similar to: MASS fitdistr call in plyr help!

Displaying 20 results from an estimated 2000 matches similar to: "MASS fitdistr call in plyr help!"

2011 Apr 27
3
MASS fitdistr with plyr or data.table?
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))
2008 Sep 19
0
Re lative Novice ? "Can I get some explanation of the docs for fitdistr(MASS)?"
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",
2008 Sep 22
0
Warnings in fitdistr() from MASS.
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
2008 Sep 22
0
Re lative novice: Working with fitdistr(MASS): 3 questions
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
2003 Nov 27
2
MASS fitdistr()
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") :
2008 Oct 30
1
Is possible, on biological grounds, suggest to fitdistr (MASS library) that the estimated parameters must be between two values?
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
2011 May 03
3
fitting distributions using fitdistr (MASS)
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
2010 Jan 03
6
Help with function "fitdistr" in "MASS"
Hi, R users: I want to fit my data into a normal distribution by using the command "fitdistr" in "MASS". I changed my data class from "ts" to "numeric" by >class(mydata)="numeric" but after using "fitdistr", I got the result below >fitdistr(mydata,"normal") mean sd NA NA (NA) (NA) the help doc of
2004 Oct 27
1
Warning messages in function fitdistr (library:MASS)
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
2002 Jan 29
1
fitdistr() in MASS library
Prof Brian Ripley wrote: > > Even better, use the function fitdistr in package MASS... > I had a look in my MASS library (from the package VR_6.2-6) and couldn't find this function. Is there a newer version available? Thanks for any help. Jim This email message and any accompanying attachments may contain confidential information. If you are not the intended recipient, do not
2013 Oct 28
0
"Optimization fail" error from fitdistr (Weibull distribution)
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
2011 Oct 28
1
weibull fitdistr problem: optimization failed
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
2003 Jul 28
1
Optimization failed in fitting mixture 3-parameter Weibull distri bution using fitdistr()
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$convergence > 0) stop("optimization failed"). The procedures I tested are as following: >w3den <- function(x, a,b,c)
2010 Jan 28
4
Problems with fitdistr
Hi, I want to estimate parameters of weibull distribution. For this, I am using fitdistr() function in MASS package.But when I give fitdistr(c,"weibull") I get a Error as follows:- Error in optim(x = c(4L, 41L, 20L, 6L, 12L, 6L, 7L, 13L, 2L, 8L, 22L, : non-finite value supplied by optim Any help or suggestions are most welcomed -- View this message in context:
2008 Feb 09
1
Problem with fitdistr function while estimating parameters
Hello, I am using fitdistr function for parameter estimation. When I use fd<-fitdistr(V2,"gamma") I get following error: Error in optim(x = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, : initial value in 'vmmin' is not finite fd<-fitdistr(V2,"weibull") Error in optim(x = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, :
2003 Jul 25
5
named list 'start' in fitdistr
Hi R lovers! I'd like to know how to use the parameter 'start' in the function fitdistr() obviously I have to provide the initial value of the parameter to optimize except in the case of a certain set of given distribution Indeed according to the help file for fitdistr " For the following named distributions, reasonable starting values will be computed if `start'
2011 Jun 17
3
rle on large data . . . without a for loop!
I think need to do something like this: dat<-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000, replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000)) rle.dat<-rle(dat$state) temp<-1 out<-data.frame(id=1:length(rle.dat$length)) for(i in 1:length(rle.dat$length)){ temp2<-temp+rle.dat$length[[i]] out$V1[i]<-mean(dat$V1[temp:temp2])
2004 Feb 17
2
problem with fitdistr ?
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
2007 Sep 09
1
fitdistr()
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
2003 Jul 04
1
Problem with fitdistr for beta
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