Re: [R] "Help On optim"
Hi
I cannot help you with optim as it is beyond my expertise, but here are some
notes.
Your mail is partially scrammbled due to HTML formating, please use plain text.
You say that you do not get right results, how they differ from your
expectation?
Preferably provide some toy data as illustration.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of niharika
> singhal
> Sent: Thursday, August 10, 2017 11:54 AM
> To: r-help@r-project.org
> Subject: [R] "Help On optim"
>
> Hello,
>
>
>
> I have some parameters from Mclust function. The parameters are in the form
>
>
>
> *parametersDf *
>
>
>
> * mu_1 mu_2 var_mc1var_mc2c1
> c2*
>
> *2 1.357283 2.962736 0.4661540.1320129 0.5258975
> 0.4741025 *
>
> *21 8.357283 9.962736 0.4661540.1320129 0.5258975
> 0.4741025 *
>
>
>
> Each row in the above data frame represent the segment and each segment
> contain two-mixture component. (Latter I will increase the segment and the
> mixture component)
>
>
>
> So for every segment with n mixture component I have to find the
>
>
>
> *max f(x) with –infinity < x < infinity*
>
>
>
> *f(x)=sum_i c_i N(O,mu_i,sigma_i). *
>
>
>
> *Since i need to calculate the derivate f’ and set f’(x)=0 I thought of using
> Newton method *
>
>
>
> *In R the function I selected for my problem is optim.*
>
>
>
> *Below is my code *
>
>
>
>
>
> *l = 2 # represent the number of mixture component*
>
> *i=1# represent the segment 1 if i=2 represent the 2 segment*
>
> *parametersDf #which I have posted above*
>
> *xnorm= function(x)*
>
> *{ *
>
>
>
> * #observation= append(segment1, segment2)*
>
> * mu.vector = parametersDf[1:l] *
>
> * var.vector =parametersDf[(l+1):((2*l))]*
>
> * coeff.vector=parametersDf[(2*l+1):(3*l)]*
>
> * val_df=data.frame()*
>
> * sum_prob=0*
>
> * for (j in 1:l){*
>
> *
> val=(coeff.vector[i,j]*(dnorm(x,mu.vector[i,j],sqrt(var.vector[i,j]*
>
> *sum_prob=sum_prob+val*
>
> * } *
>
> * return(sum_prob)*
>
> *}*
>
> *#segment1*
>
>
>
> *val1=optim(parametersDf$mean_mc_1[1],** fn=xnorm,*
>
> * method = "BFGS")*
>
>
>
> *I am not getting right result from optim function. Is there something I can
> do?*
>
>
> *Thanks in Advance for the help *
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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modifica
[R] "Help On optim"
Hello,
I have some parameters from Mclust function. The parameters are in the form
*parametersDf *
* mu_1 mu_2 var_mc1var_mc2c1
c2*
*2 1.357283 2.962736 0.4661540.1320129 0.5258975
0.4741025 *
*21 8.357283 9.962736 0.4661540.1320129 0.5258975
0.4741025 *
Each row in the above data frame represent the segment and each segment
contain two-mixture component. (Latter I will increase the segment and the
mixture component)
So for every segment with n mixture component I have to find the
*max f(x) with –infinity < x < infinity*
*f(x)=sum_i c_i N(O,mu_i,sigma_i). *
*Since i need to calculate the derivate f’ and set f’(x)=0 I thought of
using Newton method *
*In R the function I selected for my problem is optim.*
*Below is my code *
*l = 2 # represent the number of mixture component*
*i=1# represent the segment 1 if i=2 represent the 2 segment*
*parametersDf #which I have posted above*
*xnorm= function(x)*
*{ *
* #observation= append(segment1, segment2)*
* mu.vector = parametersDf[1:l] *
* var.vector =parametersDf[(l+1):((2*l))]*
* coeff.vector=parametersDf[(2*l+1):(3*l)]*
* val_df=data.frame()*
* sum_prob=0*
* for (j in 1:l){*
*
val=(coeff.vector[i,j]*(dnorm(x,mu.vector[i,j],sqrt(var.vector[i,j]*
*sum_prob=sum_prob+val*
* } *
* return(sum_prob)*
*}*
*#segment1*
*val1=optim(parametersDf$mean_mc_1[1],** fn=xnorm,*
* method = "BFGS")*
*I am not getting right result from optim function. Is there something I
can do?*
*Thanks in Advance for the help *
[[alternative HTML version deleted]]
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Re: [R] Help with optim function in R, please?
