similar to: fitting weibull distribution

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:

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 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)

I am using the MASS library function fitdistr(x, dpois, list(lambda=2)) but I get Error in optim(start, mylogfn, x = x, hessian = TRUE, ...) : Function cannot be evaluated at initial parameters In addition: There were 50 or more warnings (use warnings() to see the first 50) and all the first 50 warnings say 1: non-integer x = 1.452222 etc Can anyone tell me what I am doing

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'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

--WWm7B+u2U4 Content-Type: text/plain; charset=us-ascii Content-Description: message body text Content-Transfer-Encoding: 7bit G'day all, I believe that this is related to PR#1717 (filed under not-reproducible) which was reported for a version of R that is a quite a bit older than the ones used in for this report. But I noticed this behaviour under R 2.1.1 and R 2.2.0 on my linux box and

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-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]]

Hi all, I'm using the function "fitdistr" (library MASS) to fit a distribution to given data. What I have to do further, is getting the log-Likelihood-Value from this estimation. Is there any simple possibility to realize it? Regards, Carsten

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:

Hi,Dear all R experts, As far as I know, fitdistr() is only to estimate the parameters in univariate distributions. I have a set of data (x,y) and I assume it follows a bivariate weibull distribution. Could someone tell me a function in R that is suitable for parameter estimation in multivariate cases? Thanks in advance! Cheers, YAN

I have the following problem: I have a vector x of data (0<x<=1 ) with a U-shaped histogram and try to fit a beta distribution using fitdistr. In fact, hist(rbeta(100,0.1,0.1)) looks a lot like my data. The equivalent to the example in the manual sometimes work: > a <- rbeta(100,0.1,0.1) > fitdistr(x=a, "beta", start=list(shape1=0.1,shape2=0.1))1) > shape1

Hi R-Community, I have a vector of Weibull distributed observations and I would like to estimate the parameters "shape" and "scale" of the Weibull distribution. Is there a way to do this in R? Much thanks in advance, Hagen Schm?ller -- ----------------------------------------------------------------------- Dipl.-Ing. Hagen K. Schm?ller Institut f?r Elektrische Anlagen und

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))

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'

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)

Why the warning messages (2:4)? > x <- rexp(1000,0.2) > fitdistr(x,"exponential",list(rate=1)) rate 0.219824219 (0.006951308) Warning messages: 1: one-diml optimization by Nelder-Mead is unreliable: use optimize in: optim(start, mylogfn, x = x, hessian = TRUE, ...) 2: NaNs produced in: dexp(x, 1/rate, log) 3: NaNs produced in: dexp(x, 1/rate, log) 4: NaNs

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),

This is both a specific question and a general one. First, I am running 'fitdistr' from library(MASS) and I get the following: Error in fitdistr(templist, "weibull") : optimization failed What is the cause of the error? How can I tell? Can I just catch this error, report it and move to the next set of data (eat the exception)? Thank you. Kevin