如何传递 JuMP

如何解决如何传递 JuMP

我想转换一个 SymPy 表达式，以便在 JuMP 中用作目标函数。假设我的变量涉及两个变量

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被问到：https://discourse.julialang.org/t/how-to-pass-a-sympy-expression-supported-by-jump/62109/3）

解决方法

我来自 Discourse 的回答：

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)