I haven't tried your code, but I've seen far too many attempts like this where
there is no simple call to the function with starting parameters to see if a
sensible answer is returned. If you get NaN or Inf or ... that isn't sensible,
then you know it is not a good idea to try further.
Beyond this, you don't have gradients. I think I would try nmkb (an updated
Nelder-Mead style code that allows for bounds) from dfoptim package. Note that
you cannot start on the bounds with this code. And it won't be terribly
efficient, but it
will -- if you ask for intermediate results -- give you some idea about your
function. If everything then working reasonably, you can try L-BFGS-B, possibly
with "better" starting parameters. However, I'm fairly certain you will find
something to fix before then.
It really does pay to be very pedestrian in using optimization software. I've
been
at this game (and note I wrote 3 of the 5 routines in optim(), which now all
have
replacements!) and I am constantly reminded in my own work to ALWAYS run some
simple
tests with the objective function and, if I have them, the gradients.
JN
On 2017-08-06 11:11 AM, fadeh Alanazi wrote:
Hi all,
Many thank in advance for helping me. I tried to fit Expectation Maximization
algorithm for mixture data. I must used one of numerical method to maximize my
function.
I built my code but I do not know how to make the optim function run over a
different value of the parameters. That is,
For E-step I need to get the value of mixture weights based on the current
(initial) values of the parameter of the density. Then, multiple the weight by
the logliklihood function and maximize it (M-step)
Then, I would like to take the new values of the parameter (from M-step) and
plug it in the weight, to get a new value of the weight. Then, iterate till
converges.
I tried the following code, but it does not work.
library(copula)
library(VineCopula)
## to generate the data
set.seed(123)
cp <- mixCopula(list(frankCopula(4),claytonCopula(2)))
cop <- rCopula(100,cp)
x <- pobs(cop) ## this is the data
## my function including optim function
myfunc <- function(data,copula=list(frankCopula(4,dim=2),
claytonCopula(0.5,dim=2)),maxit=200){
# copula[[1]]@parameters <- par[1]
# copula[[2]]@parameters <- par[2]
optim_1 <- function(par, data.=data, copula.=copula){
copula[[1]]@parameters <- par[1]
copula[[2]]@parameters <- par[2]
tau_1 <- par[3]*dCopula(data,copula[[1]],log = F)
tau_2 <- par[4]*dCopula(data,copula[[2]],log=F)
Tau_1 <- tau_1 / sum(tau_1 + tau_2)
Tau_2 <- tau_2 / sum(tau_1 + tau_2)
up_pi1 <- sum(Tau_1)/ 100
up_pi2 <- sum(Tau_2) / 100
# Tau <- c(Tau_1, Tau_2)
ll <- (-sum(Tau_1*dCopula(data,copula[[1]],log=T)))
#ll_1 <- -sum(Tau_2*dCopula(data,copula[[2]],log=T))
if (is.finite(ll)) {
return(ll)
} else {
if (is.na(ll)) {
message(par)
}
message(ll)
return(-3^305)
}
}
xy <- optim(par=c(2,0.5,0.2,0.2),fn=optim_1,method="L-BFGS-B",control = list(maxit= 200, trace=1),lower= c(copula[[1]]@param.lowbnd, copula[[2]]@param.lowbnd,0,0), upper = c(copula[[1]]@param.upbnd, copula[[2]]@param.upbnd,1,1))
return(xy$par)
}
Any help please?
Regards,
Fadhah
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Help with optim function in R, please?
Hi all,
Many thank in advance for helping me. I tried to fit Expectation Maximization
algorithm for mixture data. I must used one of numerical method to maximize my
function.
I built my code but I do not know how to make the optim function run over a
different value of the parameters. That is,
For E-step I need to get the value of mixture weights based on the current
(initial) values of the parameter of the density. Then, multiple the weight by
the logliklihood function and maximize it (M-step)
Then, I would like to take the new values of the parameter (from M-step) and
plug it in the weight, to get a new value of the weight. Then, iterate till
converges.
