similar to: "Optimization fail" error from fitdistr (Weibull distribution)

Hello, How can I do a test of two weibull distributions? I have two weibull distribution sets from two wind datasets in order to check whether they are same. I thought 2 sample t-test would be applicable but I couldn't find any ways to do that on the Internet. Does anyone know what type of test is applicable to my purpose? and what R function can you recommend? Plus, if it turned out that

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]]

Hello. I am not certain even how to search the archives for this particular question, so if there is an obvious answer, please smack me with a large halibut and send me to the URLs. I have been experimenting with fitting curves by using both maximum likelihood and maximum spacing estimation techniques. Originally, I have been writing distribution-specific functions in 'R' which work

In the docs I see: Usage fitdistr(x, densfun, start, ...) Arguments x A numeric vector. densfun Either a character string or a function returning a density evaluated at its first argument. Distributions "beta", "cauchy", "chi-squared", "exponential", "f", "gamma", "geometric", "log-normal", "lognormal",

Dear R developers, I have encountered strange behaviour of fitdistr for gamma in recent R build i.e. 2.2.0. I have attached the code for data at the end of this mail so you can reproduce the problem. In short, I am able to run fitdistr under 2.1.0 without problems, while I get the following error under 2.2.0 (Version 2.2.0 Patched (2005-11-15 r36348)) > fitdistr(otm, "gamma") Error

Dear R-users I've got the next problem: I've got this *function*: fitcond=function(x,densfun,pcorte,start,...){ myfn <- function(parm,x,pcorte,...) -sum(log(dens(parm,x,pcorte,...))) Call <- match.call(expand.dots = TRUE) if (missing(start)) start <- NULL dots <- names(list(...)) dots <- dots[!is.element(dots, c("upper",

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

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

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'm trying to subset a pattern in a vector. Each argument has 6 letters, and I need those that start with Z and end with Z. e.g. x <- c("ZFHSJK", "ZFHJKZ","ZIOPWE","ZLKJSD","ZKFLPZ") I've looked up other discussions but still can't seem to find the answer. Thanks. Kangmin

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, I have some data, that when plotted looks very close to a log-normal distribution. My goal is to build a regression model to test how this variable responds to several independent variables. To do this, I want to use the fitdistr tool from the MASS package to see how well my data fits the actual distribution, and also build a generalized linear model using the glm command. The summary

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

Good-day Sir, I am R.Language users but am try to? estimate parameter of beta distribution particular dataset but give this error, which is not clear to me: (Initial value in "vmmin" is not finite) beta.fit <- fitdistr(data,densfun=dbeta,shape1=value , shape2=value) kindly assist. expecting your reply:

This is probably simple, but I have a hard time finding the solution. Any help greatly appreciated.   I would like to use the results of fitdistr(z,densfun=dweibull,start=list(scale=1,shape=1)) for further processing.  How do I assign the values of scale and shape to b and a without manually entering the numbers?   TIA __________________________________________________________________

my problem actually arised with fitting the data to the weibulldistribution, where it is hard to see, if the proposed parameterestimates make sense. data1:2743;4678;21427;6194;10286;1505;12811;2161;6853;2625;14542;694;11491; ?? ?? ?? ?? ?? 14924;28640;17097;2136;5308;3477;91301;11488;3860;64114;14334 how am I supposed to know what starting values i have to take? i get different

Dear all, I would like to estimate the parameters of a mixture of two beta binomial distributions. I saw the function fitdistr from the MASS library could be a solution but I have some difficulties. Does anyone know how to specify to fitdistr another function as these which are predefined? Or is there any other solution to estimate the parameters? Thank you in advance Caroline

Hi dear, I want to estimate d.f for Chi-squared distribution: est.chi[i,]<-c( fitdistr(as.numeric(data2[,i]),"chi-squared",start=list(df=1))$estimate)Warning message:In optim(x = c(7.86755, 7.50852, 7.86342, 7.70589, 7.70153, 7.58272, : one-diml optimization by Nelder-Mead is unreliable: use "Brent" or optimize() directly Who can help me to solve this problem? Best

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am I misusing lapply? As a second solution I tried splitting only the numeric data column to yield a

Hi R lovers Here is a numerical vector test > test [1] 206 53 124 112 92 77 118 75 48 176 90 74 107 126 99 84 114 147 99 114 99 84 99 99 99 99 99 104 1 159 100 53 [33] 132 82 85 106 136 99 110 82 99 99 89 107 99 68 130 99 99 110 99 95 153 93 136 51 103 95 99 72 99 50 110 37 [65] 102 104 92 90 94 99 76 81 109 91 98 96 104 104 93 99 125 89