Python 平滑数据

Question

我有一个想要平滑的数据集。我有两个变量 y 和 x，它们的间距不均匀。 y 是因变量。但是，我不知道 x 和 y 之间的关系是什么。

我阅读了有关插值的所有内容，但插值要求我知道将 x 与 y 相关的公式。我还研究了其他平滑函数，但这些函数会导致起点和终点出现问题。

有谁知道如何： -获得将x与y联系起来的公式 -平滑数据点而不弄乱端点

我的数据如下：

import matplotlib.pyplot as plt

x = [0.0, 2.4343476531707129, 3.606959459205791, 3.9619355597454664, 4.3503348239356558, 4.6651002761894667, 4.9360228447915109, 5.1839565805565826, 5.5418099660513596, 5.7321342976055165,5.9841050994671106, 6.0478709402949216, 6.3525180590674513, 6.5181245134579893, 6.6627517592933767, 6.9217136972938444,7.103121623408132, 7.2477706136047413, 7.4502723880766748, 7.6174503055171137, 7.7451599936721376, 7.9813193157205191, 8.115292520850506,8.3312689109403202, 8.5648187916197998, 8.6728478860287623, 8.9629327234023926, 8.9974662723308612, 9.1532523634107257, 9.369326186780814, 9.5143785756455479, 9.5732694726297893, 9.8274813411538613, 10.088572892445802, 10.097305715988142, 10.229215999264703, 10.408589988296546, 10.525354763219688, 10.574678982757082, 10.885039893236041, 11.076574204171795, 11.091570626351352, 11.223859812944436, 11.391634940142225, 11.747328449715521, 11.799186895037078, 11.947711314893802, 12.240901223703657, 12.50151825769724, 12.811712563174883, 13.153496854155087, 13.978408296586579, 17.0, 25.0]
y = [0.0, 4.0, 6.0, 18.0, 30.0, 42.0, 54.0, 66.0, 78.0, 90.0, 102.0, 114.0, 126.0, 138.0, 150.0, 162.0, 174.0, 186.0, 198.0, 210.0, 222.0, 234.0, 246.0, 258.0, 270.0, 282.0, 294.0, 306.0, 318.0, 330.0, 342.0, 354.0, 366.0, 378.0, 390.0, 402.0, 414.0, 426.0, 438.0, 450.0, 462.0, 474.0, 486.0, 498.0, 510.0, 522.0, 534.0, 546.0, 558.0, 570.0, 582.0, 594.0, 600.0, 600.0]

#Smoothing here

fig, ax = plt.subplots(figsize=(8, 6))
ax.plot(x, y, color='red', label= 'Unsmoothed curve')

Answer 1

我认为平滑（即过滤）、插值和曲线拟合之间存在混淆，

• 过滤/平滑：我们对数据应用运算符，以消除高频振荡的方式修改原始

  y

  点。这可以通过例如

  scipy.signal.convolve

  、

  scipy.signal.medfilt

  、

  scipy.signal.savgol_filter

  或基于 FFT 的方法来实现。

• 插值：我们从可用数据点创建数据的连续局部表示。插值定义函数在数据点之间的行为方式，但不会修改数据点本身。例如参见

  scipy.interpolate.interp1d

  。不过，为了让事情变得更复杂，样条插值实际上也做了一些平滑处理。

• 曲线拟合：我们通过一些分析函数来拟合数据点。这允许确定数据中

  x

  和

  y

  之间的全局关系，但需要预先了解有关合适拟合函数的一些见解。参见

  scipy.optimize.curve_fit

在这种特殊情况下，我们可以使用的方法是首先在均匀网格上进行插值（如 @agomcas 的答案），然后应用 Savitzky-Golay 滤波器来平滑数据。或者，可以将数据拟合到某些分析表达式，例如基于 tanh 函数，但这需要进一步调整：

import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.interpolate import interp1d
from scipy.signal import savgol_filter
import numpy as np

x = np.array([0.0, 2.4343476531707129, 3.606959459205791, 3.9619355597454664, 4.3503348239356558, 4.6651002761894667, 4.9360228447915109, 5.1839565805565826, 5.5418099660513596, 5.7321342976055165,5.9841050994671106, 6.0478709402949216, 6.3525180590674513, 6.5181245134579893, 6.6627517592933767, 6.9217136972938444,7.103121623408132, 7.2477706136047413, 7.4502723880766748, 7.6174503055171137, 7.7451599936721376, 7.9813193157205191, 8.115292520850506,8.3312689109403202, 8.5648187916197998, 8.6728478860287623, 8.9629327234023926, 8.9974662723308612, 9.1532523634107257, 9.369326186780814, 9.5143785756455479, 9.5732694726297893, 9.8274813411538613, 10.088572892445802, 10.097305715988142, 10.229215999264703, 10.408589988296546, 10.525354763219688, 10.574678982757082, 10.885039893236041, 11.076574204171795, 11.091570626351352, 11.223859812944436, 11.391634940142225, 11.747328449715521, 11.799186895037078, 11.947711314893802, 12.240901223703657, 12.50151825769724, 12.811712563174883, 13.153496854155087, 13.978408296586579, 17.0, 25.0])
y = np.array([0.0, 4.0, 6.0, 18.0, 30.0, 42.0, 54.0, 66.0, 78.0, 90.0, 102.0, 114.0, 126.0, 138.0, 150.0, 162.0, 174.0, 186.0, 198.0, 210.0, 222.0, 234.0, 246.0, 258.0, 270.0, 282.0, 294.0, 306.0, 318.0, 330.0, 342.0, 354.0, 366.0, 378.0, 390.0, 402.0, 414.0, 426.0, 438.0, 450.0, 462.0, 474.0, 486.0, 498.0, 510.0, 522.0, 534.0, 546.0, 558.0, 570.0, 582.0, 594.0, 600.0, 600.0])


