Julia: how do I convert a symbolic expression to a function?

Question

I have created a symbolic expression using the SymPy package (https://github.com/jverzani/SymPy.jl). I want to now find the roots of that expression using the Roots package

Answer 1

With Julia 0.6 this works straight out of box:

julia> using SymPy, Roots

julia> x = Sym("x")
x

julia> expr = sin(x)
sin(x)

julia> fzeros(expr, -10, 10)
7-element Array{Float64,1}:
 -9.42478
 -6.28319
 -3.14159
  0.0
  3.14159
  6.28319
  9.42478

But it's a lot slower than pure Julia solution. Here are some benchmarks I ran with BenchmarkTools:

Naive Solution (6.8s):

fzeros(expr, -10, 10)

Pure Julia (1.5ms):

fzeros(x-> sin(x), -10, 10)

Semi-Naive (7.6ms):

fzeros(Function(expr), -10, 10)

Explicit Conversion (3.8ms):

expr2 = Function(expr)
fzeros(expr2, -10, 10)

Answer 2

[UPDATE: The discussion below has been superseded in many cases by the recently introduced lambdify function. The call lambdify(expr) creates a julia function that does not call back into SymPy to evaluate, so should be much faster. It should work for most, but certainly not all, expressions.]

It is a two step process:

convert(Function, expr)

will return a function of the free variables, x, in your case. However, the function values are still symbolic and can't be used with fzeros. The inputs can be guessed, but the type of the return value is another story. However, coercing to float will work in this case:

fzeros(x -> float(convert(Function, expr)), -10, 10)

(You could also do this with a -> float(replace(expr, x, a)).)

For this simple example solve(expr) will also work, but in general, the findroot function in SymPy is not exposed, so numeric root solving via SymPy isn't a workaround without some effort by the end-users.