Python中的二阶导数-scipy/numpy/pandas [英] Second Derivative in Python - scipy/numpy/pandas

问题描述

我正在尝试在python中使用带有两个numpy数据数组的二阶导数.

I'm trying to take a second derivative in python with two numpy arrays of data.

例如，有问题的数组如下所示:

For example, the arrays in question look like this:

import numpy as np

x = np.array([ 120. ,  121.5,  122. ,  122.5,  123. ,  123.5,  124. ,  124.5,
        125. ,  125.5,  126. ,  126.5,  127. ,  127.5,  128. ,  128.5,
        129. ,  129.5,  130. ,  130.5,  131. ,  131.5,  132. ,  132.5,
        133. ,  133.5,  134. ,  134.5,  135. ,  135.5,  136. ,  136.5,
        137. ,  137.5,  138. ,  138.5,  139. ,  139.5,  140. ,  140.5,
        141. ,  141.5,  142. ,  142.5,  143. ,  143.5,  144. ,  144.5,
        145. ,  145.5,  146. ,  146.5,  147. ])

y = np.array([  1.25750000e+01,   1.10750000e+01,   1.05750000e+01,
         1.00750000e+01,   9.57500000e+00,   9.07500000e+00,
         8.57500000e+00,   8.07500000e+00,   7.57500000e+00,
         7.07500000e+00,   6.57500000e+00,   6.07500000e+00,
         5.57500000e+00,   5.07500000e+00,   4.57500000e+00,
         4.07500000e+00,   3.57500000e+00,   3.07500000e+00,
         2.60500000e+00,   2.14500000e+00,   1.71000000e+00,
         1.30500000e+00,   9.55000000e-01,   6.65000000e-01,
         4.35000000e-01,   2.70000000e-01,   1.55000000e-01,
         9.00000000e-02,   5.00000000e-02,   2.50000000e-02,
         1.50000000e-02,   1.00000000e-02,   1.00000000e-02,
         1.00000000e-02,   1.00000000e-02,   1.00000000e-02,
         1.00000000e-02,   1.00000000e-02,   5.00000000e-03,
         5.00000000e-03,   5.00000000e-03,   5.00000000e-03,
         5.00000000e-03,   5.00000000e-03,   5.00000000e-03,
         5.00000000e-03,   5.00000000e-03,   5.00000000e-03,
         5.00000000e-03,   5.00000000e-03,   5.00000000e-03,
         5.00000000e-03,   5.00000000e-03])

我目前有f(x) = y，我想要d^2 y / dx^2.

从数字上讲，我知道我可以对函数进行插值并进行解析求导，也可以使用更高阶有限差分.我认为，如果一个或另一个被认为更快，更准确等，则有足够的数据来使用.

Numerically, I know I can either interpolate the function and take the derivative analytically or use higher order finite-differences. I think that there is enough data to use either, if one or the other is considered faster, more accurate, etc.

我看过np.interp()和scipy.interpolate都没有成功，因为这使我得到了拟合的(线性或三次)样条曲线，但是不知道如何得到该点的导数.

I have looked at np.interp() and scipy.interpolate with no success, as this returns me a fitted (linear or cubic) spline, but don't know how to get the derivative at that point.

非常感谢任何指导.

推荐答案

您可以使用scipy的一维

You can interpolate your data using scipy's 1-D Splines functions. The computed spline has a convenient derivative method for computing derivatives.

对于您的示例数据，使用UnivariateSpline可以得到以下拟合结果

For the data of your example, using UnivariateSpline gives the following fit

import matplotlib.pyplot as plt
from scipy.interpolate import UnivariateSpline

y_spl = UnivariateSpline(x,y,s=0,k=4)

plt.semilogy(x,y,'ro',label = 'data')
x_range = np.linspace(x[0],x[-1],1000)
plt.semilogy(x_range,y_spl(x_range))

看起来合适，至少在视觉上是合适的.您可能要尝试使用UnivariateSpline使用的参数.

The fit seems reasonably good, at least visually. You might want to experiment with the parameters used by UnivariateSpline.

样条拟合的二阶导数可以简单地获得为

The second derivate of the spline fit can be simply obtained as

y_spl_2d = y_spl.derivative(n=2)

plt.plot(x_range,y_spl_2d(x_range))

结果看起来有些不自然(以防您的数据与某个物理过程相对应).您可能想要更改样条曲线拟合参数，改善数据(例如，提供更多样本，执行较少的噪声测量)，或者决定使用分析函数对数据建模并进行曲线拟合(例如，使用sicpy的 curve_fit )

The outcome appears somewhat unnatural (in case your data corresponds to some physical process). You may either want to change the spline fit parameters, improve your data (e.g., provide more samples, perform less noisy measurements), or decide on an analytic function to model your data and perform a curve fit (e.g., using sicpy's curve_fit)

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