如何传递支持JuMP的SymPy表达式

Question

我想将一个SymPy表达式转换为JuMP中的目标函数，假设该变量涉及两个变量。

x = Sym("x")
y = Sym("y")
expr = x^2 + y
fn = lambdify( expr )

"和模型"

model = Model(Ipopt.Optimizer)
l = zeros(1,2)
@variable(model, x[ j = 1:2] >= 0 )

register(model, :fn,2, fn, autodiff = true)

#obj function
@NLobjective(model,Min,fn)
print(model)
Unexpected object #99 (of type SymPy.var"#99#100"{SymPy.var"###792"} in nonlinear expression.

Stacktrace:
 [1] error(::String) at .\error.jl:33
 [2] _parse_NL_expr_runtime(::Model, ::Function, ::Array{JuMP._Derivatives.NodeData,1}, ::Int64, ::Array{Float64,1}) at C:\Users\xxxxx\.julia\packages\JuMP\e0Uc2\src\parse_nlp.jl:223
 [3] top-level scope at C:\Users\xxxxx\.julia\packages\JuMP\e0Uc2\src\parse_nlp.jl:247
 [4] top-level scope at C:\Users\xxxxx\.julia\packages\JuMP\e0Uc2\src\macros.jl:1368
 [5] top-level scope at In[92]:8
 [6] include_string(::Function, ::Module, ::String, ::String) at .\loading.jl:1091

这个问题也在另一个平台上提出过：https://discourse.julialang.org/t/how-to-pass-a-sympy-expression-supported-by-jump/62109/3。

- Fam

Answer 1

我的Discourse回答：

using JuMP, SymPy, Ipopt
x = Sym("x")
y = Sym("y")
expr = x^2 + y
fn = lambdify(expr)
model = Model(Ipopt.Optimizer)
@variable(model, x[1:2] >= 0)
register(model, :fn, 2, fn; autodiff = true)
@NLobjective(model, Min, fn(x[1], x[2]))
optimize!(model)

- Oscar Dowson