数据集调整算法以及探索参数如何影响结果非常有用。

上面的截图有些过时;从那以后，我添加了两个原始算法中没有的实验性选项:

反向处理数据集的选项(似乎至少改善了功率谱的结果)。 选项，为峰值设置硬性最小阈值。

我还在窗口中间添加了一个笨拙的缩放/平移条，只需用鼠标拖动它来缩放和平移。

模糊的构建指令:

在发布页面上有一个Windows安装程序(64位)，但如果你想从源代码构建它，要点是:

安装Qt的构建工具，然后将qmake && make放在与.pro文件相同的目录下，或者 安装Qt Creator，打开.pro文件，选择任何默认的构建配置，然后按下构建和/或运行按钮(Creator的左下角)。

我只测试过Qt5。我有91%的信心，如果你手动配置组件，Qt Creator安装程序会让你安装Qt5(如果你手动配置组件，你还需要确认是否安装了Qt Charts)。Qt6可能是一个流畅的构建，也可能不是。有一天，我将测试Qt4和Qt6，使这些文档更好。也许吧。

下面是在Golang中实现的Smoothed z-score算法(上图)。它假设一个[]int16 (PCM 16bit样本)的切片。你可以在这里找到要点。

/*
Settings (the ones below are examples: choose what is best for your data)
set lag to 5;          # lag 5 for the smoothing functions
set threshold to 3.5;  # 3.5 standard deviations for signal
set influence to 0.5;  # between 0 and 1, where 1 is normal influence, 0.5 is half
*/

// ZScore on 16bit WAV samples
func ZScore(samples []int16, lag int, threshold float64, influence float64) (signals []int16) {
    //lag := 20
    //threshold := 3.5
    //influence := 0.5

    signals = make([]int16, len(samples))
    filteredY := make([]int16, len(samples))
    for i, sample := range samples[0:lag] {
        filteredY[i] = sample
    }
    avgFilter := make([]int16, len(samples))
    stdFilter := make([]int16, len(samples))

    avgFilter[lag] = Average(samples[0:lag])
    stdFilter[lag] = Std(samples[0:lag])

    for i := lag + 1; i < len(samples); i++ {

        f := float64(samples[i])

        if float64(Abs(samples[i]-avgFilter[i-1])) > threshold*float64(stdFilter[i-1]) {
            if samples[i] > avgFilter[i-1] {
                signals[i] = 1
            } else {
                signals[i] = -1
            }
            filteredY[i] = int16(influence*f + (1-influence)*float64(filteredY[i-1]))
            avgFilter[i] = Average(filteredY[(i - lag):i])
            stdFilter[i] = Std(filteredY[(i - lag):i])
        } else {
            signals[i] = 0
            filteredY[i] = samples[i]
            avgFilter[i] = Average(filteredY[(i - lag):i])
            stdFilter[i] = Std(filteredY[(i - lag):i])
        }
    }

    return
}

// Average a chunk of values
func Average(chunk []int16) (avg int16) {
    var sum int64
    for _, sample := range chunk {
        if sample < 0 {
            sample *= -1
        }
        sum += int64(sample)
    }
    return int16(sum / int64(len(chunk)))
}

函数scipy.signal。Find_peaks，顾名思义，在这方面很有用。但是要得到好的峰值提取，必须了解其参数宽度、阈值、距离和突出度。

根据我的测试和文档，突出的概念是“有用的概念”，可以保留好的峰值，丢弃噪声峰值。

什么是(地形)突出?它是“从山顶下降到任何更高地形所需的最低高度”，如下图所示:

这个想法是:

