这是一个修改后的Fortran版本的z-score算法。 它是专门针对频率空间中传递函数的峰值(共振)检测进行修改的(每个更改在代码中都有一个小注释)。

如果在输入向量的下界附近存在共振，则第一个修改会向用户发出警告，该共振由高于某个阈值的标准偏差表示(在本例中为10%)。这仅仅意味着信号不够平坦，不足以使检测正确地初始化滤波器。

第二种修改是只将峰值的最大值添加到已找到的峰值中。这是通过将每个发现的峰值与其(滞后)前辈及其(滞后)后继者的大小进行比较来达到的。

第三个变化是尊重共振峰通常在共振频率周围表现出某种形式的对称性。因此，围绕当前数据点对称地计算平均值和std是很自然的(而不仅仅是针对之前的数据点)。这将导致更好的峰值检测行为。

这些修改的效果是，整个信号必须事先被函数知道，这是共振检测的通常情况(类似于Jean-Paul的Matlab示例，其中数据点是动态生成的，这是行不通的)。

function PeakDetect(y,lag,threshold, influence)
    implicit none
    ! Declaring part
    real, dimension(:), intent(in) :: y
    integer, dimension(size(y)) :: PeakDetect
    real, dimension(size(y)) :: filteredY, avgFilter, stdFilter
    integer :: lag, ii
    real :: threshold, influence

    ! Executing part
    PeakDetect = 0
    filteredY = 0.0
    filteredY(1:lag+1) = y(1:lag+1)
    avgFilter = 0.0
    avgFilter(lag+1) = mean(y(1:2*lag+1))
    stdFilter = 0.0
    stdFilter(lag+1) = std(y(1:2*lag+1))

    if (stdFilter(lag+1)/avgFilter(lag+1)>0.1) then ! If the coefficient of variation exceeds 10%, the signal is too uneven at the start, possibly because of a peak.
        write(unit=*,fmt=1001)
1001        format(1X,'Warning: Peak detection might have failed, as there may be a peak at the edge of the frequency range.',/)
    end if
    do ii = lag+2, size(y)
        if (abs(y(ii) - avgFilter(ii-1)) > threshold * stdFilter(ii-1)) then
            ! Find only the largest outstanding value which is only the one greater than its predecessor and its successor
            if (y(ii) > avgFilter(ii-1) .AND. y(ii) > y(ii-1) .AND. y(ii) > y(ii+1)) then
                PeakDetect(ii) = 1
            end if
            filteredY(ii) = influence * y(ii) + (1 - influence) * filteredY(ii-1)
        else
            filteredY(ii) = y(ii)
        end if
        ! Modified with respect to the original code. Mean and standard deviation are calculted symmetrically around the current point
        avgFilter(ii) = mean(filteredY(ii-lag:ii+lag))
        stdFilter(ii) = std(filteredY(ii-lag:ii+lag))
    end do
end function PeakDetect

real function mean(y)
    !> @brief Calculates the mean of vector y
    implicit none
    ! Declaring part
    real, dimension(:), intent(in) :: y
    integer :: N
    ! Executing part
    N = max(1,size(y))
    mean = sum(y)/N
end function mean

real function std(y)
    !> @brief Calculates the standard deviation of vector y
    implicit none
    ! Declaring part
    real, dimension(:), intent(in) :: y
    integer :: N
    ! Executing part
    N = max(1,size(y))
    std = sqrt((N*dot_product(y,y) - sum(y)**2) / (N*(N-1)))
end function std

对于我的应用程序，算法的工作就像一个魅力!

在计算拓扑学中，持久同调的思想导致一个有效的 -快如排序数字-解决方案。它不仅检测峰值，还以一种自然的方式量化峰值的“重要性”，使您能够选择对您重要的峰值。

算法的总结。 在一维设置(时间序列，实值信号)中，算法可以简单地描述为下图:

Think of the function graph (or its sub-level set) as a landscape and consider a decreasing water level starting at level infinity (or 1.8 in this picture). While the level decreases, at local maxima islands pop up. At local minima these islands merge together. One detail in this idea is that the island that appeared later in time is merged into the island that is older. The "persistence" of an island is its birth time minus its death time. The lengths of the blue bars depict the persistence, which is the above mentioned "significance" of a peak.

