How to copy Mathematica expression into a Julia function

Specific Domains Modelling & Simulations
,
1

Very similar questions were asked before in the discourse (see e.g. this, this, and this). Answers point to MathLink where one can plug an expression like this

using MathLink
sqx = W`Sqrt[x]`

and then evaluate this function like so

weval(sqx; x=2.0)
# 1.4142135623730951

I would like to define a normal Julia function like

f(x) = sqrt(x)

using sqx above. Is there a way to do that?

My end goal is to be able to take the string output from Mathematica and edit some of the variables to define such a function f(x).

2

Why not simply

using MathLink
const sqx = W`Sqrt[x]`
f(x) = weval(sqx; x=x)

?

3

I assume you mean?

f(x) = weval(sqx; x)

When I plug a number to this f I get the following type

typeof(f(2))
# MathLink.WExpr

I would like to get just a Float64.

4 
f(x) = weval(Float64, sqx; x)

?

1 Like
5

I am reviving this post with a follow-up question.

Is there a way to get an expression from mathematical, for example Sqrt[x y] to be typed in julia like

Sqrt[x y]

would be

sqrt(x*y)

Thanks!

6

See: Best Practices for converting a Mathematica expression to Julia - #3 by stevengj

1 Like
7

For completeness, the answer I was looking for was given in this post and this is the solution (for more details read the thread in the other post).

using SymPy
const sympy_parsing_mathematica = SymPy.PyCall.pyimport("sympy.parsing.mathematica")

s = "(Sqrt[Pi])/(2*EllipticE[m])"
ex = sympy_parsing_mathematica.mathematica(s, Dict("EllipticE[x]"=>"elliptic_e(x)"))

# option 1
SymPy.walk_expression(ex, fns=Dict("Pow"=>:^)) 
#:((1 / 2) * pi ^ (1 / 2) * elliptic_e(m) ^ -1)

# option 2 (should work soon after some updates of SymPy)
SymPy.convert_expr(ex, use_julia_code=true) 
# :(sqrt(pi) ./ (2 * elliptic_e(m)))
8

https://github.com/JuliaInterop/MathLink.jl/issues/85

1 Like