xx = np.linspace(x.min(),x.max(), 1000)

# interpolate + smooth
itp = interp1d(x,y, kind='linear')
window_size, poly_order = 101, 3
yy_sg = savgol_filter(itp(xx), window_size, poly_order)


# or fit to a global function
def func(x, A, B, x0, sigma):
    return A+B*np.tanh((x-x0)/sigma)

fit, _ = curve_fit(func, x, y)
yy_fit = func(xx, *fit)

fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(x, y, 'r.', label= 'Unsmoothed curve')
ax.plot(xx, yy_fit, 'b--', label=r"$f(x) = A + B \tanh\left(\frac{x-x_0}{\sigma}\right)$")
ax.plot(xx, yy_sg, 'k', label= "Smoothed curve")
plt.legend(loc='best')

Answer 2

插值不要求您知道与 x 和 y 相关的公式。

import matplotlib.pyplot as plt
from scipy import interpolate
import numpy as np

x = [0.0, 2.4343476531707129, 3.606959459205791, 3.9619355597454664, 4.3503348239356558, 4.6651002761894667, 4.9360228447915109, 5.1839565805565826, 5.5418099660513596, 5.7321342976055165,5.9841050994671106, 6.0478709402949216, 6.3525180590674513, 6.5181245134579893, 6.6627517592933767, 6.9217136972938444,7.103121623408132, 7.2477706136047413, 7.4502723880766748, 7.6174503055171137, 7.7451599936721376, 7.9813193157205191, 8.115292520850506,8.3312689109403202, 8.5648187916197998, 8.6728478860287623, 8.9629327234023926, 8.9974662723308612, 9.1532523634107257, 9.369326186780814, 9.5143785756455479, 9.5732694726297893, 9.8274813411538613, 10.088572892445802, 10.097305715988142, 10.229215999264703, 10.408589988296546, 10.525354763219688, 10.574678982757082, 10.885039893236041, 11.076574204171795, 11.091570626351352, 11.223859812944436, 11.391634940142225, 11.747328449715521, 11.799186895037078, 11.947711314893802, 12.240901223703657, 12.50151825769724, 12.811712563174883, 13.153496854155087, 13.978408296586579, 17.0, 25.0]
y = [0.0, 4.0, 6.0, 18.0, 30.0, 42.0, 54.0, 66.0, 78.0, 90.0, 102.0, 114.0, 126.0, 138.0, 150.0, 162.0, 174.0, 186.0, 198.0, 210.0, 222.0, 234.0, 246.0, 258.0, 270.0, 282.0, 294.0, 306.0, 318.0, 330.0, 342.0, 354.0, 366.0, 378.0, 390.0, 402.0, 414.0, 426.0, 438.0, 450.0, 462.0, 474.0, 486.0, 498.0, 510.0, 522.0, 534.0, 546.0, 558.0, 570.0, 582.0, 594.0, 600.0, 600.0]


f = interpolate.interp1d(x, y, kind="linear")
x_int = np.linspace(x[0],x[-1], 20)
y_int = f(x_int)

#Smoothing here

fig, ax = plt.subplots(figsize=(8, 6))
ax.plot(x, y, color='red', label= 'Unsmoothed curve')
ax.plot(x_int, y_int, color="blue", label= "Interpolated curve")

Answer 3

我创建了自己的简单移动平均平滑函数：

import numpy as np

def MySmooth(in_array, subset_len):
    array_len = len(in_array)
    start_sum = np.sum(in_array[0 : subset_len-1])
    end_mean = np.mean(in_array[array_len - subset_len : ])
    out_array = np.zeros(array_len)
    out_array[0 : subset_len-1] = start_sum/subset_len
    out_array[array_len-1 - subset_len+1 : ] = end_mean
    index = int(subset_len/2)
    for i in range(subset_len, array_len):
        start_sum -= in_array[i - subset_len]
        start_sum += in_array[i]
        out_array[index] = start_sum/subset_len
        index +=1
    # Here I might calculate the accumulated float-operation error by comparing start_sum and end_mean
    return out_array