Lambdify getting error if expression is function of time

General Usage
,
1 
using SymPy
using PyCall
sp = pyimport("sympy")
@vars t
x = sp.Function("x")(t)
y = sp.Function("y")(t)
z = x^2+y^2
hz = lambdify(z,(x,y))
# hz = lambdify(z,(x,y))
hz(1,1)
2

It is a bit confusing what is desired here. As is you have three variables: t, x, and y and are passing in two values.

Maybe this is what you want:

julia> @syms t x() y()
(t, x, y)

julia> z = x(t)^2 + y(t)^2
 2       2   
x (t) + y (t)

julia> f = lambdify(z, (t, x, y))
#118 (generic function with 1 method)

julia> f(t, x, y)
 2       2   
x (t) + y (t)

julia> f(pi/4, sin, cos)
1.0
3

Thank you.
but if I try like this, I’m getting error. my whole code is written in this method. please provide solution.
Thank You.

using SymPy
using PyCall
sp = pyimport("sympy")
@vars t
x = sp.Function("x")(t)
y = sp.Function("y")(t)
z = x^2+y^2
hz = lambdify(z,(t,x,y))
hz(1,1,1)

even this one also getting error

using SymPy
@vars t
x = SymFunction("x")(t)
y = SymFunction("y")(t)
z = x^2+y^2
hz = lambdify(z,(x,y,t))
hz(1,1,1)
4

Sorry, lambdify doesn’t correctly identify the variables x(t) and y(t) in your approach. The unevaluated function approach allows a function to be passed in; in your approach your are trying to pass in a value of x(1) to be 1 making me unclear why you are using a function there.

5

I have done kinematics and dynamics in this format. can you suggest any other alternative for converting symbolic expression to numerical function.
Thanks.

6

You can use substitution. It is more flexible, but not as performant as creating a julia function.

7

Thanks :+1: