问Optim.jl在每次迭代中进行许多冗余的函数调用
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Stack Overflow用户

提问于 2019-01-19 20:29:58

回答 1查看 270关注 0票数 2

Optim.jl做了许多冗余的函数调用。对于一个有6个变量的函数和方法LBFGS() (没有提供梯度-我的函数是一个不动点问题的解决方案，不容易计算梯度和ForwardDiff和ReverseDiff，由于某些原因，不能使用我的代码)。

它在第一次迭代中调用函数39次。更重要的是，这些函数调用中的许多都是在完全相同的输入上进行计算的。这似乎非常低效--我做错了什么吗？如果没有，我可以做些什么来提高效率？

我的代码目前如下-如果你需要更多的答案，请告诉我。

function f(x::Vector{Float64})

    modelPar.ρ = x[1]
    modelPar.χI = x[2]
    modelPar.χS = x[3]
    modelPar.χE = x[4] * x[3]
    modelPar.λ = x[5]
    modelPar.ν = x[6]

    f = open("figures/log.txt","a")

    write(f,"Iteration: ρ = $(x[1]); χI = $(x[2]); χS = $(x[3]); 
                        χE = $(x[3] * x[4]); λ = $(x[5]); ν = $(x[6])\n")

    close(f)

    output = computeScore(algoPar,modelPar,guess,targets,weights)

end

initial_x = [ modelPar.ρ;
              modelPar.χI;
              modelPar.χS;
              modelPar.χE / modelPar.χS;
              modelPar.λ;
              modelPar.ν ]

lower = [0.01, 0.1, 0.1, 0.01, 1.001, 0.01]
upper = [0.1, 6, 6, 0.99, 1.5, 0.5]

inner_optimizer = LBFGS()

results = optimize(f,lower,upper,initial_x,Fminbox(inner_optimizer),
         Optim.Options(iterations = 0, store_trace = true, show_trace = true))

跟踪如下所示

Results of Optimization Algorithm
* Algorithm: Fminbox with L-BFGS
* Starting Point: [0.04,4.0,2.0,0.5,1.05,0.05]
* Minimizer: [0.04,4.0,2.0,0.5,1.05,0.05]
* Minimum: 2.069848e-02
* Iterations: 1
* Convergence: true
  * |x - x'| ≤ 0.0e+00: true 
    |x - x'| = 0.00e+00 
  * |f(x) - f(x')| ≤ 0.0e+00 |f(x)|: true
    |f(x) - f(x')| = 0.00e+00 |f(x)|
  * |g(x)| ≤ 1.0e-08: false 
    |g(x)| = 1.63e-01 
  * Stopped by an increasing objective: false
  * Reached Maximum Number of Iterations: true
* Objective Calls: 1
* Gradient Calls: 1

但是，文件log.txt长39行，包含以下内容：

    Iteration: ρ = 0.04000605545445239; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.03999394454554761; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.000024221817809; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 3.9999757781821903; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0000121109089046; χE = 1.0000060554544523; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 1.9999878890910952; χE = 0.9999939445455476; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0000121109089048; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 0.9999878890910953; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.050006358227175; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.049993641772825; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05000605545445239
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.04999394454554761
    Iteration: ρ = 0.04000605545445239; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.03999394454554761; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.000024221817809; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 3.9999757781821903; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0000121109089046; χE = 1.0000060554544523; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 1.9999878890910952; χE = 0.9999939445455476; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0000121109089048; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 0.9999878890910953; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.050006358227175; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.049993641772825; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05000605545445239
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.04999394454554761
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04000605545445239; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.03999394454554761; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.000024221817809; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 3.9999757781821903; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0000121109089046; χE = 1.0000060554544523; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 1.9999878890910952; χE = 0.9999939445455476; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0000121109089048; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 0.9999878890910953; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.050006358227175; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.049993641772825; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05000605545445239
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.04999394454554761
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05
    Iteration: ρ = 0.04; χI = 4.0; χS = 2.0; χE = 1.0; λ = 1.05; ν = 0.05

这里发生什么事情？

julia

nonlinear-optimization

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回答 1

Stack Overflow用户

发布于 2019-02-26 18:37:43

1.0e-08相当低。您可以尝试使用g_tol=1.0e-6来查看它的收敛情况吗？

results = optimize(f, lower, upper, initial_x, Fminbox(inner_optimizer), Optim.Options(iterations=0, store_trace=true, show_trace=true, g_tol=1.0e-6))