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.

这种z-scores方法在峰值检测方面非常有效，也有助于异常值的去除。异常值对话经常讨论每个点的统计价值和变化数据的伦理。

但是，在来自易出错的串行通信或易出错的传感器的重复错误传感器值的情况下，错误或虚假读数中没有统计值。它们需要被识别并移除。

从视觉上看，错误是显而易见的。下图中的直线显示了需要删除的内容。但是用算法识别和消除错误是相当具有挑战性的。z分数效果很好。

下图是通过串行通信从传感器获得的值。偶尔的串行通信错误，传感器错误或两者都导致重复的，明显错误的数据点。

z-score峰值检测器能够在虚假数据点上发出信号，并生成一个干净的结果数据集，同时保留正确数据的特征:

另外，这个算法对我来说也很好…

sensitivity = 4; dwindow = 4; k = dwindow; 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>=Math.max(...c.slice(b-k,b))?a**3:0); var data2 = [{ name: 'filtered source', y: filtered }]; Plotly.newPlot('stage2', data2, { title: 'Filtered data<br>aₙ = aₙ³', 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: 'Maximum in a window of 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: 'Maximum in a window of 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>

一个python/numpy的迭代版本的答案https://stackoverflow.com/a/22640362/6029703在这里。对于大数据(100000+)，此代码比计算平均和标准偏差的速度更快。

def peak_detection_smoothed_zscore_v2(x, lag, threshold, influence):
    '''
    iterative smoothed z-score algorithm
    Implementation of algorithm from https://stackoverflow.com/a/22640362/6029703
    '''
    import numpy as np
    labels = np.zeros(len(x))
    filtered_y = np.array(x)
    avg_filter = np.zeros(len(x))
    std_filter = np.zeros(len(x))
    var_filter = np.zeros(len(x))

    avg_filter[lag - 1] = np.mean(x[0:lag])
    std_filter[lag - 1] = np.std(x[0:lag])
    var_filter[lag - 1] = np.var(x[0:lag])
    for i in range(lag, len(x)):
        if abs(x[i] - avg_filter[i - 1]) > threshold * std_filter[i - 1]:
            if x[i] > avg_filter[i - 1]:
                labels[i] = 1
            else:
                labels[i] = -1
            filtered_y[i] = influence * x[i] + (1 - influence) * filtered_y[i - 1]
        else:
            labels[i] = 0
            filtered_y[i] = x[i]
        # update avg, var, std
        avg_filter[i] = avg_filter[i - 1] + 1. / lag * (filtered_y[i] - filtered_y[i - lag])
        var_filter[i] = var_filter[i - 1] + 1. / lag * ((filtered_y[i] - avg_filter[i - 1]) ** 2 - (
            filtered_y[i - lag] - avg_filter[i - 1]) ** 2 - (filtered_y[i] - filtered_y[i - lag]) ** 2 / lag)
        std_filter[i] = np.sqrt(var_filter[i])

    return dict(signals=labels,
                avgFilter=avg_filter,
                stdFilter=std_filter)

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.

下面是@Jean-Paul为Arduino微控制器设计的平滑z分数的C语言实现，用于获取加速度计读数，并判断撞击的方向是来自左边还是右边。这表现得非常好，因为这个设备返回一个反弹信号。这是设备对峰值检测算法的输入-显示了来自右边的冲击，然后是来自左边的冲击。你可以看到最初的峰值然后传感器的振荡。

#include <stdio.h>
#include <math.h>
#include <string.h>


#define SAMPLE_LENGTH 1000

float stddev(float data[], int len);
float mean(float data[], int len);
void thresholding(float y[], int signals[], int lag, float threshold, float influence);


void thresholding(float y[], int signals[], int lag, float threshold, float influence) {
    memset(signals, 0, sizeof(int) * SAMPLE_LENGTH);
    float filteredY[SAMPLE_LENGTH];
    memcpy(filteredY, y, sizeof(float) * SAMPLE_LENGTH);
    float avgFilter[SAMPLE_LENGTH];
    float stdFilter[SAMPLE_LENGTH];

    avgFilter[lag - 1] = mean(y, lag);
    stdFilter[lag - 1] = stddev(y, lag);

    for (int i = lag; i < SAMPLE_LENGTH; i++) {
        if (fabsf(y[i] - avgFilter[i-1]) > threshold * stdFilter[i-1]) {
            if (y[i] > avgFilter[i-1]) {
                signals[i] = 1;
            } else {
                signals[i] = -1;
            }
            filteredY[i] = influence * y[i] + (1 - influence) * filteredY[i-1];
        } else {
            signals[i] = 0;
        }
        avgFilter[i] = mean(filteredY + i-lag, lag);
        stdFilter[i] = stddev(filteredY + i-lag, lag);
    }
}

float mean(float data[], int len) {
    float sum = 0.0, mean = 0.0;

    int i;
    for(i=0; i<len; ++i) {
        sum += data[i];
    }

    mean = sum/len;
    return mean;


}

float stddev(float data[], int len) {
    float the_mean = mean(data, len);
    float standardDeviation = 0.0;

    int i;
    for(i=0; i<len; ++i) {
        standardDeviation += pow(data[i] - the_mean, 2);
    }

    return sqrt(standardDeviation/len);
}

int main() {
    printf("Hello, World!\n");
    int lag = 100;
    float threshold = 5;
    float influence = 0;
    float y[]=  {1,1,1.1,1,0.9,1,1,1.1,1,0.9,1,1.1,1,1,0.9,1,1,1.1,1,1,1,1,1.1,0.9,1,1.1,1,1,0.9,
  ....
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,       2.6,4,3,3.2,2,1,1,0.8,4,4,2,2.5,1,1,1,1.2,1,1.5,1,3,2,5,3,2,1,1,1,0.9,1,1,3,
       2.6,4,3,3.2,2,1,1,0.8,4,4,2,2.5,1,1,1}

    int signal[SAMPLE_LENGTH];

    thresholding(y, signal,  lag, threshold, influence);

    return 0;
}

她的结果是影响= 0

不是很好，但这里的影响力= 1

这很好。