Substitute a SciPy spline function into a SymPy expression

[Maksim Surov]

I would like to lambdify a sympy expression, and substitute an implementation of an abstract function into the expression (as usual the implementation is a scipy.interpolate.spline function). The typical problem is the expression contains derivatives of the function. For example:

import sympy as sy
import numpy as np
import scipy.interpolate as interp

# the expression:
t = sy.Symbol('t')
f = sy.Function('f')(t)
expr = (f + f**2 / 2).diff()

# implementation:
f_spline = interp.splrep(np.linspace(0, 1, 100), np.sin(np.linspace(0, 1, 100)))

def df_impl(x):
 print '[trace] f_der(' + str(x) + ')'
 return interp.splev(x, f_spline, der=1)

def f_impl(x):
 print '[trace] f(' + str(x) + ')'
 return interp.splev(x, f_spline)

expr_impl = sy.lambdify(t, expr, {'Derivative(f(t), t)': df_impl, 'f': f_impl})
print expr_impl(0.1)

This latest line throws the error:

File "<string>", line 1, in <lambda>
NameError: global name 'Derivative' is not defined

It's clear because lambdify can't substitute the the derivative string. There is a workaround, I can do 2 subs:

expr = expr.subs({'Derivative(f(t), t)': 'g(t)'})
expr_impl = sy.lambdify(t, expr, {'g': df_impl, 'f': f_impl})
print expr_impl(0.1)

But the solution looks ugly. Any suggestions?

[Aaron Meurer]

I would use subs to replace the Derivative with a different Symbol
before using lambdify.

Aaron Meurer

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