similar to: Fitdistr() versus nls()

Dear all, Sorry for bringing up an old issue: >pexp(50, 0.5) [1] 1 In some cases, pexp() gives CDF=1. I read some discussion in 2002 saying it has been patched. However it's not working in "R2.4.1Patched". Could anyone help me out? Thanks a lot, Jeann _________________________________________________________________ Fine Dining & Fancy Food. Check Out This

Hi all, Using the exponential distribution to exemplify: The dexp function is the PDF (1) and pexp is the CDF (2), that is obtained integrating the PDF. How can I get the qexp and the rexp? Considering that I have the PDF, how this two are mathematically related to the PDF? (1) ke^{-kx} (2) 1-e^{kx} Thanks in advance.

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

Hi everybody, while performing ks.test for a standard exponential distribution on samples of dimension 2500, generated everytime as new, i had this strange behaviour: >data<-rexp(2500,0.4) >ks.test(data,"pexp",0.4) One-sample Kolmogorov-Smirnov test data: data D = 0.0147, p-value = 0.6549 alternative hypothesis: two.sided >data<-rexp(2500,0.4)

Hi, >I would like to rank a time-series of data, extract the top ten data items from this series, determine the >corresponding row numbers for each value in the sample, and take a mean of these *row numbers* (not the data). >I would like to do this in R, rather than pre-process the data on the UNIX command line if possible, as I need to >calculate other statistics for the series.

Dear R People: Hope you're having a nice day. Here is a character vector: > yz [1] "pexp(3.2,rate=1)" > str(yz) chr "pexp(3.2,rate=1)" > And I would like to evaluate that vector. I tried: > eval(as.expression(yz)) [1] "pexp(3.2,rate=1)" > But that doesn't work. Any suggestions would be most welcome. I have a feeling that it's quite

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",

Hi, I'm trying fitdistr but I'm getting some errors > fitdistr(rnorm(100),"Normal") Error in fitdistr(rnorm(100), "Normal") : 'start' must be a named list > fitdistr(rnorm(100),"Normal",start=list(mean=0,sd=1)) Error in fitdistr(rnorm(100), "Normal", start = list(mean = 0, sd = 1)) : supplying pars for the Normal is not

Dear all, I have wrote some sample code that would allow me easier fit fast many distributions and check which of the fits performs better. My sample code (that you can of course execute it looks like that) distrList<-list(   "exponential",  "geometric", "log-normal",  "normal", "Poisson") fitfunction<-function(Type,x){     return

For a lark, I experimented a bit with the data from Ted Byers' recent postings. The result of fitdistr() seemed sensible, but I was bothered by the warnings about NaNs that arose. Warnings always make me nervous. Explicitly this is what I did: TXT <- "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

OK, I am now at the point where I can use fitdistr to obtain a fit of one of the standard distributions to mydata. It is quite remarkable how different the parameters are for different samples through from the same system. Clearly the system itself is not stationary. Anyway, question 1: I require a visual perspective of the fit I get. I can use hist.scott to get a hisogram (and just have to

I am trying to fit the chi-squared distribution to a set of data using the fitdistr function found in the MASS4 library, the data set is called ONES3, I have loaded it using the command ONES3<-read.table("ONES3.pdf",header=TRUE,na="NA") I print out the dataset ONES3 to the screen to make sure it has loaded Then I try to fit this data using the command

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

Hello everyone, This is Kangmin. I am trying to produce shape and scale of my wind data. My data is based on wind speed frequency with 1km/hr increment. data is described below. Windspeed (km/h) Frequency 1 351 2 147 3 317 4 378 5 527 6 667 7 865 8 970 9 987 10 907 11 905 12 642 13 1000 14 983 15 847 16 842 17 757 18 698 19 632 20 626 21 599 22 529 23 325 24 391

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)

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

Dear R experts, I am trying to use the R MASS library fitdistr() to fit the following list: k21stsList<-c(0.76697,0.57642,0.75938,0.82616,0.93706,0.77377,0.58923,0.37157,0.60796,1.00070,0.97529,0.62858,0.63504,0.68697,0.61714,0.75227,1.16390,0.66702,0.83578) as follows, library(MASS) fitdistr(k21stsList, "normal") But, I get Error in fitdistr(k21stsList, "normal") :

Dear developers, There's a small bug in print.fitdistr that can cause output to be printed twice, but only if print is called explicitly: > fit<-fitdistr(rt(1000,3),"t") There were 11 warnings (use warnings() to see them) > fit m s df -0.02181723 1.00145296 3.13723878 ( 0.03865057) ( 0.03999447) ( 0.33298377) > print(fit)

Sorry if it is a silly question, I haven't found documentation on this and I don't know if it is possible. library(MASS) ## for fitdistr library(msm) ## for dtnorm #prepare truncated normal distribution dtnorm0 <- function(x, mean, sd , log = FALSE) { dtnorm(x, mean, sd, 105, 135, log) } set.seed(1) #Generate normal distribution with the TRUE population mean (day 106 of the

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: