# Best Practices for converting a Mathematica expression to Julia

Are there suggestions how to convert a rather long Mathematica expression to Julia?

With Matlab I use http://library.wolfram.com/infocenter/MathSource/577/

I also would appreciate suggestions for the same problem with Sage Math

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The closest thing to it is this

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Use SymPy.jl. SymPy can parse Mathematica expressions, and SymPy.jl can convert SymPy expressions to Julia expressions. For example, if we define:

``````using SymPy
const sympy_parsing_mathematica = SymPy.PyCall.pyimport("sympy.parsing.mathematica")
mathematica2julia(s::AbstractString, substitutions::Pair{<:AbstractString,<:AbstractString}...) =
SymPy.lambdify_expr(sympy_parsing_mathematica["mathematica"](s, Dict(substitutions...))).args
``````

then we get:

``````julia> mathematica2julia("Log[Log[x]]")
:(log(log(x)))

julia> mathematica2julia("Exp[Log3[x]]", "Log3[x]"=>"log(x,3)")
:(x ^ (log(3) ^ -1))
``````
11 Likes

Works charmingly well… thank you!

`SymataSyntax` has not been upgraded for v1.0.

Ah, this stops working now for SymPy v1.0.10.

The updated code should be:

``````using SymPy
const sympy_parsing_mathematica = SymPy.PyCall.pyimport("sympy.parsing.mathematica")
mathematica2julia(s::AbstractString, substitutions::Pair{<:AbstractString,<:AbstractString}...) =
SymPy.walk_expression(sympy_parsing_mathematica."mathematica"(s, Dict(substitutions...)))
``````
3 Likes

How to replace the unknown function?Naively the following code does not work.

``````mathematica2julia("(Sqrt[Pi])/(2*Gamma[23/10])","Gamma"=>"gamma")
``````

PyError (\$(Expr(:escape, :(ccall(#= /Users/gangchen/.julia/packages/PyCall/ttONZ/src/pyfncall.jl:44 =# @pysym(:PyObject_Call), PyPtr, (PyPtr, PyPtr, PyPtr), o, pyargsptr, kw))))) <class ‘ValueError’>
ValueError("‘Gamma’ function form is invalid.")
File “/Users/gangchen/.julia/conda/3/lib/python3.7/site-packages/sympy/parsing/mathematica.py”, line 17, in mathematica
File “/Users/gangchen/.julia/conda/3/lib/python3.7/site-packages/sympy/parsing/mathematica.py”, line 154, in init
File “/Users/gangchen/.julia/conda/3/lib/python3.7/site-packages/sympy/parsing/mathematica.py”, line 187, in _compile_dictionary
raise ValueError(err)

The dictionary has to map a Mathematica function-call expression to a Python/SymPy function call:

``````julia> mathematica2julia("(Sqrt[Pi])/(2*Gamma[23/10])","Gamma[x]"=>"gamma(x)")
:(__prod__(1 / 2, pi ^ (1 / 2), gamma(23//10) ^ -1))
``````

See the documentation for the `sympy.parsing.mathematica.mathematica` function that is being called here.

Thanks. That works well

I got a Julia expression if I do this. How to turn expression that I get into a function?

you can use SyntaxTree.jl. The genfun is useful. You can also see the examples in my UseMathLink.jl

I eventually ended up using sage and copy paste to Julia since the syntax are quite similar. The only problem I found so far is that sage convert `Exp(2)` from Mathematica to `e^2`. But I often want it written as `exp(2)`.