Second Derivative in Python - scipy/numpy/pandas

问题:

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

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])

I currently then have f(x) = y, and I want d^2 y / dx^2.

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.

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.

Any guidance is much appreciated.

回答1:

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

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))

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

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))

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)

回答2:

By finite differences, the first order derivative of y for each mean value of x over your array is given by :

dy=np.diff(y,1)
dx=np.diff(x,1)
yfirst=dy/dx

And the corresponding values of x are :

xfirst=0.5*(x[:-1]+x[1:])

For the second order, do the same process again :

dyfirst=np.diff(yfirst,1)
dxfirst=np.diff(xfirst,1)
ysecond=dyfirst/dxfirst

xsecond=0.5*(xfirst[:-1]+xfirst[1:])