similar to: dweibull retuns NaN instead of Inf (PR#9080)

Displaying 20 results from an estimated 1000 matches similar to: "dweibull retuns NaN instead of Inf (PR#9080)"

2012 Feb 21
5
help error: In dweibull(x, shape, scale, log) : NaNs produzidos
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
2008 Aug 21
1
pnmath compilation failure; dylib issue?
(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after
2000 Jan 27
1
error in dweibull (PR#405)
Full_Name: R. Woodrow Setzer, Jr. Version: 0.90.1 OS: linux - Redhat 6.1 Submission from: (NULL) (165.247.155.206) dweibull(0,1,1) evaluates to 0; it should be 1. Note that dweibull(.Machine$double.eps) evaluates to 1. > dweibull(.01,1,1) [1] 0.9900498 > dweibull(.00001,1,1) [1] 0.99999 > dweibull(.Machine$double.eps,1,1) [1] 1 > dweibull(0,1,1) [1] 0
2006 Jun 02
2
Problem with mle
R 2.3.0 Linux, SuSE 10.0 Hi I have two problems with mle - probably I am using it the wrong way so please let me know. I want to fit different distributions to an observed count of seeds and in the next step use AIC or BIC to identify the best distribution. But when I run the script below (which is part of my original script), I get one error message for the first call of mle: Error in
2009 Mar 17
3
R does not compile any more on FreeBSD 8.0-CURRENT
On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include
2003 Oct 20
1
Fitting a Weibull/NaNs
I'm trying to fit a Weibull distribution to some data via maximum likelihood estimation. I'm following the procedure described by Doug Bates in his "Using Open Source Software to Teach Mathematical Statistics" but I keep getting warnings about NaNs being converted to maximum positive value: > llfunc <- function (x) { -sum(dweibull(AM,shape=x[1],scale=x[2], log=TRUE))} >
2011 Mar 15
3
fitting a distribution to a ecdf plot
Dear all, I need to plot an cumulative distribution plot of a variable and then to fit a distribution to that, probably a weibull or lognormal. I have plotted the ecdf as > plot(ecdf(x)) but I haven't managed to fit the distribution. I have as well attached the data. I would appreciate if you could help me on that. Thank you. Kind regards Maria -------------- next part --------------
2005 Jul 22
1
multiplicate 2 functions
Thks for your answer, here is an exemple of what i do with the errors in french... > tmp [1] 200 150 245 125 134 345 320 450 678 > beta18 Erreur : Objet "beta18" not found //NORMAL just to show it > eta [1] 500 > func1<-function(beta18) dweibull(tmp[1],beta18,eta) > func1<-func1(beta18) * function(beta18) dweibull(tmp[2],beta18,eta) Erreur dans dweibull(tmp[1],
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))
2005 Jul 22
1
Generate a function
hi all, I need to generate a function inside a loop: tmp is an array for (i in 1:10) { func<- func * function(beta1) dweibull(tmp[i],beta1,eta) } because then i need to integrate this function on beta. I could have written this : func<-function(beta1) prod(dweibull(tmp,beta1,eta)) (with eta and beta1 set) but it is unplottable and no integrable... i could make it a bit different but
2009 Nov 20
2
How to use results of distribution fitting for further processing?
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 __________________________________________________________________
2005 Dec 28
1
NaN in R distribution functions
Dear R developers, I noticed that core R distribution functions return NaN, when parameter values are out of parameter space. I have looked in source code and found that warnings and return of NaN are done internally in C code. For dgamma.c the line 49 is: if (shape <= 0 || scale <= 0) ML_ERR_return_NAN; OK. How should this be implemented if distribution functions are written
2012 Feb 23
3
why is generating the same graph???
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
2004 Jan 14
1
estimation of lambda and gamma with std errors for a weibull model
Dear R experts, How should lambda and gamma (with std.errors) be calculated for a weibull model with age as an independent predictor? I have assumed that this can be done with survreg with e. g. (summary(survreg(Surv(time, status) ~ age, dist = 'weibull')) ) and predict.survreg with e.g. (predict(model, se.fit = T, newdata = data.frame(age = seq(50, 80, 5)) but unfortunately I'm
2011 Nov 03
1
Fit continuous distribution to truncated empirical values
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)
2005 Sep 06
2
(no subject)
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
1999 Dec 09
1
nlm() problem or MLE problem?
I am trying to do a MLE fit of the weibull to some data, which I attach. fitweibull<-function() { rt<-scan("r/rt/data2/triam1.dat") rt<-sort(rt) plot(rt,ppoints(rt)) a<-9 b<-.27 fn<-function(p) -sum( log(dweibull(rt,p[1],p[2])) ) cat("starting -log like=",fn(c(a,b)),"\n") out<-nlm(fn,p=c(a,b), hessian=TRUE)
2018 Feb 09
1
R Compilation gets stuck on Windows 64
Please note that building R on Windows is documented in "R Installation and Administration", including links to external software. Particularly there is a link to texinfo which is part of Rtools. The documentation is maintained and it is a sufficient source of information for building R on Windows. https://cran.r-project.org/doc/manuals/r-release/R-admin.html
2013 Apr 09
5
Error when using fitdist function in R
Hello everyone, I was trying to do some distribution fitting with a numerical field called Tolls. The sample size = 999 rows. Basically I assigned the Toll data to a new variable K by doing: k<-dtest$Toll After that, tried to fit a gamma distribution by doing: fitG<-fitdist(k, "gamma") Then the following messages showed (oh and I checked for empty rows before doing this):
2001 Nov 16
6
case conversion and/or string comparison
This is no doubt trivial but after searching the help files and the web, I cannot seem to find it. 1) How do I convert 'hgt' into 'HGT' in R? 2) How should I have used the help facilities to find this? At the end of the day, all I want to do is case insensitive string matching... i.e. 'if ("HGT" == 'hgt') print('this should be true')' I tried