突出位置越高，山峰就越“重要”。

我在我的机器人项目中需要这样的东西。我想我可以归还Kotlin实现。

/**
* Smoothed zero-score alogrithm shamelessly copied from https://stackoverflow.com/a/22640362/6029703
* Uses a rolling mean and a rolling deviation (separate) to identify peaks in a vector
*
* @param y - The input vector to analyze
* @param lag - The lag of the moving window (i.e. how big the window is)
* @param threshold - The z-score at which the algorithm signals (i.e. how many standard deviations away from the moving mean a peak (or signal) is)
* @param influence - The influence (between 0 and 1) of new signals on the mean and standard deviation (how much a peak (or signal) should affect other values near it)
* @return - The calculated averages (avgFilter) and deviations (stdFilter), and the signals (signals)
*/
fun smoothedZScore(y: List<Double>, lag: Int, threshold: Double, influence: Double): Triple<List<Int>, List<Double>, List<Double>> {
    val stats = SummaryStatistics()
    // the results (peaks, 1 or -1) of our algorithm
    val signals = MutableList<Int>(y.size, { 0 })
    // filter out the signals (peaks) from our original list (using influence arg)
    val filteredY = ArrayList<Double>(y)
    // the current average of the rolling window
    val avgFilter = MutableList<Double>(y.size, { 0.0 })
    // the current standard deviation of the rolling window
    val stdFilter = MutableList<Double>(y.size, { 0.0 })
    // init avgFilter and stdFilter
    y.take(lag).forEach { s -> stats.addValue(s) }
    avgFilter[lag - 1] = stats.mean
    stdFilter[lag - 1] = Math.sqrt(stats.populationVariance) // getStandardDeviation() uses sample variance (not what we want)
    stats.clear()
    //loop input starting at end of rolling window
    (lag..y.size - 1).forEach { i ->
        //if the distance between the current value and average is enough standard deviations (threshold) away
        if (Math.abs(y[i] - avgFilter[i - 1]) > threshold * stdFilter[i - 1]) {
            //this is a signal (i.e. peak), determine if it is a positive or negative signal
            signals[i] = if (y[i] > avgFilter[i - 1]) 1 else -1
            //filter this signal out using influence
            filteredY[i] = (influence * y[i]) + ((1 - influence) * filteredY[i - 1])
        } else {
            //ensure this signal remains a zero
            signals[i] = 0
            //ensure this value is not filtered
            filteredY[i] = y[i]
        }
        //update rolling average and deviation
        (i - lag..i - 1).forEach { stats.addValue(filteredY[it]) }
        avgFilter[i] = stats.getMean()
        stdFilter[i] = Math.sqrt(stats.getPopulationVariance()) //getStandardDeviation() uses sample variance (not what we want)
        stats.clear()
    }
    return Triple(signals, avgFilter, stdFilter)
}

带有验证图的示例项目可以在github上找到。

假设你的数据来自传感器(所以算法不可能知道未来的任何事情)，

我做了这个算法，它与我在自己的项目中获得的数据非常好。

该算法有2个参数:灵敏度和窗口。

最后，只需一行代码就可以得到你的结果:

detected=data.map((a, b, c) => (a > 0) ? c[b] ** 4 * c[b - 1] ** 3 : -0).map((a, b, c) => a > Math.max(...c.slice(2)) / sensitivity).map((a, b, c) => (b > dwindow) && c.slice(b - dwindow, b).indexOf(a) == -1);

因为我是程序员而不是数学家，所以我不能更好地解释它。但我相信有人可以。

sensitivity = 20; dwindow = 4; data = [1., 1., 1., 1., 1., 1., 1., 1.1, 1., 0.8, 0.9, 1., 1.2, 0.9, 1., 1., 1.1, 1.2, 1., 1.5, 1., 3., 2., 5., 3., 2., 1., 1., 1., 0.9, 1., 1., 3., 2.6, 4., 3., 3.2, 2., 1., 1., 1., 1., 1. ]; //data = data.concat(data); //data = data.concat(data); var data1 = [{ name: 'original source', y: data }]; Plotly.newPlot('stage1', data1, { title: 'Sensor data', yaxis: { title: 'signal' } }); filtered = data.map((a, b, c) => (a > 0) ? c[b] ** 4 * c[b - 1] ** 3 : -0); var data2 = [{ name: 'filtered source', y: filtered }]; Plotly.newPlot('stage2', data2, { title: 'Filtered data<br>aₙ = aₙ⁴ * aₙ₋₁³', yaxis: { title: 'signal' } }); dwindow = 6; k = dwindow; detected = filtered.map((a, b, c) => a > Math.max(...c.slice(2)) / sensitivity).map((a, b, c) => (b > k) && c.slice(b - k, b).indexOf(a) == -1) var data3 = [{ name: 'detected peaks', y: detected }]; Plotly.newPlot('stage3', data3, { title: 'Window 6', yaxis: { title: 'signal' } }); dwindow = 10; k = dwindow; detected = filtered.map((a, b, c) => a > Math.max(...c.slice(2)) / 20).map((a, b, c) => (b > k) && c.slice(b - k, b).indexOf(a) == -1) var data4 = [{ name: 'detected peaks', y: detected }]; Plotly.newPlot('stage4', data4, { title: 'Window 10', yaxis: { title: 'signal' } }); <script src="https://cdn.jsdelivr.net/npm/plotly.js@2.16.5/dist/plotly.min.js"></script> <div id="stage1"></div> <div id="stage2"></div> <div id="stage3"></div> <div id="stage4"></div>