python - 计算时间序列中的峰值

Question

我正在计算 numpy 数组中峰和谷的数量。

我有一个像这样的 numpy 数组:

stack = np.array([0,0,5,4,1,1,1,5,1,1,5,1,1,1,5,1,1,5,1,1,5,1,1,5,1,1,5,1,1])

绘图后，这些数据看起来像这样:

我要查找此时间序列中的峰值数量:

这是我的代码，它适用于像这样的示例，其中时间序列表示中有明显的波峰和波谷。我的代码返回已找到峰值的数组索引。

#example
import numpy as np
from scipy.signal import argrelextrema

stack = 
np.array([0,0,5,4,1,1,1,5,1,1,5,1,1,1,5,1,1,5,1,1,5,1,1,5,1,1,5,1,1])

# for local maxima
y = argrelextrema(stack, np.greater)

print(y)

结果:

(array([ 2,  7, 10, 14, 17, 20, 23, 26]),)

已发现并能正确计数的清晰峰有8个。

我的解决方案似乎无法很好地处理不太清晰且比较困惑的数据。

下面的数组不能很好地工作并且找不到我需要的峰:

array([ 0.        ,  5.70371806,  5.21210157,  3.71144767,  3.9020162 ,
    3.87735984,  3.89030171,  6.00879918,  4.91964227,  4.37756275,
    4.03048542,  4.26943028,  4.02080471,  7.54749062,  3.9150576 ,
    4.08933851,  4.01794766,  4.13217794,  4.15081972,  8.11213474,
    4.6561735 ,  4.54128693,  3.63831552,  4.3415324 ,  4.15944019,
    8.55171441,  4.86579459,  4.13221943,  4.487663  ,  3.95297979,
    4.35334706,  9.91524674,  4.44738182,  4.32562141,  4.420753  ,
    3.54525697,  4.07070637,  9.21055852,  4.87767969,  4.04429321,
    4.50863677,  3.38154581,  3.73663523,  3.83690315,  6.95321174,
    5.11325128,  4.50351938,  4.38070175,  3.20891173,  3.51142661,
    7.80429569,  3.98677631,  3.89820773,  4.15614576,  3.47369797,
    3.73355768,  8.85240649,  6.0876192 ,  3.57292324,  4.43599135,
    3.77887259,  3.62302175,  7.03985076,  4.91916556,  4.22246518,
    3.48080777,  3.26199699,  2.89680969,  3.19251448])

绘图，此数据如下所示:

同样的代码返回:

(array([ 1,  4,  7, 11, 13, 15, 19, 23, 25, 28, 31, 34, 37, 40, 44, 50, 53,
   56, 59, 62]),)

此输出错误地将数据点计为峰值。

理想输出

理想的输出应该返回清晰峰的数量，在本例中为 11 个，它们位于索引处:

[1,7,13,19,25,31,37,44,50,56,62]

我认为我的问题是由于 argrelextrema 函数的聚合性质造成的。

Answer 1

你应该在 scipy.signal 模块中尝试 find_peaks

您的数据示例

import numpy as np
from scipy.signal import find_peaks
import matplotlib.pyplot as plt

a = np.array([ 0.        ,  5.70371806,  5.21210157,  3.71144767, 3.9020162 ,  3.87735984,  3.89030171,  6.00879918,  4.91964227,  4.37756275,
4.03048542,  4.26943028,  4.02080471,  7.54749062,  3.9150576 ,
4.08933851,  4.01794766,  4.13217794,  4.15081972,  8.11213474,
4.6561735 ,  4.54128693,  3.63831552,  4.3415324 ,  4.15944019,
8.55171441,  4.86579459,  4.13221943,  4.487663  ,  3.95297979,
4.35334706,  9.91524674,  4.44738182,  4.32562141,  4.420753  ,
3.54525697,  4.07070637,  9.21055852,  4.87767969,  4.04429321,
4.50863677,  3.38154581,  3.73663523,  3.83690315,  6.95321174,
5.11325128,  4.50351938,  4.38070175,  3.20891173,  3.51142661,
7.80429569,  3.98677631,  3.89820773,  4.15614576,  3.47369797,
3.73355768,  8.85240649,  6.0876192 ,  3.57292324,  4.43599135,
3.77887259,  3.62302175,  7.03985076,  4.91916556,  4.22246518,
3.48080777,  3.26199699,  2.89680969,  3.19251448])

# Here you should fine tune parameters to get what you want
peaks = find_peaks(a, prominence=1.5)
print("Peaks position:", peaks[0])

# Plotting
plt.plot(a)
plt.title("Finding Peaks")

[plt.axvline(p, c='C3', linewidth=0.3) for p in peaks[0]]

plt.show()

情节

输出

# Peaks position: [ 1  7 13 19 25 31 37 44 50 56 62]