[英]Passing parameters to nloptr objective function - R
I intend to use nloptr
package in a for
loop as below: 我打算在
for
循环中使用nloptr
包,如下所示:
for(n in 1:ncol(my.data.matrix.prod))
{
alpha.beta <- as.vector(Alpha.beta.Matrix.Init[,n])
opts = list("algorithm"="NLOPT_LN_COBYLA",
"xtol_rel"=1.0e-8, "maxeval"= 2000)
lb = vector("numeric",length= length(alpha.beta))
result <- nloptr(alpha.beta,eval_f = Error.func.oil,lb=lb,
ub = c(Inf,Inf),eval_g_ineq=Const.func.oil,
opts = opts)
Final.Alpha.beta.Matrix[,n] <- result$solution
}
Apart from passing the "optimization parameters: alpha.beta
" to the error function(minimization function) , I also would like to send n
from the for
loop. 除了将“优化参数:
alpha.beta
”传递给错误函数(最小化函数)之外,我还想从for
循环中发送n
。 Is there anyway to do this? 反正有这样做吗?
The error func is defined as: 错误函数定义为:
Error.func.oil <- function(my.data.var,n)
{
my.data.var.mat <- matrix(my.data.var,nrow = 2,ncol = ncol(my.data.matrix.prod) ,byrow = TRUE)
qo.est.matrix <- Qo.Est.func(my.data.var.mat)
diff.values <- well.oilprod-qo.est.matrix #FIND DIFFERENCE BETWEEN CAL. MATRIX AND ORIGINAL MATRIX
Error <- ((colSums ((diff.values^2), na.rm = FALSE, dims = 1))/nrow(well.oilprod))^0.5 #sum of square root of the diff
Error[n]
}
The constraint function is simple and defined as: 约束函数很简单,定义为:
Const.func.oil <- function(alpha.beta)
{
cnst <- alpha.beta[2]-1
cnst
}
So, when I run the above code, I get an error 因此,当我运行上面的代码时,出现错误
Error in .checkfunargs(eval_f, arglist, "eval_f") : eval_f requires argument 'n' but this has not been passed to the 'nloptr' function.
.checkfunargs(eval_f,arglist,“ eval_f”)中的错误:eval_f需要参数'n',但尚未传递给'nloptr'函数。
How do I pass "n" to the error function? 如何将“ n”传递给错误函数? note that "n" is not to be optimized.
注意,“ n”将不被优化。 It's just an index.
这只是一个索引。
Okay. 好的。 I read some examples online and found out that I can probably mention "n" in the definition of
nloptr
itself as: 我在网上阅读了一些示例,发现我可以在
nloptr
本身的定义中提及“ n”:
for(n in 1:ncol(my.data.matrix.prod))
{
alpha.beta <- as.vector(Alpha.beta.Matrix.Init[,n])
opts = list("algorithm"="NLOPT_LN_COBYLA",
"xtol_rel"=1.0e-8, "maxeval"= 5000)
lb = c(0,0)
result <- nloptr(alpha.beta,eval_f = Error.func.oil,lb=lb,
ub = c(Inf,Inf),
opts = opts, n=n) #Added 'n' HERE
Final.Alpha.beta.Matrix[,n] <- result$solution
}
This seems to have worked for me. 这似乎对我有用。 Hence, I am setting this as closed.
因此,我将其设置为关闭。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.