similar to: Problem with Newton_Raphson

Hello, When i run the code below from Weibull distribution with 30% censoring by using optim i get an error form R, which states that Error in optim(start, fn = z, data = q, hessian = T) :? ? objective function in optim evaluates to length 25 not 1 can somebody?help me remove this error. Is my censoring approach correct. n=25;rr=1000 p=1.5;b=1.2 for (i in 1:rr){ q<-c(t,cen)

Your function is giving NaN's during the optimization. The R-forge version of optimx() has functionality specifically intended to deal with this. NOTE: the CRAN version does not, and the R-forge version still has some glitches! However, I easily ran the code you supplied by changing optim to optimx in the penultimate line. Here's the final output. KKT condition testing Number of

Hello, I wish to?censor 10% of my sample units of 50 from a Weibull distribution. Below is the code for it. I will need to know whether what i have done is correct and if not, can i have any suggestion to improve it? Thank you ?p=2;b=120 n=50 r=45 t<-rweibull(r,shape=p,scale=b) meantrue<-gamma(1+(1/p))*b meantrue cen<- runif(n-r,min=0,max=meantrue) cen Chris Guure Researcher,

Hello, I want to estimate the exponential parameter by using?optim?with the following input, where t contains 40% of the data and q contains 60% of the data within an interval. In implementing the code command for optim i want it to contain both the t and q data so i can obtain the correct estimate. Is there any suggestion as to how this can be done. I have tried h<-c(t,q) but it is not working

Hi , I am trying to get some estimator based on lognormal distribution when we have left,interval, and right censored data. Since, there is now avalible pakage in R can help me in this, I had to write my own code using Newton Raphson method which requires first and second derivative of log likelihood but my problem after runing the code is the estimators were too high. with this email ,I provide

Dear All, Can you help me, with the code below how do I obtain the fisher information from it. Is my q<-replicate(1000,x) the right way to do simulation. thank you. x<-rweibull(100,0.8,1.5) q<-replicate(1000,x) z<-function(p){ beta<-p[1] eta<-p[2] log1<-(n*log(beta)-n*beta*log(eta)+(beta-1)*sum(log(x))-sum((x/eta)^beta)) return(-log1) } zz<-optim(c(0.5,0.5),z) zz Chris

Dear all, I am a newcomer to R programming, I met the problem: I have a lot of .txt files in my directory. Firstly, I check whether the file satisfies the conditions: 1.empty 2.the "Rep" column of the file has no "useractivity_idle" or "useractivity_act" 3.even The "rep" has both of them, numbers of "useractivity_idle"==numbers of

Dear All, I have many files in my directory. I want to transfer each data into one which is readable. They have so many possibilities, i have collected(manually and visually) all possibilities and represent them as different numbers. Rep[grep('context_log',log1$Remain[1:length(log1$Date)]),]<-"2" Rep[grep('gs',log1$Remain[1:length(log1$Date)]),]<-"5"

Dear all, I am a newcomer to R programming, I met the problem: I have a lot of .txt files in my directory. Firstly, I check whether the file satisfies the conditions: 1.empty 2.the "Rep" column of the file has no "useractivity_idle" or "useractivity_act" 3.even The "rep" has both of them, numbers of "useractivity_idle"==numbers of

I have not included the previous postings because they came out very strangely on my mail reader. However, the question concerned the choice of minimizer for the zeroinfl() function, which apparently allows any of the current 6 methods of optim() for this purpose. The original poster wanted to use Newton-Raphson. Newton-Raphson (or just Newton for simplicity) is commonly thought to be the

hello list, I'm trying to write a script that will search through a directory of trace logs for an oracle database. From what I understand new files are always being created in the directory and it's not possible to know the exact names of the files before they are created. The purpose of this is to create service checks in nagios. Because you don't know the names of the files ahead

Dear R-devel During the last half a year I have several times encountered the following problem with optim() when using method= "L-BFGS-B". The function return a value which is clearly not the maximum (seen from printing the value each time the function is called). Some output is shown below. A few things I have observed (as I remember it): a. The problem seems to occur when the

Hi, I'm analyzing a large number of large simulation datasets, and I've isolated one of the bottlenecks. Any help in speeding it up would be appreciated. `dat` is a dataframe of samples from a regular grid. The first two columns are the spatial coordinates of the samples, the remaining 20 columns are the abundances of species in each cell. I need to calculate the species richness in

Hi I have the following formula (I hope it is clear - if no, I can try to do better the next time) h(x, a, b) = integral(0 to pi/2) ( ( integral(D/sin(alpha) to Inf) ( ( f(x, a, b) ) dx ) dalpha ) and I want to do an mle with it. I know how to use mle() and I also know about integrate(). My problem is to give the parameter values a and b to the

Dear list, i am a student of psychology and have to do a multilevelanalysis on some data. About that i have one general and one specific question. This is what i have copied from the help-file on lme: data(bdf) fm <- lme(langPOST ~ IQ.ver.cen + avg.IQ.ver.cen, data = bdf, random = ~ IQ.ver.cen | schoolNR) summary(fm) after summary(fm) i get the following error:

I'm trying to calculate the maximum likelihood estimate for a binomial distribution. Here is my code: y <- c(2, 4, 2, 4, 5, 3) n <- length(y) binomial.ll <- function (pi, y, n) { ## define log-likelihood output <- y*log(pi)+(n-y)*(log(1-pi)) return(output) } binomial.mle <- optim(0.01, ## starting value binomial.ll,

Hi, there: It's my first time to post question in this forum, so thanks for your tolerance if my question is too naive. I am using a nonparametric smoothing procedure in sm package to generate smoothed survival curves for continuous covariate. I want to truncate the suvival curve and only display the part with covariate value between 0 and 7. The following is the code I wrote:

Hello all, I am trying to use stepwise procedure to select covariates in Cox model and use bootstrap to repeat stepwise selection, then record how many times variables are chosen by step() in bootstrap replications. When I use step() (or stepAIC) to do model selection, I got errors. Here is the part of my code for (j in 1:mm){ #<--mm=10 for (b in 1:nrow(reg.bs)){ #<--bootstrap 10

Hi all, I was using the Newton-Raphson method to estimate paremeters in the model developed by my supervisor. However, when I interatively computed theta(t+1)=theta(t) - solve(H)*s (where the Hessian matrix and score vector were explicitely derived), I got the error message: Error in solve.default(H) : system is computationally singular: reciprocal condition number = 1.70568e-032. Assume my score

Hi all - (this is posted to r-help and R-SIG-MAC) OSX 10.3.7, R 2.0.1, lme4/Matrix/latticeExtra latest, fresh install of R. MASS loaded (or not). I am getting an error message for the print() and summary() commands with all lme models I try and run in lme4 (GLMM's work fine). Using the example from the lme help, summary and print produce the following errors, despite the model being