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)))
}

如果你的数据在一个数据库表中，这里是一个简单的z-score算法的SQL版本:

with data_with_zscore as (
    select
        date_time,
        value,
        value / (avg(value) over ()) as pct_of_mean,
        (value - avg(value) over ()) / (stdev(value) over ()) as z_score
    from {{tablename}}  where datetime > '2018-11-26' and datetime < '2018-12-03'
)


-- select all
select * from data_with_zscore 

-- select only points greater than a certain threshold
select * from data_with_zscore where z_score > abs(2)

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:

我为Jean-Paul最受欢迎的答案写了一个Go包。它假设y值的类型为float64。

github.com/MicahParks/peakdetect

下面的示例使用了这个包，并基于上面提到的流行答案中的R示例。它在编译时没有任何依赖关系，试图保持较低的内存占用，并且在有新数据点进入时不重新处理过去的点。该项目有100%的测试覆盖率，主要来自上述R示例的输入和输出。但是，如果有人发现任何错误，请打开一个GitHub问题。

编辑:我对v0.0.5进行了性能改进，似乎快了10倍!它使用Welford的方法进行初始化，并使用类似的方法计算滞后期(滑动窗口)的平均值和总体标准偏差。特别感谢另一个帖子的回答:https://stackoverflow.com/a/14638138/14797322

下面是基于R例子的Golang例子:

package main

import (
    "fmt"
    "log"

    "github.com/MicahParks/peakdetect"
)

// This example is the equivalent of the R example from the algorithm's author.
// https://stackoverflow.com/a/54507329/14797322
func main() {
    data := []float64{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}

    // Algorithm configuration from example.
    const (
        lag       = 30
        threshold = 5
        influence = 0
    )

    // Create then initialize the peak detector.
    detector := peakdetect.NewPeakDetector()
    err := detector.Initialize(influence, threshold, data[:lag]) // The length of the initial values is the lag.
    if err != nil {
        log.Fatalf("Failed to initialize peak detector.\nError: %s", err)
    }

    // Start processing new data points and determine what signal, if any they produce.
    //
    // This method, .Next(), is best for when data is being processed in a stream, but this simply iterates over a slice.
    nextDataPoints := data[lag:]
    for i, newPoint := range nextDataPoints {
        signal := detector.Next(newPoint)
        var signalType string
        switch signal {
        case peakdetect.SignalNegative:
            signalType = "negative"
        case peakdetect.SignalNeutral:
            signalType = "neutral"
        case peakdetect.SignalPositive:
            signalType = "positive"
        }

        println(fmt.Sprintf("Data point at index %d has the signal: %s", i+lag, signalType))
    }

    // This method, .NextBatch(), is a helper function for processing many data points at once. It's returned slice
    // should produce the same signal outputs as the loop above.
    signals := detector.NextBatch(nextDataPoints)
    println(fmt.Sprintf("1:1 ratio of batch inputs to signal outputs: %t", len(signals) == len(nextDataPoints)))
}

下面是这个答案的平滑z-score算法的c++实现

std::vector<int> smoothedZScore(std::vector<float> input)
{   
    //lag 5 for the smoothing functions
    int lag = 5;
    //3.5 standard deviations for signal
    float threshold = 3.5;
    //between 0 and 1, where 1 is normal influence, 0.5 is half
    float influence = .5;

    if (input.size() <= lag + 2)
    {
        std::vector<int> emptyVec;
        return emptyVec;
    }

    //Initialise variables
    std::vector<int> signals(input.size(), 0.0);
    std::vector<float> filteredY(input.size(), 0.0);
    std::vector<float> avgFilter(input.size(), 0.0);
    std::vector<float> stdFilter(input.size(), 0.0);
    std::vector<float> subVecStart(input.begin(), input.begin() + lag);
    avgFilter[lag] = mean(subVecStart);
    stdFilter[lag] = stdDev(subVecStart);

    for (size_t i = lag + 1; i < input.size(); i++)
    {
        if (std::abs(input[i] - avgFilter[i - 1]) > threshold * stdFilter[i - 1])
        {
            if (input[i] > avgFilter[i - 1])
            {
                signals[i] = 1; //# Positive signal
            }
            else
            {
                signals[i] = -1; //# Negative signal
            }
            //Make influence lower
            filteredY[i] = influence* input[i] + (1 - influence) * filteredY[i - 1];
        }
        else
        {
            signals[i] = 0; //# No signal
            filteredY[i] = input[i];
        }
        //Adjust the filters
        std::vector<float> subVec(filteredY.begin() + i - lag, filteredY.begin() + i);
        avgFilter[i] = mean(subVec);
        stdFilter[i] = stdDev(subVec);
    }
    return signals;
}

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.