I adapted a selfStart function and the lower bounds are not working. The
parameter "b" is negative, whereas I would like the lower bound to be zero.
Any ideas? Thanks.
Here is my code (I am still figuring out how to easily make replicable
examples):
A<-1.75
mu<-.2
l<-2
b<-0
x<-seq(0,18,.25)
create.y<-function(x){
y<-b+A/(1+exp(4*mu/A*(l-x)+2))
return(y)
}
ys<-create.y(x)
yvec<-(rep(ys,5))*(.95+runif(length(x)*5)/10)
Trt<-factor(c(rep("A1",length(x)),rep("A2",length(x)),rep("A3",length(x)),rep("A4",length(x)),rep("A5",length(x))))
Data<-data.frame(Trt,rep(x,5),yvec)
names(Data)<-c("Trt","x","y")
NewData<-groupedData(y~x|Trt,data=Data)
powrDpltInit <-
function(mCall, LHS, data) {
xy <- sortedXyData(mCall[["x"]],LHS,data)
A.s <- max(xy$y)-min(xy$y)
mu.s <- A.s/7.5
l.s <- 0
b.s <- max(min(xy$y),0.00001)
value <- c(A.s, l.s, mu.s, b.s)
#function to optimize
func1 <- function(value) {
A.s <- value[1]
mu.s <- value[2]
l.s <- value[3]
b.s <- value[4]
y1<-rep(0,length(xy$x)) # generate vector for predicted y (y1) to evaluate
against observed y
for(cnt in 1:length(xy$x)){
y1[cnt]<- b.s+A.s/(1+exp(4*mu.s/A.s*(l.s-x[cnt])+2))} #predicting y1 for
values of y
evl<-sum((xy$y-y1)^2) #sum of squares is function to minimize
return(evl)}
#optimizing
oppar<-optim(c(A.s , mu.s , l.s , b.s),func1,method="L-BFGS-B",
lower=c(0.0001,0.0,0.0,0.0),
control=list(maxit=2000,trace=TRUE))
#saving optimized parameters
value<-c(oppar$par[1L],oppar$par[2L],oppar$par[3L],oppar$par[4L])
names(value) <- mCall[c("A","mu","l","b")]
value
}
SSpowrDplt<-selfStart(~b+A/(1+exp(4*mu/A*(l-x)+2)),initial=powrDpltInit,
parameters=c("A","mu","l","b"))
test1<-nlsList(SSpowrDplt,NewData)
coef(test1)
-----
In theory, practice and theory are the same. In practice, they are not - Albert Einstein
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Lower bounds on selfStart function not working
4 messages · Schatzi
I tested the "optim" function and that is returning non-negative parameter values (meaning they are bound by the lower limits), but I think those are the starting estimates for the nlsList model which is then finding negative values for the solution. ----- In theory, practice and theory are the same. In practice, they are not - Albert Einstein -- View this message in context: http://r.789695.n4.nabble.com/Lower-bounds-on-selfStart-function-not-working-tp3999231p4001986.html Sent from the R help mailing list archive at Nabble.com.
I ran the code again and got an error saying that the "x" was unknown. I
don't know why I hadn't seen that error before. Anyway, I made the edits to
"func1" so instead of "x", it is "xy$x."
#function to optimize
func1 <- function(value) {
A.s <- value[1]
mu.s <- value[2]
l.s <- value[3]
b.s <- value[4]
y1<-rep(0,length(xy$x)) # generate vector for predicted y (y1) to evaluate
against observed y
for(cnt in 1:length(xy$x)){
y1[cnt]<- b.s+A.s/(1+exp(4*mu.s/A.s*(l.s-xy$x[cnt])+2))} #predicting y1
for values of y
evl<-sum((xy$y-y1)^2) #sum of squares is function to minimize
return(evl)}
There is another place where there is an "x" in the selfStart function:
SSpowrDplt<-selfStart(~b+A/(1+exp(4*mu/A*(l-x)+2)),initial=powrDpltInit,
parameters=c("A","mu","l","b"))
I don't know why that is working fine or how it knows that my "x" is that
specific one. It seems that I am not fully understanding how this is
working.
-----
In theory, practice and theory are the same. In practice, they are not - Albert Einstein
--
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7 days later
I was able to solve this problem by going back to nls and obtaining the
initial parameter estimates through optim. When I used nlsList with my
dataset, it took 2 minutes to solve and was not limited by the bounds. Now I
have the bounds working and it takes 45 seconds to solve. Here is the new
code:
A<-1.75
mu<-.2
l<-2
b<-0
x<-seq(0,18,.25)
create.y<-function(x){
y<-b+A/(1+exp(4*mu/A*(l-x)+2))
return(y)
}
ys<-create.y(x)
yvec<-(rep(ys,5))*(.9+runif(length(x)*5)/5)
Trt<-factor(c(rep("A1",length(x)),rep("A2",length(x)),rep("A3",length(x)),rep("A4",length(x)),rep("A5",length(x))))
Data<-data.frame(Trt,rep(x,5),yvec)
names(Data)<-c("Trt","x","y")
NewData<-groupedData(y~x|Trt,data=Data)
ids<-levels(factor(xy$Trt))
output<-matrix(0,length(ids),4)
modeltest<- function(A,mu,l,b,x){
out<-vector(length=length(x))
for (i in 1:length(x)) {
out[i]<-b+A/(1+exp(4*mu/A*(l-x[i])+2))
}
return(out)
}
lower.bound<-list(A=.01,mu=0,l=0,b=0)
for (i in 1:length(ids)){
xy<-subset(NewData,Trt==ids[i])
y<-xy$y
x<-xy$x
A.s <- max(xy$y)-min(xy$y)
mu.s <- A.s/7.5
l.s <- 0
b.s <- max(min(xy$y),0.00001)
value <- c(A.s, l.s, mu.s, b.s)
#function to optimize
func1 <- function(value) {
A.s <- value[1]
mu.s <- value[2]
l.s <- value[3]
b.s <- value[4]
y1<-rep(0,length(xy$x)) # generate vector for predicted y (y1) to evaluate
against observed y
for(cnt in 1:length(xy$x)){
y1[cnt]<- b.s+A.s/(1+exp(4*mu.s/A.s*(l.s-x[cnt])+2))} #predicting y1 for
values of y
evl<-sum((xy$y-y1)^2) #sum of squares is function to minimize
return(evl)}
#optimizing
oppar<-optim(c(A.s , mu.s , l.s , b.s),func1,method="L-BFGS-B",
lower=c(0.0001,0.0,0.0,0.0),
control=list(maxit=2000))
#saving optimized parameters
value<-c(oppar$par[1L],oppar$par[2L],oppar$par[3L],oppar$par[4L])
names(value) <- c("A","mu","l","b")
try(nmodel<-nls(y~modeltest(A,mu,l,b,x),data=xy,
start=value,
lower=lower.bound,
algorithm="port")
)
coefv<-coef(nmodel)
output[i,]<-coefv
}
-----
In theory, practice and theory are the same. In practice, they are not - Albert Einstein
--
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