使用两个变量最小化函数

Question

我想用两个变量最小化一个函数.

首先,我已经创建了一个函数(rba),在(kvasum)我需要最小化的函数内部需要它.最小化的值是其中的一部分rba.

# Data
vpk = data.frame(V1 =c(3650000000, 19233, 2211.2, 479.47, 168.46, 83.447, 52.349, 38.738,
                     32.34, 29.588), V2 = 1:10)
n = nrow(vpk)

# functions to minimize 

# This function returns a vector with 10 values

rba = function(par){
  v <- matrix(ncol = 1, nrow = 10)
  for (p in 1:10){
    k<- ifelse (par[1] < 1-1/p && par[1]>0 && p > par[2] && 
par[2]>0 && par[2]<2, par[2]*p,
                    ifelse(par[1] < 1-1/par[2] && par[1] > 0 && 
p < par[2] && par[2]>0 && par[2]<2, -1+(par[1]+1/par[2]),
                           ifelse(par[1] > (1 - 1 / max(p,par[2])) && 
par[2]>0 && par[2]<2, -1+p, "error")))
    v[p] <- k 
  }
  return(v)
}

# This function uses the function rba, and returns a value

kvasum = function(par){
  sum( (log(vpk$V1)/log(1/n) - rba(par) )^2)
}

# what I would I to do is to find par[1] and par[2] such that kvasum is minimized

m1 = optim(par=c(0.1,0.4),kvasum, lower=0)

我试过使用optim函数,但是我无法使用它.我得到一个非数字参数,并尝试了我能想到的一切.任何帮助表示赞赏.

Answer 1

您的整个过程存在一些问题,可能会导致问题.

首先作为@ user227710的提到了应更换的意见&&与&.这些有不同的含义.

现在为优化者

看起来您想要为参数设置限制(即所谓的框约束).为了做到这一点并因此使用lower参数,您需要使用该L-BFGS-B方法.使用它时,您还需要指定upper参数.

您收到的错误,您收到它是因为您的ifelse语句仅在值大致介于0和1之间时才起作用.否则,k变量获取值error(如果ifelse语句中的所有条件都为FALSE 则返回的值)这就是为什么你得到的

Error in log(vpk$V1)/log(1/n) - rba(par) : 
  non-numeric argument to binary operator

错误.

因此,如果你相应地指定你的盒子约束(或者你可能看看你的ifelse语句,因为你可能编码错了),这似乎是完美的:

# Data
vpk = data.frame(V1 =c(3650000000, 19233, 2211.2, 479.47, 168.46, 83.447, 52.349, 38.738,
                       32.34, 29.588), V2 = 1:10)
n = nrow(vpk)

# functions to minimize 

# This function returns a vector with 10 values

rba = function(par){
  v <- matrix(ncol = 1, nrow = 10)
  for (p in 1:10){
    k<- ifelse (par[1] < 1-1/p & par[1]>0 & p > par[2] & 
                  par[2]>0 & par[2]<2, par[2]*p,
                ifelse(par[1] < 1-1/par[2] & par[1] > 0 & 
                         p < par[2] & par[2]>0 & par[2]<2, -1+(par[1]+1/par[2]),
                       ifelse(par[1] > (1 - 1 / max(p,par[2])) & 
                                par[2]>0 & par[2]<2, -1+p, "error")))
    #I am adding a line here so that you know why the optim failed
    if(k=='error') stop('your ifelse function returned an error')
    v[p] <- k 
  }
  return(v)
}

# This function uses the function rba, and returns a value

kvasum = function(par){
  sum( (log(vpk$V1)/log(1/n) - rba(par) )^2)
}

# what I would I to do is to find par[1] and par[2] such that kvasum is minimized

m1 = optim(par=c(0.1,0.4),kvasum, method='L-BFGS-B', lower= c(0.1,0.1), upper=c(0.9,0.9))

输出:

> m1
$par
[1] 0.1 0.1

$value
[1] 171.5774

$counts
function gradient 
       2        2 

$convergence
[1] 0

$message
[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"