使用 sympy 获取 function 的导数作为 function （用于以后的评估和替换）

Question

I want to work with generic functions as long as possible, and only substitute functions at the end. I'd like to define a function as the derivative of another one, define a generic expression with the function and its derivative, and substitute the function at the end. Right now my attempts is as follows, but I get the error 'Derivative' object is not callable :

from sympy import Function
x, y, z = symbols('x y z')
f  = Function('f') 
df = f(x).diff(x)  # <<< I'd like this to be a function of dummy variable x
expr = f(x) * df(z)  + df(y) + df(0) # df is unfortunately not callable
# At the end, substitute with example function
expr.replace(f, Lambda(X, cos(X))) # should return: -cos(x)*sin(z) - sin(y) - sin(0)

I think I got it to work with integrals as follows: I= Lambda( x, integrate( f(y), (y, 0, x))) but that won't work for derivatives. If that helps, I'm fine restricting myself to functions of a single variable for now.

As a bonus, I'd like to get this to work with any combination (products, derivatives, integrals) of the original function.

Answer 1

It's pretty disappointing that f.diff(x) doesn't work, as you say. Maybe someone will create support it sometime in the future. In the mean time, there are 2 ways to go about it: either substitute x for your y, z, ... OR lambdify df .

I think the first option will work more consistently in the long run (for example, if you decide to extend to multivariate calculus). But the expr in second option is far more natural.

Using substitution:

from sympy import *

x, y, z = symbols('x y z')
X = Symbol('X')

f = Function('f')
df = f(x).diff(x)

expr = f(x) * df.subs(x, z) + df.subs(x, y) + df.subs(x, 0)
print(expr.replace(f, Lambda(X, cos(X))).doit())

Lambdifying df :

from sympy import *

x, y, z = symbols('x y z')
X = Symbol('X')

f = Function('f')
df = lambda t: f(t).diff(t) if isinstance(t, Symbol) else f(X).diff(X).subs(X, t)

expr = f(x) * df(z) + df(y) + df(0)
print(expr.replace(f, Lambda(X, cos(X))).doit())

Both give the desired output.