下面是平滑z-score算法的Python / numpy实现(见上面的答案)。你可以在这里找到要点。

#!/usr/bin/env python
# Implementation of algorithm from https://stackoverflow.com/a/22640362/6029703
import numpy as np
import pylab

def thresholding_algo(y, lag, threshold, influence):
    signals = np.zeros(len(y))
    filteredY = np.array(y)
    avgFilter = [0]*len(y)
    stdFilter = [0]*len(y)
    avgFilter[lag - 1] = np.mean(y[0:lag])
    stdFilter[lag - 1] = np.std(y[0:lag])
    for i in range(lag, len(y)):
        if abs(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]
            avgFilter[i] = np.mean(filteredY[(i-lag+1):i+1])
            stdFilter[i] = np.std(filteredY[(i-lag+1):i+1])
        else:
            signals[i] = 0
            filteredY[i] = y[i]
            avgFilter[i] = np.mean(filteredY[(i-lag+1):i+1])
            stdFilter[i] = np.std(filteredY[(i-lag+1):i+1])

    return dict(signals = np.asarray(signals),
                avgFilter = np.asarray(avgFilter),
                stdFilter = np.asarray(stdFilter))

下面是在同一个数据集上的测试，它产生的图与R/Matlab的原始答案相同

# Data
y = np.array([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])

# Settings: lag = 30, threshold = 5, influence = 0
lag = 30
threshold = 5
influence = 0

# Run algo with settings from above
result = thresholding_algo(y, lag=lag, threshold=threshold, influence=influence)

# Plot result
pylab.subplot(211)
pylab.plot(np.arange(1, len(y)+1), y)

pylab.plot(np.arange(1, len(y)+1),
           result["avgFilter"], color="cyan", lw=2)

pylab.plot(np.arange(1, len(y)+1),
           result["avgFilter"] + threshold * result["stdFilter"], color="green", lw=2)

pylab.plot(np.arange(1, len(y)+1),
           result["avgFilter"] - threshold * result["stdFilter"], color="green", lw=2)

pylab.subplot(212)
pylab.step(np.arange(1, len(y)+1), result["signals"], color="red", lw=2)
pylab.ylim(-1.5, 1.5)
pylab.show()

一种方法是根据以下观察来检测峰:

时间t是一个峰值(y (t) > y (t - 1)) & & ((t) > y (t + 1))

它通过等待上升趋势结束来避免误报。它并不完全是“实时”的，因为它会比峰值差一个dt。灵敏度可以通过要求比较的裕度来控制。在噪声检测和时延检测之间存在一种折衷。 您可以通过添加更多参数来丰富模型:

峰如果y (y (t) - (t-dt) > m) && (y (t) - y (t + dt) > m)

dt和m是控制灵敏度和延时的参数

这是你用上述算法得到的结果:

下面是在python中重现图的代码:

import numpy as np
import matplotlib.pyplot as plt
input = np.array([ 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. ])
signal = (input > np.roll(input,1)) & (input > np.roll(input,-1))
plt.plot(input)
plt.plot(signal.nonzero()[0], input[signal], 'ro')
plt.show()

通过设置m = 0.5，你可以得到一个更清晰的信号，只有一个假阳性:

我们尝试在我们的数据集上使用平滑的z-score算法，这导致了过度敏感或不敏感(取决于参数如何调整)，几乎没有中间地带。在我们站点的交通信号中，我们观察到一个低频基线，它代表了每天的周期，即使有最好的可能参数(如下所示)，它仍然在第4天下降，特别是因为大多数数据点被认为是异常的。

在原始z-score算法的基础上，我们提出了一种通过反向滤波来解决这个问题的方法。改进后的算法及其在电视商业流量归因中的应用详见我们的团队博客。

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.

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>

下面是平滑z-score算法的Groovy (Java)实现(见上面的答案)。

/**
 * "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)
 */

public HashMap<String, List<Object>> thresholdingAlgo(List<Double> y, Long lag, Double threshold, Double influence) {
    //init stats instance
    SummaryStatistics stats = new SummaryStatistics()

    //the results (peaks, 1 or -1) of our algorithm
    List<Integer> signals = new ArrayList<Integer>(Collections.nCopies(y.size(), 0))
    //filter out the signals (peaks) from our original list (using influence arg)
    List<Double> filteredY = new ArrayList<Double>(y)
    //the current average of the rolling window
    List<Double> avgFilter = new ArrayList<Double>(Collections.nCopies(y.size(), 0.0d))
    //the current standard deviation of the rolling window
    List<Double> stdFilter = new ArrayList<Double>(Collections.nCopies(y.size(), 0.0d))
    //init avgFilter and stdFilter
    (0..lag-1).each { stats.addValue(y[it as int]) }
    avgFilter[lag - 1 as int] = stats.getMean()
    stdFilter[lag - 1 as int] = Math.sqrt(stats.getPopulationVariance()) //getStandardDeviation() uses sample variance (not what we want)
    stats.clear()
    //loop input starting at end of rolling window
    (lag..y.size()-1).each { i ->
        //if the distance between the current value and average is enough standard deviations (threshold) away
        if (Math.abs((y[i as int] - avgFilter[i - 1 as int]) as Double) > threshold * stdFilter[i - 1 as int]) {
            //this is a signal (i.e. peak), determine if it is a positive or negative signal
            signals[i as int] = (y[i as int] > avgFilter[i - 1 as int]) ? 1 : -1
            //filter this signal out using influence
            filteredY[i as int] = (influence * y[i as int]) + ((1-influence) * filteredY[i - 1 as int])
        } else {
            //ensure this signal remains a zero
            signals[i as int] = 0
            //ensure this value is not filtered
            filteredY[i as int] = y[i as int]
        }
        //update rolling average and deviation
        (i - lag..i-1).each { stats.addValue(filteredY[it as int] as Double) }
        avgFilter[i as int] = stats.getMean()
        stdFilter[i as int] = Math.sqrt(stats.getPopulationVariance()) //getStandardDeviation() uses sample variance (not what we want)
        stats.clear()
    }

    return [
        signals  : signals,
        avgFilter: avgFilter,
        stdFilter: stdFilter
    ]
}

下面是同一个数据集上的测试，其结果与上面的Python / numpy实现相同。

    // Data
    def y = [1d, 1d, 1.1d, 1d, 0.9d, 1d, 1d, 1.1d, 1d, 0.9d, 1d, 1.1d, 1d, 1d, 0.9d, 1d, 1d, 1.1d, 1d, 1d,
         1d, 1d, 1.1d, 0.9d, 1d, 1.1d, 1d, 1d, 0.9d, 1d, 1.1d, 1d, 1d, 1.1d, 1d, 0.8d, 0.9d, 1d, 1.2d, 0.9d, 1d,
         1d, 1.1d, 1.2d, 1d, 1.5d, 1d, 3d, 2d, 5d, 3d, 2d, 1d, 1d, 1d, 0.9d, 1d,
         1d, 3d, 2.6d, 4d, 3d, 3.2d, 2d, 1d, 1d, 0.8d, 4d, 4d, 2d, 2.5d, 1d, 1d, 1d]

    // Settings
    def lag = 30
    def threshold = 5
    def influence = 0


    def thresholdingResults = thresholdingAlgo((List<Double>) y, (Long) lag, (Double) threshold, (Double) influence)

    println y.size()
    println thresholdingResults.signals.size()
    println thresholdingResults.signals

    thresholdingResults.signals.eachWithIndex { x, idx ->
        if (x) {
            println y[idx]
        }
    }