下面是我尝试为“Smoothed z-score算法”创建一个Ruby解决方案:

module ThresholdingAlgoMixin
  def mean(array)
    array.reduce(&:+) / array.size.to_f
  end

  def stddev(array)
    array_mean = mean(array)
    Math.sqrt(array.reduce(0.0) { |a, b| a.to_f + ((b.to_f - array_mean) ** 2) } / array.size.to_f)
  end

  def thresholding_algo(lag: 5, threshold: 3.5, influence: 0.5)
    return nil if size < lag * 2
    Array.new(size, 0).tap do |signals|
      filtered = Array.new(self)

      initial_slice = take(lag)
      avg_filter = Array.new(lag - 1, 0.0) + [mean(initial_slice)]
      std_filter = Array.new(lag - 1, 0.0) + [stddev(initial_slice)]
      (lag..size-1).each do |idx|
        prev = idx - 1
        if (fetch(idx) - avg_filter[prev]).abs > threshold * std_filter[prev]
          signals[idx] = fetch(idx) > avg_filter[prev] ? 1 : -1
          filtered[idx] = (influence * fetch(idx)) + ((1-influence) * filtered[prev])
        end

        filtered_slice = filtered[idx-lag..prev]
        avg_filter[idx] = mean(filtered_slice)
        std_filter[idx] = stddev(filtered_slice)
      end
    end
  end
end

以及示例用法:

test_data = [
  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
].extend(ThresholdingAlgoMixin)

puts test_data.thresholding_algo.inspect

# Output: [
#   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
#   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0,
#   0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1,
#   1, 1, 0, 0, 0, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0
# ]

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:

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]
        }
    }

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

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.

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”。