similar to: Determining cause of error?

This may be a begining question. If so, please bear with me. If I have some data that based on the historgram and other plots it "looks" like a beta distribution. Is there a function or functions within R to help me determine the model parameters for such a distirbution? Similarily for other "common" distirbutions, Poisson(lambda), Chi-Square(degrees of freedom, chi-square

Hi, I want to estimate parameters of weibull distribution. For this, I am using fitdistr() function in MASS package.But when I give fitdistr(c,"weibull") I get a Error as follows:- Error in optim(x = c(4L, 41L, 20L, 6L, 12L, 6L, 7L, 13L, 2L, 8L, 22L, : non-finite value supplied by optim Any help or suggestions are most welcomed -- View this message in context:

GUYS, I NEED HELP WITH ERROR: library(MASS) > dados<-read.table("mediaRGinverno.txt",header=FALSE) > vento50<-fitdistr(dados[[1]],densfun="weibull") Erro em fitdistr(dados[[1]], densfun = "weibull") : Weibull values must be > 0 WHY RETURN THIS ERROR? WHAT CAN I DO? BEST REGARDS [[alternative HTML version deleted]]

Dear All, I have two questions regarding distribution fitting. I have several datasets, all left-truncated at x=1, that I am attempting to fit distributions to (lognormal, weibull and exponential). I had been using fitdistr in the MASS package as follows: fitdistr<-(x,"weibull") However, this does not take into consideration the truncation at x=1. I read another posting in this

I am trying to extract the shape and scale parameters of a wind speed distribution for different sites. I can do this in a clunky way, but I was hoping to find a way using data.table or plyr. However, when I try I am met with the following: set.seed(144) weib.dist<-rweibull(10000,shape=3,scale=8) weib.test<-data.table(cbind(1:10,weib.dist))

Dear All; I tried to use fitdistr() in the MASS library to fit a mixture distribution of the 3-parameter Weibull, but the optimization failed. Looking at the source code, it seems to indicate the error occurs at if (res$convergence > 0) stop("optimization failed"). The procedures I tested are as following: >w3den <- function(x, a,b,c)

Hi R lovers! I'd like to know how to use the parameter 'start' in the function fitdistr() obviously I have to provide the initial value of the parameter to optimize except in the case of a certain set of given distribution Indeed according to the help file for fitdistr " For the following named distributions, reasonable starting values will be computed if `start'

I'm getting errors when running what seems to be a simple Weibull distribution function: This works: x <- c(23,19,37,38,40,36,172,48,113,90,54,104,90,54,157,51,77,78,144,34,29,45,16,15,37,218,170,44,121) rate <- c(.01,.02,.04,.05,.1,.2,.3,.4,.5,.8,.9) year <- c(100,50,25,20,10,5,3.3,2.5,2,1.2,1.1) library(MASS) x <- sort(x) tryCatch( f<-fitdistr(x, 'weibull'), error

Hi all, I am trying to fit a distribution to some data about survival times. I am interested only in a specific interval, e.g., while the data lies in the interval (0,...., 600), I want the best for the interval (0,..., 24). I have tried both fitdistr (MASS package) and fitdist (from the fitdistrplus package), but I could not get them working, e.g. fitdistr(left, "weibull", upper=24)

Dear R users, This is a basic question. I want to fit a Weibull distribution. fitdistr(data, "weibull") works and it is a maximum likelihood fitting. Is it a good method ? Or is it better to write a function for the log-likelihood and the gradient and to use a numerical routine ? Fitdistr works for uncensored data, but what can I use for censored (and uncensored) data ? Thank you

Dear R gurus, I have the following code, but I still not know how to estimate and extract confidence intervals (95%CI) from resampling. Thanks! ~Adriana #data penta<-c(770,729,640,486,450,410,400,340,306,283,278,260,253,242,240,229,201,198,190,186,180,170,168,151,150,148,147,125,117,110,107,104,85,83,80,74,70,66,54,46,45,43,40,38,10) x<-log(penta+1) plot(ecdf(x),

Dear R-users I would like to fit weibull parameters using "Method of moments" in order to provide the inital values of the parameter to de function 'fitdistr' . I don`t have much experience with maths and I don't know how to do it. Can anyone please put me in the rigth direction? Borja [[alternative HTML version deleted]]

Is there a likelihod function for the Weibull distribution in 'R'? I found the following reference: http://www.weibull.com/LifeDataWeb/weibull_log_likelihood_functions_and_their_partials.htm But I had a hard time understanding the parameters required Particularly 'number of groups of times-to-failure data points", "number of groups of suspension data points", and

Guys, I'm having an error when I use the command: library(MASS)> dados<-read.table("inverno.txt",header=FALSE)> vento50<-fitdistr(dados[[1]],densfun="weibull")Mensagens de aviso perdidas:1: In dweibull(x, shape, scale, log) : NaNs produzidos2: In dweibull(x, shape, scale, log) : NaNs produzidos3: In dweibull(x, shape, scale, log) : NaNs produzidos4: In

I am having a hard time understanding what is happening with ifelse. Let me illustrate: h <- numeric(5) p <- 1:5 j <- floor(j) x <- 1:1000 ifelse(h == 0, x[j+2], 1:5) [1] 2 3 4 5 6 My question is, "shouldn't this be retruning 25 numbers?" It seems that the ifelse should check 5 values of h for zero. For each of the 5 values I am thinking it should return an array of 5

Dear all, I get the following error message. And I cannot quite work out what is wrong. I think the optim gets infinite values. Certainly my data do not have any infinite values. How can I solve this? fitdistr(A1, "weibull") Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : non-finite value supplied by optim I am using R version 1.9.1 on RedHat Linux, Kernel 2.6.8.

Dear R Users, I am trying to obtain a best-fit analytic distribution for a dataset with 11535459 entries. The data range in value from 1 to 300000000. I use: fitdistr(data, "gamma") to obtain mle's for the parameters. I get the following error: Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : non-finite finite-difference value [1] And the following warnings:

Please guide me through to resolve the error message that I get this is what i have done. >x1<- rnorm(100,2,1) >x1fitbeta<-fitdistr(x1,"beta") Error in fitdistr(x1, "beta") : 'start' must be a named list Yes, I do understand that sometime for the distribution to converge to the given set of data, it requires initial parameters of the distribution, to

Hello, I am a new user of R and found the function dtnorm() in the package msm. My problem now is, that it is not possible for me to get the mean and sd out of a sample when I want a left-truncated normal distribution starting at "0". fitdistr(x,dtnorm, start=list(mean=0, sd=1)) returns the error message "Fehler in "[<-"(`*tmp*`, x >= lower & x <= upper,

Hi, why my script iss always generating the same graph?when I change the parameters and the name of text file? library(MASS) dados<-read.table("inverno.txt",header=FALSE) vento50<-fitdistr(dados[[1]],densfun="weibull") png(filename="invernoRG.png",width=800,height=600) hist(dados[[1]], seq(0, 18, 0.5), prob=TRUE, xlab="Velocidade