重建线性回归的R nlotr问题:优化过早停止

Question

我试图用nloptr重建多元线性回归，但优化停止得太早。为什么会这样？(我知道nloptr适用于非线性模型，但我仍然希望得到正确的解决方案，例如optim包(

model <- lm(mpg ~ cyl + hp + wt, data = mtcars)
summary(model)
model$coefficients

Call:
lm(formula = mpg ~ cyl + hp + wt, data = mtcars)
Residuals:
Min      1Q  Median      3Q     Max 
-3.9290 -1.5598 -0.5311  1.1850  5.8986 
Coefficients:
Estimate Std. Error t value Pr(>|t|)    
(Intercept) 38.75179    1.78686  21.687  < 2e-16 ***
cyl         -0.94162    0.55092  -1.709 0.098480 .  
hp          -0.01804    0.01188  -1.519 0.140015    
wt          -3.16697    0.74058  -4.276 0.000199 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.512 on 28 degrees of freedom
Multiple R-squared:  0.8431,    Adjusted R-squared:  0.8263 
F-statistic: 50.17 on 3 and 28 DF,  p-value: 2.184e-11

因此，对于经典lm，R平方为0.8431，现在用nloptr:重建它

preds <- c('cyl','hp','wt')
opty_func <- function(para){
mtcars[,paste0(preds,'_pred')] <- mapply(function(preds,b)
{preds*b}, 
mtcars[,preds], 
b = para[2:4])

mtcars[,'predsum'] <- + para[1] + rowSums(mtcars[,c(paste0(preds,'_pred'))])

sst <- sum((mtcars$mpg - mean(mtcars$mpg))^2)
sse <- sum((mtcars$predsum - mtcars$mpg)^2)
rsq <- 1 - (sse / sst)
-rsq
}

#To prove that the function is correct run
opty_func(model$coefficients)
[1] -0,84315

optmodel <- nloptr::nloptr(x0=rep(0,4),                         
eval_f=opty_func,
lb = rep(-Inf,4),
ub = rep(Inf,4), 
opts = list("algorithm"="NLOPT_LN_AUGLAG",
"xtol_rel"=1.0e-10,"ftol_abs"=1.0e-10,
"maxeval" = 1000,
"local_opts" = list("algorithm" = "NLOPT_LN_AUGLAG", "xtol_rel" = 1.0e-10,"ftol_abs"=1.0e-10)))
optmodel

结果：

Call:
nloptr::nloptr(x0 = rep(0, 4), eval_f = opty_func, lb = rep(-Inf, 
4), ub = rep(Inf, 4), opts = list(algorithm = "NLOPT_LN_AUGLAG", 
xtol_rel = 1e-10, ftol_abs = 1e-10, maxeval = 1000, local_opts = list(algorithm = "NLOPT_LN_AUGLAG", 
xtol_rel = 1e-10, ftol_abs = 1e-10)))

Minimization using NLopt version 2.4.2 
NLopt solver status: 3 ( NLOPT_FTOL_REACHED: Optimization stopped because ftol_rel or ftol_abs (above) was reached. )
Number of Iterations....: 102 
Termination conditions:  xtol_rel: 1e-10    ftol_abs: 1e-10 maxeval: 1000 
Number of inequality constraints:  0 
Number of equality constraints:    0 
Optimal value of objective function:  2,4169018891219 
Optimal value of controls: 1,64369 1,088852 0,02969905 1,021684

如何强制nloptr运行更多迭代？状态信息令人困惑，因为无法达到公差，还是在2,41与-0.83之间？BTW:optim包找到正确的R-Square(但速度较慢(。

Answer 1

我发现我的问题的答案是：使用"；算法"="NLOT_ LN_；