更新:到目前为止表现最好的算法是这个。


这个问题探讨了在实时时间序列数据中检测突然峰值的稳健算法。

考虑以下示例数据:

这个数据的例子是Matlab格式的(但这个问题不是关于语言,而是关于算法):

p = [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];

你可以清楚地看到有三个大峰和一些小峰。这个数据集是问题所涉及的时间序列数据集类的一个特定示例。这类数据集有两个一般特征:

有一种具有一般平均值的基本噪声 有很大的“峰值”或“更高的数据点”明显偏离噪声。

让我们假设以下情况:

峰的宽度不能事先确定 峰的高度明显偏离其他值 算法实时更新(因此每个新数据点都会更新)

对于这种情况,需要构造一个触发信号的边值。但是,边界值不能是静态的,必须通过算法实时确定。


我的问题是:什么是实时计算这些阈值的好算法?有没有针对这种情况的特定算法?最著名的算法是什么?


健壮的算法或有用的见解都受到高度赞赏。(可以用任何语言回答:这是关于算法的)


当前回答

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)))
}

其他回答

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

以下是平滑z-score算法的Scala版本(非惯用):

/**
  * 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)
  */
private def smoothedZScore(y: Seq[Double], lag: Int, threshold: Double, influence: Double): Seq[Int] = {
  val stats = new SummaryStatistics()

  // the results (peaks, 1 or -1) of our algorithm
  val signals = mutable.ArrayBuffer.fill(y.length)(0)

  // filter out the signals (peaks) from our original list (using influence arg)
  val filteredY = y.to[mutable.ArrayBuffer]

  // the current average of the rolling window
  val avgFilter = mutable.ArrayBuffer.fill(y.length)(0d)

  // the current standard deviation of the rolling window
  val stdFilter = mutable.ArrayBuffer.fill(y.length)(0d)

  // init avgFilter and stdFilter
  y.take(lag).foreach(s => stats.addValue(s))

  avgFilter(lag - 1) = stats.getMean
  stdFilter(lag - 1) = Math.sqrt(stats.getPopulationVariance) // getStandardDeviation() uses sample variance (not what we want)

  // loop input starting at end of rolling window
  y.zipWithIndex.slice(lag, y.length - 1).foreach {
    case (s: Double, i: Int) =>
      // if the distance between the current value and average is enough standard deviations (threshold) away
      if (Math.abs(s - 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 (s > avgFilter(i - 1)) 1 else -1
        // filter this signal out using influence
        filteredY(i) = (influence * s) + ((1 - influence) * filteredY(i - 1))
      } else {
        // ensure this signal remains a zero
        signals(i) = 0
        // ensure this value is not filtered
        filteredY(i) = s
      }

      // update rolling average and deviation
      stats.clear()
      filteredY.slice(i - lag, i).foreach(s => stats.addValue(s))
      avgFilter(i) = stats.getMean
      stdFilter(i) = Math.sqrt(stats.getPopulationVariance) // getStandardDeviation() uses sample variance (not what we want)
  }

  println(y.length)
  println(signals.length)
  println(signals)

  signals.zipWithIndex.foreach {
    case(x: Int, idx: Int) =>
      if (x == 1) {
        println(idx + " " + y(idx))
      }
  }

  val data =
    y.zipWithIndex.map { case (s: Double, i: Int) => Map("x" -> i, "y" -> s, "name" -> "y", "row" -> "data") } ++
    avgFilter.zipWithIndex.map { case (s: Double, i: Int) => Map("x" -> i, "y" -> s, "name" -> "avgFilter", "row" -> "data") } ++
    avgFilter.zipWithIndex.map { case (s: Double, i: Int) => Map("x" -> i, "y" -> (s - threshold * stdFilter(i)), "name" -> "lower", "row" -> "data") } ++
    avgFilter.zipWithIndex.map { case (s: Double, i: Int) => Map("x" -> i, "y" -> (s + threshold * stdFilter(i)), "name" -> "upper", "row" -> "data") } ++
    signals.zipWithIndex.map { case (s: Int, i: Int) => Map("x" -> i, "y" -> s, "name" -> "signal", "row" -> "signal") }

  Vegas("Smoothed Z")
    .withData(data)
    .mark(Line)
    .encodeX("x", Quant)
    .encodeY("y", Quant)
    .encodeColor(
      field="name",
      dataType=Nominal
    )
    .encodeRow("row", Ordinal)
    .show

  return signals
}

下面是一个测试,返回与Python和Groovy版本相同的结果:

val y = List(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)

val lag = 30
val threshold = 5d
val influence = 0d

smoothedZScore(y, lag, threshold, influence)

这里的要点

我认为delica的Python回答器有一个bug。我不能评论他的帖子,因为我没有代表来做这件事,编辑队列已经满了,所以我可能不是第一个注意到它的人。

avgFilter[lag - 1]和stdFilter[lag - 1]在init中设置,然后在lag == i时再次设置,而不是改变[lag]值。这个结果使得第一个信号总是1。