I tried the following code, but it does not work.
library(copula)
library(VineCopula)
## to generate the data
set.seed(123)
cp <- mixCopula(list(frankCopula(4),claytonCopula(2)))
cop <- rCopula(100,cp)
x <- pobs(cop) ## this is the data
## my function including optim function
myfunc <- function(data,copula=list(frankCopula(4,dim=2),
claytonCopula(0.5,dim=2)),maxit=200){
# copula[[1]]@parameters <- par[1]
# copula[[2]]@parameters <- par[2]
optim_1 <- function(par, data.=data, copula.=copula){
copula[[1]]@parameters <- par[1]
copula[[2]]@parameters <- par[2]
tau_1 <- par[3]*dCopula(data,copula[[1]],log = F)
tau_2 <- par[4]*dCopula(data,copula[[2]],log=F)
Tau_1 <- tau_1 / sum(tau_1 + tau_2)
Tau_2 <- tau_2 / sum(tau_1 + tau_2)
up_pi1 <- sum(Tau_1)/ 100
up_pi2 <- sum(Tau_2) / 100
# Tau <- c(Tau_1, Tau_2)
ll <- (-sum(Tau_1*dCopula(data,copula[[1]],log=T)))
#ll_1 <- -sum(Tau_2*dCopula(data,copula[[2]],log=T))
if (is.finite(ll)) {
return(ll)
} else {
if (is.na(ll)) {
message(par)
}
message(ll)
return(-3^305)
}
}
xy <- optim(par=c(2,0.5,0.2,0.2),fn=optim_1,method="L-BFGS-B",control =
list(maxit= 200, trace=1),lower= c(copula[[1]]@param.lowbnd,
copula[[2]]@param.lowbnd,0,0), upper = c(copula[[1]]@param.upbnd,
copula[[2]]@param.upbnd,1,1))
return(xy$par)
}
Any help please?
Regards,
Fadhah
[[alternative HTML version deleted]]
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Re: [R] Help with optim() to maximize log-likelihood
1) It helps to include the require statements for those of us who work
outside your particular box.
lme4 and (as far as I can guess) fastGHQuad
are needed.
2) Most nonlinear functions have domains where they cannot be
evaluated. I'd be richer than Warren Buffett if I got $5 for
each time someone said your optimizer doesn't work and I
found f(start, ...) was NaN or Inf, as in this case, i.e.,
start - c(plogis(sum(Y/m)),log(sigma2H))
cat(starting params:)
print(start)
tryf0 - ll(start,Y,m)
print(tryf0)
It really is worthwhile actually computing your function at the initial
parameters EVERY time. (Or turn on the trace etc.)
JN
On 15-03-10 07:00 AM, r-help-requ...@r-project.org wrote:
Message: 12
Date: Mon, 9 Mar 2015 16:18:06 +0200
From: Sophia Kyriakou sophia.kyriako...@gmail.com
To: r-help@r-project.org
Subject: [R] Help with optim() to maximize log-likelihood
Message-ID:
cao4ga+qokumhozwvbu7ey3xabbo2lnqjrcwxqkzchm3u9oz...@mail.gmail.com
Content-Type: text/plain; charset=UTF-8
hello, I am using the optim function to maximize the log likelihood of a
generalized linear mixed model and I am trying to replicate glmer's
estimated components. If I set both the sample and subject size to q=m=100
I replicate glmer's results for the random intercept model with parameters
beta=-1 and sigma^2=1. But if I change beta to 2 glmer works and optim
gives me the error message function cannot be evaluated at initial
parameters.
If anyone could please help?