效率。 在对函数值进行排序之后，找到一个在线性时间内运行的实现并不难——实际上它是一个单一的、简单的循环。因此，这种实现在实践中应该是快速的，也很容易实现。

参考文献 一篇关于整个故事的文章和对持久同调(计算代数拓扑中的一个领域)动机的引用可以在这里找到: https://www.sthu.org/blog/13-perstopology-peakdetection/index.html

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>

在Palshikar(2009)中发现了另一个算法:

Palshikar, G.(2009)。时间序列中峰值检测的简单算法。在Proc. 1st Int。高级数据分析，商业分析和智能(卷122)。

论文可以从这里下载。

算法是这样的:

algorithm peak1 // one peak detection algorithms that uses peak function S1 

input T = x1, x2, …, xN, N // input time-series of N points 
input k // window size around the peak 
input h // typically 1 <= h <= 3 
output O // set of peaks detected in T 

begin 
O = empty set // initially empty 

    for (i = 1; i < n; i++) do
        // compute peak function value for each of the N points in T 
        a[i] = S1(k,i,xi,T); 
    end for 

    Compute the mean m' and standard deviation s' of all positive values in array a; 

    for (i = 1; i < n; i++) do // remove local peaks which are “small” in global context 
        if (a[i] > 0 && (a[i] – m') >( h * s')) then O = O + {xi}; 
        end if 
    end for 

    Order peaks in O in terms of increasing index in T 

    // retain only one peak out of any set of peaks within distance k of each other 

    for every adjacent pair of peaks xi and xj in O do 
        if |j – i| <= k then remove the smaller value of {xi, xj} from O 
        end if 
    end for 
end

优势

本文提出了5种不同的峰值检测算法 算法在原始时间序列数据上工作(不需要平滑)

缺点

很难事先确定k和h 峰不能是平的(就像我测试数据中的第三个峰)

例子:

根据@Jean-Paul提出的解决方案，我用c#实现了他的算法

public class ZScoreOutput
{
    public List<double> input;
    public List<int> signals;
    public List<double> avgFilter;
    public List<double> filtered_stddev;
}

public static class ZScore
{
    public static ZScoreOutput StartAlgo(List<double> input, int lag, double threshold, double influence)
    {
        // init variables!
        int[] signals = new int[input.Count];
        double[] filteredY = new List<double>(input).ToArray();
        double[] avgFilter = new double[input.Count];
        double[] stdFilter = new double[input.Count];

        var initialWindow = new List<double>(filteredY).Skip(0).Take(lag).ToList();

        avgFilter[lag - 1] = Mean(initialWindow);
        stdFilter[lag - 1] = StdDev(initialWindow);

        for (int i = lag; i < input.Count; i++)
        {
            if (Math.Abs(input[i] - avgFilter[i - 1]) > threshold * stdFilter[i - 1])
            {
                signals[i] = (input[i] > avgFilter[i - 1]) ? 1 : -1;
                filteredY[i] = influence * input[i] + (1 - influence) * filteredY[i - 1];
            }
            else
            {
                signals[i] = 0;
                filteredY[i] = input[i];
            }

            // Update rolling average and deviation
            var slidingWindow = new List<double>(filteredY).Skip(i - lag).Take(lag+1).ToList();

            var tmpMean = Mean(slidingWindow);
            var tmpStdDev = StdDev(slidingWindow);

            avgFilter[i] = Mean(slidingWindow);
            stdFilter[i] = StdDev(slidingWindow);
        }

        // Copy to convenience class 
        var result = new ZScoreOutput();
        result.input = input;
        result.avgFilter       = new List<double>(avgFilter);
        result.signals         = new List<int>(signals);
        result.filtered_stddev = new List<double>(stdFilter);

        return result;
    }

    private static double Mean(List<double> list)
    {
        // Simple helper function! 
        return list.Average();
    }

    private static double StdDev(List<double> values)
    {
        double ret = 0;
        if (values.Count() > 0)
        {
            double avg = values.Average();
            double sum = values.Sum(d => Math.Pow(d - avg, 2));
            ret = Math.Sqrt((sum) / (values.Count() - 1));
        }
        return ret;
    }
}

使用示例:

var input = new List<double> {1.0, 1.0, 1.1, 1.0, 0.9, 1.0, 1.0, 1.1, 1.0, 0.9, 1.0,
    1.1, 1.0, 1.0, 0.9, 1.0, 1.0, 1.1, 1.0, 1.0, 1.0, 1.0, 1.1, 0.9, 1.0, 1.1, 1.0, 1.0, 0.9,
    1.0, 1.1, 1.0, 1.0, 1.1, 1.0, 0.8, 0.9, 1.0, 1.2, 0.9, 1.0, 1.0, 1.1, 1.2, 1.0, 1.5, 1.0,
    3.0, 2.0, 5.0, 3.0, 2.0, 1.0, 1.0, 1.0, 0.9, 1.0, 1.0, 3.0, 2.6, 4.0, 3.0, 3.2, 2.0, 1.0,
    1.0, 0.8, 4.0, 4.0, 2.0, 2.5, 1.0, 1.0, 1.0};

int lag = 30;
double threshold = 5.0;
double influence = 0.0;

var output = ZScore.StartAlgo(input, lag, threshold, influence);

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。