Julia+JuMP:函数的可变参数数量 [英] Julia+JuMP: variable number of arguments to function

问题描述

我正在尝试使用 JuMP 来解决非线性问题，其中变量的数量由用户决定 - 也就是说，在编译时不知道.

I'm trying to use JuMP to solve a non-linear problem, where the number of variables are decided by the user - that is, not known at compile time.

为此，@NLobjective 行如下所示:

@eval @JuMP.NLobjective(m, Min, $(Expr(:call, :myf, [Expr(:ref, :x, i) for i=1:n]...)))

例如，如果 n=3，编译器会将该行解释为与以下内容相同:

Where, for instance, if n=3, the compiler interprets the line as identical to:

@JuMP.NLobjective(m, Min, myf(x[1], x[2], x[3]))

问题是 @eval 只能在全局范围内工作，当包含在函数中时，会抛出错误.

The issue is that @eval works only in the global scope, and when contained in a function, an error is thrown.

我的问题是:我怎样才能完成同样的功能——让 @NLobjective 用可变数量的 调用 myf>x[1],...,x[n] 参数——在函数的本地、编译时未知范围内?

My question is: how can I accomplish this same functionality -- getting @NLobjective to call myf with a variable number of x[1],...,x[n] arguments -- within the local, not-known-at-compilation scope of a function?

def testme(n)
    myf(a...) = sum(collect(a).^2)

    m = JuMP.Model(solver=Ipopt.IpoptSolver())

    JuMP.register(m, :myf, n, myf, autodiff=true)
    @JuMP.variable(m, x[1:n] >= 0.5)

    @eval @JuMP.NLobjective(m, Min, $(Expr(:call, :myf, [Expr(:ref, :x, i) for i=1:n]...)))
    JuMP.solve(m)
end

testme(3)

谢谢！

推荐答案

如 http://jump.readthedocs.io/en/latest/nlp.html#raw-expression-input ，可以在没有宏的情况下给出目标函数.相关表达:

As explained in http://jump.readthedocs.io/en/latest/nlp.html#raw-expression-input , objective functions can be given without the macro. The relevant expression:

    JuMP.setNLobjective(m, :Min, Expr(:call, :myf, [x[i] for i=1:n]...))

甚至比基于 @eval 的更简单，并且可以在函数中使用.代码是:

is even simpler than the @eval based one and works in the function. The code is:

using JuMP, Ipopt

function testme(n)
    myf(a...) = sum(collect(a).^2)

    m = JuMP.Model(solver=Ipopt.IpoptSolver())

    JuMP.register(m, :myf, n, myf, autodiff=true)
    @JuMP.variable(m, x[1:n] >= 0.5)

    JuMP.setNLobjective(m, :Min, Expr(:call, :myf, [x[i] for i=1:n]...))
    JuMP.solve(m)
    return [getvalue(x[i]) for i=1:n]
end

testme(3)

然后它返回:

julia> testme(3)

:

 EXIT: Optimal Solution Found.
3-element Array{Float64,1}:
 0.5
 0.5
 0.5

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