Thanks
# likelihood function
ll - function(x,Y,m){
beta - x[1]
psi - x[2]
q - length(Y)
p - 20
rule20 - gaussHermiteData(p)
wStar - exp(rule20$x * rule20$x + log(rule20$w))
# Integrate over(-Inf, +Inf) using adaptive Gauss-Hermite quadrature
g - function(alpha, beta, psi, y, m) {-y+m*exp(alpha + beta)/(1 +
exp(alpha + beta)) + alpha/exp(psi)}
DDfLik - deriv(expression(-y+m*exp(alpha + beta)/(1 + exp(alpha + beta))
+ alpha/exp(psi)),
namevec = alpha, func = TRUE,function.arg = c(alpha, beta, psi,
y, m))
int0 - rep(NA,q)
piYc_ir - matrix(NA,q,p)
for (i in 1:q){
muHat - uniroot(g, c(-10, 10),extendInt =yes, beta = beta, psi = psi, y
= Y[i], m = m)$root
jHat - attr(DDfLik(alpha = muHat, beta, psi, Y[i], m), gradient)
sigmaHat - 1/sqrt(jHat)
z - muHat + sqrt(2) * sigmaHat * rule20$x
piYc_ir[i,] -
choose(m,Y[i])*exp(Y[i]*(z+beta))*exp(-z^2/(2*exp(psi)))/((1+exp(z+beta))^m*sqrt(2*pi*exp(psi)))
int0[i] - sqrt(2)*sigmaHat*sum(wStar*piYc_ir[i,])
}
ll - -sum(log(int0))
ll
}
beta - 2
sigma2 - 1
m - 100
q - 100
cl - seq.int(q)
tot - rep(m,q)
set.seed(123)
alpha - rnorm(q, 0, sqrt(sigma2))
Y - rbinom(q,m,plogis(alpha+beta))
dat - data.frame(y = Y, tot = tot, cl = cl)
f1 - glmer(cbind(y, tot - y) ~ 1 + (1 | cl), data = dat,family =
binomial(),nAGQ = 20)
betaH - summary(f1)$coefficients[1]
sigma2H - as.numeric(summary(f1)$varcor)
thetaglmer - c(betaH,sigma2H)
logL - function(x) ll(x,Y,m)
thetaMLb - optim(c(plogis(sum(Y/m)),log(sigma2H)),fn=logL)$par
Error in optim(c(plogis(sum(Y/m)), log(sigma2H)), fn = logL) : function
cannot be evaluated at initial parameters
thetaglmer
[1] 2.1128529 0.8311484
(thetaML - c(thetaMLb[1],exp(thetaMLb[2])))
[[alternative HTML version deleted]]
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[R] Help with optim() to maximize log-likelihood
hello, I am using the optim function to maximize the log likelihood of a
generalized linear mixed model and I am trying to replicate glmer's
estimated components. If I set both the sample and subject size to q=m=100
I replicate glmer's results for the random intercept model with parameters
beta=-1 and sigma^2=1. But if I change beta to 2 glmer works and optim
gives me the error message function cannot be evaluated at initial
parameters.
If anyone could please help?
Thanks
# likelihood function
ll - function(x,Y,m){
beta - x[1]
psi - x[2]
q - length(Y)
p - 20
rule20 - gaussHermiteData(p)
wStar - exp(rule20$x * rule20$x + log(rule20$w))
# Integrate over(-Inf, +Inf) using adaptive Gauss-Hermite quadrature
g - function(alpha, beta, psi, y, m) {-y+m*exp(alpha + beta)/(1 +
exp(alpha + beta)) + alpha/exp(psi)}
DDfLik - deriv(expression(-y+m*exp(alpha + beta)/(1 + exp(alpha + beta))
+ alpha/exp(psi)),
namevec = alpha, func = TRUE,function.arg = c(alpha, beta, psi,
y, m))
int0 - rep(NA,q)
piYc_ir - matrix(NA,q,p)
for (i in 1:q){
muHat - uniroot(g, c(-10, 10),extendInt =yes, beta = beta, psi = psi, y
= Y[i], m = m)$root
jHat - attr(DDfLik(alpha = muHat, beta, psi, Y[i], m), gradient)
sigmaHat - 1/sqrt(jHat)
z - muHat + sqrt(2) * sigmaHat * rule20$x
piYc_ir[i,] -
choose(m,Y[i])*exp(Y[i]*(z+beta))*exp(-z^2/(2*exp(psi)))/((1+exp(z+beta))^m*sqrt(2*pi*exp(psi)))
int0[i] - sqrt(2)*sigmaHat*sum(wStar*piYc_ir[i,])
}
ll - -sum(log(int0))
ll
}
beta - 2
sigma2 - 1
m - 100
q - 100
cl - seq.int(q)
tot - rep(m,q)
set.seed(123)
alpha - rnorm(q, 0, sqrt(sigma2))
Y - rbinom(q,m,plogis(alpha+beta))
dat - data.frame(y = Y, tot = tot, cl = cl)
f1 - glmer(cbind(y, tot - y) ~ 1 + (1 | cl), data = dat,family =
binomial(),nAGQ = 20)
betaH - summary(f1)$coefficients[1]
sigma2H - as.numeric(summary(f1)$varcor)
thetaglmer - c(betaH,sigma2H)
logL - function(x) ll(x,Y,m)
thetaMLb - optim(c(plogis(sum(Y/m)),log(sigma2H)),fn=logL)$par
Error in optim(c(plogis(sum(Y/m)), log(sigma2H)), fn = logL) : function
cannot be evaluated at initial parameters
thetaglmer
[1] 2.1128529 0.8311484
(thetaML - c(thetaMLb[1],exp(thetaMLb[2])))
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with optim() to maximize log-likelihood
