similar to: L-BFGS-B needs finite values of 'fn'

Hi all~ I am wondering if it is possible to alter the stopping precision for parameters estimated using the 'optim'? If it helps, I am minimizing the log-likelihood of a function using constraints (i.e. L-BFG-S). -Jeremy

Dear R-Help List, I am using optim, with method=L-BFGS-B, to maximize a likelihood inside a large simulation exercise. This runs fine for most simulated data sets, but for some reason, about 1 out of 100 times, optim will just hang. Using a dumb approach to the problem (i.e. printing the parameter values each time the function being maximized is evaluated), I tracked down when this happens,

Hi, When using method L-BFGS-B along with a parscale argument, should the lower and upper bounds provided be on the scaled or unscaled values? Thanks. Cheers, -- Seb

Hi, I am using optim with a L-BFGS-B method to minimize a function. As I've understood, the way to specify a tolerance for stopping optimization is through "factr" argument. My function, is by construction, minimal when equal to 1. I wonder if there is any way to pass this info to "optim". If not, how "factr" argument works (I am quite confused about the

Hi! I'm trying to figure out how to use optim... I get some really strange results, so I guess I got something wrong. I defined the following function which should be minimized: errorFunction <- function(localShifts,globalShift,fileName,experimentalPI,lambda) { lambda <- 1/sqrt(147) # error <- abs(errHuber(localShifts,globalShift, #

Hi all! I'm running some simulations and I need to estimate some paramaters with optim( ), in some cases optim stops with the next message: "L-BFGS-B needs finite values of 'fn' " I would like to know how to include and "if" condition when this happen, could it be something like: myfun <- optim(....) # run my function

Hi, When I run the following code, r <- c(3,4,4,3,5,4,5,9,8,11,12,13) n <- rep(15,12) x <- c(0, 1.1, 1.3, 2.0, 2.2, 2.8, 3.7, 3.9, 4.4, 4.8, 5.9, 6.8) x <- log10(x) fr <- function(c, alpha, beta) { P <- c + (1-c) * pnorm(alpha + beta * x) P <- pmax(pmin(P,1),0) -(sum(log(choose(n,r))) + sum(r * log(P)) + sum((n -r)* log(1-P))) } fit <- mle((fr), start = list(c

Dear all, when using optim() for a function that uses the parent environment, I see the following unexpected behavior: makeFn <- function(){ ??? xx <- ret <- NA ??? fn <- function(x){ ?????? if(!is.na(xx) && x==xx){ ?????????? cat("x=", xx, ", ret=", ret, " (memory)", fill=TRUE, sep="") ?????????? return(ret) ?????? } ?????? xx

On 03/05/2019 10:31, Serguei Sokol wrote: > On 02/05/2019 21:35, Florian Gerber wrote: >> Dear all, >> >> when using optim() for a function that uses the parent environment, I >> see the following unexpected behavior: >> >> makeFn <- function(){ >> ???? xx <- ret <- NA >> ???? fn <- function(x){ >> ??????? if(!is.na(xx)

Can anyone help me with lbfgsb (function for optimization)? It takes the following parameters: void lbfgsb (int n, int lmm, double *x, double *lower, double *upper, int *nbd, double *Fmin, optimfn fn, optimgr gr, int *fail, void *ex, double factr, double pgtol, int *fncount, int *grcount, int maxit, char *msg, int trace, int nREPORT); What do I put for parameter ex (11th parameter)? I looked at

Dear R users, I am trying to figure out the control parameter in "optim," especially, "fnscale" and "parscale." In the R docu., ------------------------------------------------------ fnscale An overall scaling to be applied to the value of fn and gr during optimization. If negative, turns the problem into a maximization problem. Optimization is performed on

Dear all, Firstly, I do apologize if my question is simple and posted in the wrong place but I had no reply from the R-help mailing list (maybe it is too simple!). I was wondering why parscale is set to 20 in the "wild" function example used in ?optim. This function has only one parameter and if we set parscale equal to 1 then the solution near the global minimum is not found. I

Yes, I think you are right. I was at first confused by the fact that after the optim() call, > environment(fn)$xx [1] 10 > environment(fn)$ret [1] 100.02 so not 9.999, but this could come from x being assigned the final value without calling fn. -pd > On 3 May 2019, at 11:58 , Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > > Your results below make it look like a

R CMD check tells me * checking R code for possible problems ... NOTE agexact.fit.rds: ... may be used in an incorrect context: ?optim(init, agfitfn, ...)? Warning: <anonymous>: ... may be used in an incorrect context: ?optim(init, agfitfn, ...)? Can anyone tell me what this message means? My searches haven't turned up anything useful. This is with R 2.7 and 2.9. The message

In trying to streamline various optimization functions, I would like to have a scratch pad of working data that is shared across a number of functions. These can be called from different levels within some wrapper functions for maximum likelihood and other such computations. I'm sure there are other applications that could benefit from this. Below are two approaches. One uses the <<-

Hi, in R-3.0 I get the following error when calling lsfit with more observations than variables, which seems to come from an error in the formatting of the error message (note that this was not happening in 2.15.3): > nobs <- 5; nvar <- 6; lsfit(matrix(runif(nobs*nvar), ncol=nvar), runif(nobs), intercept=FALSE) Error in sprintf(ngettext(nry, "%d response", "%d

Hi, I'm currently using the R package e1071 to train naive bayes classifiers and came across a bug: When the posterior probabilities of all classes are small, the result from the predict.naiveBayes function become NaNs. This is an issue with the treatment of the log-transformed probabilities inside the predict.naiveBayes function. Here is an example to demonstrate the problem (you might need

Optim's Nelder-Mead works correctly for this example. > optim(par=10, fn=fn, method="Nelder-Mead") x=10, ret=100.02 (memory) x=11, ret=121 (calculate) x=9, ret=81 (calculate) x=8, ret=64 (calculate) x=6, ret=36 (calculate) x=4, ret=16 (calculate) x=0, ret=0 (calculate) x=-4, ret=16 (calculate) x=-4, ret=16 (memory) x=2, ret=4 (calculate) x=-2, ret=4 (calculate) x=1, ret=1

Dear R community members and R experts I am stuck at a point and I tried with my colleagues and did not get it out. Sorry, I need your help. Here my data (just created to show the example): # generating a dataset just to show how my dataset look like, here I have x variables # x1 .........to X1000 plus ind and y ind <- c(1:100) y <- rnorm(100, 10,2) set.seed(201) P <-

Hallo All, I would like to apply a function to all permutations of variables in a dataframe (except the first). What is the best way to achieve this? I produce the permutations using: nvar <- ncol(dat) - 1 perms <- as.matrix( expand.grid(rep( list(1:0) , nvar ))[ , nvar:1] ) Thanks in advance Serguei Test-dataframe, comma-delimited: code,wav,w,area,gdp,def,pop,coast,milspend,agr