define numerical evaluation of a derivative of a sympy function

问题

How can I define the numerical evaluation of a derivative of a function in sympy? I have some functions I can describe with splines for the function and it's derivative using scipy.interpolate. I want to manipulate some expressions with this function and then evaluate the expressions with the splines.

I can use lambdify to make a sympy function evaluate numerically as a spline. But how can I define the derivative of a sympy function to evaluate numerically as a spline?

E.g.

import sympy as sp
import numpy as np
from scipy.interpolate import InterpolatedUnivariateSpline
from sympy.ultilitis.lambdify import implemented_function, lambdify

r = sp.symbols('r')
B = sp.symbols('B', cls=sp.Function)

B_spline = InterpolatedUnivariateSpline([1,2,3,4],[1,4,9,16])
B_der_spline = InterpolatedUnivariateSpline([1,2,3,4],[2,4,6,8])
B = implemented_function(B, lambda r: B_spline(r))

class A(sp.Function):
    nargs = 2

    @classfunction
    def eval(cls, r, B):
        return r**2*B(r)

 A_eval = lambdify(r, A(r,B))
 A_eval(3)
 >>> 81.0
 A_diff_eval = lambdify(r, sp.diff(A(r,B)))
 A_diff_eval(3)
 >>> NameError: global name 'Derivative' is not defined

回答1:

SymPy doesn't know how to take the derivative of the spline function, since it only has the numeric version of it from scipy.

Also, A here could just be a Python function, since you never don't evaluate it. That also makes more sense in that passing a function as an argument to a SymPy function is a bit odd.

All implemented_function does is symfunc._imp_ = staticmethod(implementation) (here symfunc = B and implementation = lambda r: B_spline(r)). You will also need to add fdiff so that it returns a new SymPy Function for B_der_spline. Something like

class B_spline_sym(Function):
    _imp_ = staticmethod(B_spline)

    def fdiff(self, argindex=1):
        return B_der_spline_sym(self.args[0])

class B_der_spline_sym(Function):
    _imp_ = staticmethod(B_der_spline)

def A(r, B):
    return r**2*B(r)

Giving

In [87]: B = B_spline_sym

In [88]:  A_eval = lambdify(r, A(r,B))

In [89]:  A_eval(3)
Out[89]: 81.0

In [91]:  A_diff_eval = lambdify(r, sp.diff(A(r,B)))

In [92]:  A_diff_eval(3)
Out[92]: 108.0

来源：https://stackoverflow.com/questions/24718313/define-numerical-evaluation-of-a-derivative-of-a-sympy-function

标签

python

scipy

sympy

spline