python – 定义sympy函数导数的数值计算

如何在sympy中定义函数导数的数值计算？
我有一些功能,我可以用函数的样条线来描述,它是使用scipy.interpolate的派生.
我想用这个函数操作一些表达式,然后用样条曲线计算表达式.

我可以使用lambdify使一个sympy函数作为样条数值进行评估.
但是,如何定义sympy函数的导数以数值方式作为样条函数进行求值？

例如.

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

解决方法:

SymPy不知道如何获取样条函数的导数,因为它只有scipy的数字版本.

此外,A这里可能只是一个Python函数,因为你永远不会评估它.这也更有意义,因为将函数作为参数传递给SymPy函数有点奇怪.

所有implements_function都是symfunc._imp_ = staticmethod(implementation)(这里symfunc = B,implementation = lambda r：B_spline(r)).您还需要添加fdiff,以便为B_der_spline返回一个新的SymPy函数.就像是

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)

给予

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

标签：spline,python,scipy,sympy
来源： https://codeday.me/bug/20191007/1864825.html