You are doing yourself no favors by posting HTML format email... the code
shown below has extra characters in it that you probably did not put
there. This is a plain text mailing list, so you need to send plain text
email. Read the Posting Guide mentioned in the footer of this and every
R-help posting.
On top of that, your code is not presented as a stand-alone reproducible
example. Read [1], [2], and in particular [3]. Failing to make it
reproducible means most readers will just ignore it and move on.
I cannot figure out why you have "seg" as an optimized parameter... in
particular, you are trying to index one-dimensional vectors with two
indexes, one of which is the floating point variable "seg".
Anyway, despite the above, here is my take on your problem.
####
myfunc <- function( x ) {
meanval <- c( 506.8644, 672.8448, 829.902 )
sigmaval <- c( 61.02859, 9.149168, 74.84682 )
coeffval <- c( 0.1241933, 0.6329082, 0.2428986 )
sapply( x
, function( xi )
sum( coeffval * dnorm( xi
, meanval
, sigmaval
)
)
)
}
plot( 400:1000, myfunc( 400:1000 ) )
#' 
# fooled by local maximum
val1 <- optim( par = c( x = 800 )
, fn = myfunc
, method = "BFGS"
, control = list( fnscale= -1
, parscale = 1/0.025
)
)
val1
#> $par
#> x
#> 829.5249
#>
#> $value
#> [1] 0.001294662
#>
#> $counts
#> function gradient
#> 5 4
#>
#> $convergence
#> [1] 0
#>
#> $message
#> NULL
val2 <- optim( par = c( x = 1000 )
, fn = myfunc
, method = "SANN"
, control = list( fnscale= -1
, parscale = 1/0.025
)
)
val2
#> $par
#> x
#> 672.8166
#>
#> $value
#> [1] 0.02776057
#>
#> $counts
#> function gradient
#> 10000 NA
#>
#> $convergence
#> [1] 0
#>
#> $message
#> NULL
####
[1]
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
[2] http://adv-r.had.co.nz/Reproducibility.html
[3] https://cran.r-project.org/web/packages/reprex/index.html (read the
vignette)
On Thu, 24 Aug 2017, niharika singhal wrote:
Hello,
I am facing a problem with optimization in R from 2-3 weeks.
I have some Gaussian mixtures parameters and I want to find the maximum in
that
*Parameters are in the form *
mean1 mean2 mean3 sigma1 sigma2 sigma3 c1 c2 c3
506.8644 672.8448 829.902 61.02859 9.149168 74.84682 0.1241933
0.6329082 0.2428986
I have used optima and optimx to find the maxima, but it gives me value
near by the highest mean as an output, for example 830 in the above
parameters. The code for my optim function is
val1=*optim*(par=c(X=1000, *seg=1*),fn=xnorm, opt=c(NA,seg=1),
method="BFGS",lower=-Inf,upper=+Inf, control=list(fnscale=-1))
*I am running the optim function in a loop and for different initial value
and taking the val1$par[1] as the best value.*
*I am taking this parameter since I want to run it on n number of segments
latter*
and xnorm is simply calculating the dnorm
*xnorm*=function(param, opt = rep(NA, 2)){
if (any(!sapply(opt, is.na))) {
i = !sapply(opt, is.na)
# Fix non-NA values
param[i] <- opt[i]
}
xval= param[1]
seg <- param[2]
sum_prob=0
val=0
l=3
meanval=c(506.8644, 672.8448, 829.902)
sigmaval=c(61.02859, 9.149168, 74.84682)
coeffval(0.1241933, 0.6329082, 0.2428986)
for(n in 1 :l)
{
mu=meanval[seg,n]
sg=sigmaval[seg,n]
cval=coeffval[seg,n]
val=cval*(dnorm(xval,mu,sg))
#print(paste0("The dnorm value for x is.: ", val))
sum_prob=sum_prob+val
}
sum_prob
}
The output is not correct. Since I check my data using*
UnivarMixingDistribution* from distr package and according to this the max
should lie somewhere between 600-800
Code I used to check
mc0=c( 0.6329082,0.6329082,0.2428986)
rv
<-UnivarMixingDistribution(Norm(506.8644,61.02859),Norm(672.8448,9.149168),Norm(
829.902,74.84682), mixCoeff=mc0/sum(mc0))
plot(rv, to.draw.arg="d")
Can someone please help how I can solve this problem?
Thanks & Regards
Niharika Singhal
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