Fitting distributions
On Fri, 7 Sep 2001, kjetil halvorsen wrote:
Hola! I think the original posters problem was to fit a weibull. this could be done with: library(gnlm)
x <- rweibull(100, 2, 3) y <- rep(1,100)
fit.dist(x,y,"Weibull")
Weibull distribution, n = 100
mean variance alpha.hat mu.hat
2.438586 1.840581 1.890002 6.782797
-log likelihood AIC
-296.4451 -294.4451
yi ni pi.hat pi.tilde like.comp resid
1 1.8744688 1 0.01 0.300583890 -3.403142 -5.300156
From the help page again:
breaks: If TRUE, `y' contains breaks between categories instead of
mid-points.
so according the help page it is assuming grouped observations
specified by midpoints. That is not the same problem, as I did say.
Who knows if the help page is right?
Kjetil Halvorsen Prof Brian Ripley wrote:
On Thu, 6 Sep 2001, kjetil halvorsen wrote:
Hola! I think some of this is avaliable as fit.dist in Jim Lindsays package gnlm.
Not according to his help page:
\name{fit.dist}
\title{Fit Probability Distributions to Frequency Data}
Not the same problem.
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272860 (secr)
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Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272860 (secr) Oxford OX1 3TG, UK Fax: +44 1865 272595 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._