以下是带有轻微修正的代码:

import numpy as np

class real_time_peak_detection():
    def __init__(self, array, lag, threshold, influence):
        self.y = list(array)
        self.length = len(self.y)
        self.lag = lag
        self.threshold = threshold
        self.influence = influence
        self.signals = [0] * len(self.y)
        self.filteredY = np.array(self.y).tolist()
        self.avgFilter = [0] * len(self.y)
        self.stdFilter = [0] * len(self.y)
        self.avgFilter[self.lag - 1] = np.mean(self.y[0:self.lag]).tolist()
        self.stdFilter[self.lag - 1] = np.std(self.y[0:self.lag]).tolist()

    def thresholding_algo(self, new_value):
        self.y.append(new_value)
        i = len(self.y) - 1
        self.length = len(self.y)
        if i < self.lag:
            return 0
        elif i == self.lag:
            self.signals = [0] * len(self.y)
            self.filteredY = np.array(self.y).tolist()
            self.avgFilter = [0] * len(self.y)
            self.stdFilter = [0] * len(self.y)
            self.avgFilter[self.lag] = np.mean(self.y[0:self.lag]).tolist()
            self.stdFilter[self.lag] = np.std(self.y[0:self.lag]).tolist()
            return 0

        self.signals += [0]
        self.filteredY += [0]
        self.avgFilter += [0]
        self.stdFilter += [0]

        if abs(self.y[i] - self.avgFilter[i - 1]) > self.threshold * self.stdFilter[i - 1]:
            if self.y[i] > self.avgFilter[i - 1]:
                self.signals[i] = 1
            else:
                self.signals[i] = -1

            self.filteredY[i] = self.influence * self.y[i] + (1 - self.influence) * self.filteredY[i - 1]
            self.avgFilter[i] = np.mean(self.filteredY[(i - self.lag):i])
            self.stdFilter[i] = np.std(self.filteredY[(i - self.lag):i])
        else:
            self.signals[i] = 0
            self.filteredY[i] = self.y[i]
            self.avgFilter[i] = np.mean(self.filteredY[(i - self.lag):i])
            self.stdFilter[i] = np.std(self.filteredY[(i - self.lag):i])

        return self.signals[i]

这是一个Python实现的鲁棒峰值检测算法算法。

初始化和计算部分被分开,只有filtered_y数组被保留,它的最大大小等于延迟,因此内存没有增加。(结果与上述答案相同)。 为了绘制图形,还保留了标签数组。

我做了一个github要点。

import numpy as np
import pylab

def init(x, lag, threshold, influence):
    '''
    Smoothed z-score algorithm
    Implementation of algorithm from https://stackoverflow.com/a/22640362/6029703
    '''

    labels = np.zeros(lag)
    filtered_y = np.array(x[0:lag])
    avg_filter = np.zeros(lag)
    std_filter = np.zeros(lag)
    var_filter = np.zeros(lag)

    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])

    return dict(avg=avg_filter[lag - 1], var=var_filter[lag - 1],
                std=std_filter[lag - 1], filtered_y=filtered_y,
                labels=labels)


def add(result, single_value, lag, threshold, influence):
    previous_avg = result['avg']
    previous_var = result['var']
    previous_std = result['std']
    filtered_y = result['filtered_y']
    labels = result['labels']

    if abs(single_value - previous_avg) > threshold * previous_std:
        if single_value > previous_avg:
            labels = np.append(labels, 1)
        else:
            labels = np.append(labels, -1)

        # calculate the new filtered element using the influence factor
        filtered_y = np.append(filtered_y, influence * single_value
                               + (1 - influence) * filtered_y[-1])
    else:
        labels = np.append(labels, 0)
        filtered_y = np.append(filtered_y, single_value)

    # update avg as sum of the previuos avg + the lag * (the new calculated item - calculated item at position (i - lag))
    current_avg_filter = previous_avg + 1. / lag * (filtered_y[-1]
            - filtered_y[len(filtered_y) - lag - 1])

    # update variance as the previuos element variance + 1 / lag * new recalculated item - the previous avg -
    current_var_filter = previous_var + 1. / lag * ((filtered_y[-1]
            - previous_avg) ** 2 - (filtered_y[len(filtered_y) - 1
            - lag] - previous_avg) ** 2 - (filtered_y[-1]
            - filtered_y[len(filtered_y) - 1 - lag]) ** 2 / lag)  # the recalculated element at pos (lag) - avg of the previuos - new recalculated element - recalculated element at lag pos ....

    # calculate standard deviation for current element as sqrt (current variance)
    current_std_filter = np.sqrt(current_var_filter)

    return dict(avg=current_avg_filter, var=current_var_filter,
                std=current_std_filter, filtered_y=filtered_y[1:],
                labels=labels)

lag = 30
threshold = 5
influence = 0

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])

# Run algo with settings from above
result = init(y[:lag], lag=lag, threshold=threshold, influence=influence)

i = open('quartz2', 'r')
for i in y[lag:]:
    result = add(result, i, lag, threshold, influence)

# Plot result
pylab.subplot(211)
pylab.plot(np.arange(1, len(y) + 1), y)
pylab.subplot(212)
pylab.step(np.arange(1, len(y) + 1), result['labels'], color='red',
           lw=2)
pylab.ylim(-1.5, 1.5)
pylab.show()

This problem looks similar to one I encountered in a hybrid/embedded systems course, but that was related to detecting faults when the input from a sensor is noisy. We used a Kalman filter to estimate/predict the hidden state of the system, then used statistical analysis to determine the likelihood that a fault had occurred. We were working with linear systems, but nonlinear variants exist. I remember the approach being surprisingly adaptive, but it required a model of the dynamics of the system.