问题
I have been doing some data analysis in R and I am trying to figure out how to fit my data to a 3 parameter Weibull distribution. I found how to do it with a 2 parameter Weibull but have come up short in finding how to do it with a 3 parameter.
Here is how I fit the data using the fitdistr
function from the MASS
package:
y <- fitdistr(x[[6]], 'weibull')
x[[6]]
is a subset of my data and y is where I am storing the result of the fitting.
回答1:
First, you might want to look at FAdist package. However, that is not so hard to go from rweibull3
to rweibull
:
> rweibull3
function (n, shape, scale = 1, thres = 0)
thres + rweibull(n, shape, scale)
<environment: namespace:FAdist>
and similarly from dweibull3
to dweibull
> dweibull3
function (x, shape, scale = 1, thres = 0, log = FALSE)
dweibull(x - thres, shape, scale, log)
<environment: namespace:FAdist>
so we have this
> x <- rweibull3(200, shape = 3, scale = 1, thres = 100)
> fitdistr(x, function(x, shape, scale, thres)
dweibull(x-thres, shape, scale), list(shape = 0.1, scale = 1, thres = 0))
shape scale thres
2.42498383 0.85074556 100.12372297
( 0.26380861) ( 0.07235804) ( 0.06020083)
Edit: As mentioned in the comment, there appears various warnings when trying to fit the distribution in this way
Error in optim(x = c(60.7075705026659, 60.6300379017397, 60.7669410153573, :
non-finite finite-difference value [3]
There were 20 warnings (use warnings() to see them)
Error in optim(x = c(60.7075705026659, 60.6300379017397, 60.7669410153573, :
L-BFGS-B needs finite values of 'fn'
In dweibull(x, shape, scale, log) : NaNs produced
For me at first it was only NaNs produced
, and that is not the first time when I see it so I thought that it isn't so meaningful since estimates were good. After some searching it seemed to be quite popular problem and I couldn't find neither cause nor solution. One alternative could be using stats4
package and mle()
function, but it seemed to have some problems too. But I can offer you to use a modified version of code by danielmedic which I have checked a few times:
thres <- 60
x <- rweibull(200, 3, 1) + thres
EPS = sqrt(.Machine$double.eps) # "epsilon" for very small numbers
llik.weibull <- function(shape, scale, thres, x)
{
sum(dweibull(x - thres, shape, scale, log=T))
}
thetahat.weibull <- function(x)
{
if(any(x <= 0)) stop("x values must be positive")
toptim <- function(theta) -llik.weibull(theta[1], theta[2], theta[3], x)
mu = mean(log(x))
sigma2 = var(log(x))
shape.guess = 1.2 / sqrt(sigma2)
scale.guess = exp(mu + (0.572 / shape.guess))
thres.guess = 1
res = nlminb(c(shape.guess, scale.guess, thres.guess), toptim, lower=EPS)
c(shape=res$par[1], scale=res$par[2], thres=res$par[3])
}
thetahat.weibull(x)
shape scale thres
3.325556 1.021171 59.975470
回答2:
A alternative is the package "lmom". The estimative by L-moments technique
library(lmom)
thres <- 60
x <- rweibull(200, 3, 1) + thres
moments = samlmu(x, sort.data = TRUE)
log.moments <- samlmu( log(x), sort.data = TRUE )
weibull_3parml <- pelwei(moments)
weibull_3parml
zeta beta delta
59.993075 1.015128 3.246453
But I don´t know how to do some Goodness-of-fit statistics in this package or in the solution above. Others packages you can do Goodness-of-fit statistics easily. Anyway, you can use alternatives like: ks.test or chisq.test
来源:https://stackoverflow.com/questions/11817883/fitting-a-3-parameter-weibull-distribution