Yes, your function out<- epiann(f = function(a,b) log(dnorm(a)*dnorm(b))),
N = 10) works well.
But why you are changing the loglikelihood function to f = function(a,b)
log(dnorm(a)*dnorm(b))? how it is equivalent to loglikelihood? is there any
mathematical relation? I also want to see the plot of aout and bout versus
loglikelihood, and see the cooling rate in different temperature. how is
this possible?
On Mon, Dec 5, 2011 at 6:07 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> If you run
>
> out<- epiann(f = function(a,b) log(dnorm(a)*dnorm(b))), N = 10)
>
> It takes less than 0.5 seconds so there's no problem I can see:
> perhaps you want to look elsewhere to get better speed (like Rcpp or
> general vectorization), or maybe your loglikihood is not what's
> desired, but there's no problem with the loop.
>
> Michael
>
> On Mon, Dec 5, 2011 at 5:29 PM, Gyanendra Pokharel
> <gyanendra.pokha...@gmail.com> wrote:
> > Yes, I checked the acceptprob, it is very high but in my view, the while
> > loop is not stopping, so there is some thing wrong in the use of while
> loop.
> > When I removed the while loop, it returned some thing but not the result
> > what I want. When i run the while loop separately, it never stops.
> >
> >
> >
> > On Mon, Dec 5, 2011 at 5:18 PM, R. Michael Weylandt
> > <michael.weyla...@gmail.com> wrote:
> >>
> >> Your code is not reproducible nor minimal, but why don't you put a
> >> command print(acceptprob) in and see if you are getting reasonable
> >> values. If these values are extremely low it shouldn't surprise you
> >> that your loop takes a long time to run.
> >>
> >> More generally, read up on the use of print() and browser() as debugging
> >> tools.
> >>
> >> Michael
> >>
> >> On Mon, Dec 5, 2011 at 3:47 PM, Gyanendra Pokharel
> >> <gyanendra.pokha...@gmail.com> wrote:
> >> > I forgot to upload the R-code in last email, so heare is one
> >> >
> >> > epiann <- function(T0 = 1, N=1000, ainit=1, binit=1,rho = 0.99, amean
> =
> >> > 3,
> >> > bmean=1.6, avar =.1, bvar=.1, f){
> >> >
> >> > moving <- 1
> >> > count <- 0
> >> > Temp <- T0
> >> > aout <- ainit
> >> > bout <- binit
> >> > while(moving > 0){
> >> > moving <- 0
> >> > for (i in 1:N) {
> >> > aprop <- rnorm (1,amean, avar)
> >> > bprop <- rnorm (1,bmean, bvar)
> >> > if (aprop > 0 & bprop > 0){
> >> > acceptprob <- min(1,exp((f(aout, bout) -
> >> > f(aprop,bprop))/Temp))
> >> > u <- runif(1)
> >> > if(u<acceptprob){
> >> > moving <- moving +1
> >> > aout <- aprop
> >> > bout <- bprop
> >> > }
> >> > else{aprob <- aout
> >> > bprob <- bout}
> >> > }
> >> > }
> >> > Temp <- Temp*rho
> >> > count <- count +1
> >> >
> >> > }
> >> > fmin <- f(aout,bout)
> >> > return(c(aout, bout,fmin, count) )
> >> >
> >> > }
> >> > out<- epiann(f = loglikelihood)
> >> >
> >> > On Mon, Dec 5, 2011 at 3:46 PM, Gyanendra Pokharel <
> >> > gyanendra.pokha...@gmail.com> wrote:
> >> >
> >> >> Hi all,
> >> >> I have the following code,
> >> >> When I run the code, it never terminate this is because of the while
> >> >> loop
> >> >> i am using. In general, if you need a loop for which you don't know
> in
> >> >> advance how many iterations there will be, you can use the `while'
> >> >> statement so here too i don't know the number how many iterations are
> >> >> there. So Can some one suggest me whats going on?
> >> >> I am using the Metropolis simulated annealing algorithm
> >> >> Best
> >> >>
> >> >
> >> > [[alternative HTML version deleted]]
> >> >
> >> > ______________________________________________
> >> > R-help@r-project.org mailing list
> >> > https://stat.ethz.ch/mailman/listinfo/r-help
> >> > PLEASE do read the posting guide
> >> > http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> >> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
>
[[alternative HTML version deleted]]
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