Sophia Kyriakou sophia.kyriakou17 at gmail.com writes: hello, I am using the optim function to maximize the log likelihood of a generalized linear mixed model and I am trying to replicate glmer's estimated components. If I set both the sample and subject size to q=m=100 I replicate glmer's results for the random intercept model with parameters beta=-1 and sigma^2=1. But if I change beta to 2 glmer works and optim gives me the error message function cannot be evaluated at initial parameters. If anyone could please help? Thanks snip to make gmane happy. It looks like you're getting floating-point under/overflow. If you do all the computations on the log scale first and then exponentiate, it seems to work, i.e.: piYc_ir[i,] - lchoose(m,Y[i]) + Y[i]*(z+beta) + (-z^2/(2*exp(psi))) - m*(log1p(exp(z+beta))) - 0.5*(log(2*pi)+psi) piYc_ir[i,] - exp(piYc_ir[i,]) follow-ups should probably go to r-sig-mixed-mod...@r-project.org instead ... Ben Bolker __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with optim() to maximize log-likelihood
yes Ben, this works indeed! Thanks a million!! On Mon, Mar 9, 2015 at 7:17 PM, Ben Bolker bbol...@gmail.com wrote: Sophia Kyriakou sophia.kyriakou17 at gmail.com writes: hello, I am using the optim function to maximize the log likelihood of a generalized linear mixed model and I am trying to replicate glmer's estimated components. If I set both the sample and subject size to q=m=100 I replicate glmer's results for the random intercept model with parameters beta=-1 and sigma^2=1. But if I change beta to 2 glmer works and optim gives me the error message function cannot be evaluated at initial parameters. If anyone could please help? Thanks snip to make gmane happy. It looks like you're getting floating-point under/overflow. If you do all the computations on the log scale first and then exponentiate, it seems to work, i.e.: piYc_ir[i,] - lchoose(m,Y[i]) + Y[i]*(z+beta) + (-z^2/(2*exp(psi))) - m*(log1p(exp(z+beta))) - 0.5*(log(2*pi)+psi) piYc_ir[i,] - exp(piYc_ir[i,]) follow-ups should probably go to r-sig-mixed-mod...@r-project.org instead ... Ben Bolker __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help using optim function
tmuman mumantariq at gmail.com writes:
Hi, am very new to R and I've written an optim function, but can't
get it to work
least.squares.fitter-function(start.params,gr,
low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
{
result-optim(par=start.params,
method=c('Nelder-Mead'),
fn=least.squares.fit,
lower=low.constraints,
upper=high.constraints,data=data,scale=scale,ploton=ploton)
return(result)
}
least.squares.fitter(c(2,2),c(0,0),c(Inf,Inf),ricker.one.step,dat,'linear')
where least.squares.fit is the function returning the difference between
model data points and predicted points (residuals, residuals^2 and SSR),
which i am trying to minimise
and ricker.one.step is the function that generated the predictions
I've been trying to figure out whats wrong for hours but can't.
You haven't given us much to work with here -- have you
read the posting guide (URL at the bottom of every message)?
It's not reproducible ( http://tinyurl.com/reproducible-000 ),
and you haven't told us what's going wrong. Are you getting
error messages? What are they? Warnings? No errors but
you have reason to believe the answer is wrong?
A few comments:
* 'scale' and 'ploton' are not arguments to optim(),
so they'd better be accepted by your least.squares.fit() function
* the Nelder-Mead algorithm doesn't accept constraints (lower/upper)
(see the nloptr package for a Nelder-Mead implementation that does)
* except for the constraints, you might be better off with ?nls
if all you want to do is least-squares fitting; nlsLM in the
minpack.lm package does least-squares fitting with constraints
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] help using optim function
Hi, am very new to R and I've written an optim function, but can't get it to
work
least.squares.fitter-function(start.params,gr,low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
{
result-optim(par=start.params,method=c('Nelder-Mead'),fn=least.squares.fit,lower=low.constraints,upper=high.constraints,data=data,scale=scale,ploton=ploton)
return(result)
}
least.squares.fitter(c(2,2),c(0,0),c(Inf,Inf),ricker.one.step,dat,'linear')
where least.squares.fit is the function returning the difference between
model data points and predicted points (residuals, residuals^2 and SSR),
which i am trying to minimise
and ricker.one.step is the function that generated the predictions
I've been trying to figure out whats wrong for hours but can't.
Many thanks,
Tariq
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
