拟合R中的3参数Weibull分布

Question

我一直在R中进行一些数据分析，并试图找出如何将我的数据拟合成3参数威布尔分布。我发现如何使用2参数Weibull来做到这一点，但是却发现如何用3参数来做到这一点。

这里是我如何配合使用fitdistr从MASS包（）函数的数据：

y <- fitdistr(x[[6]], 'weibull') 

x[[6]]是我的数据的一个子集，y是我在哪里存储拟合的结果。

Answer 1

首先，你可能想看看FAdist package。然而，事实并非如此用心地去从rweibull3到rweibull：

> rweibull3 
function (n, shape, scale = 1, thres = 0) 
thres + rweibull(n, shape, scale) 
<environment: namespace:FAdist> 

从 dweibull3

和类似dweibull

> dweibull3 
function (x, shape, scale = 1, thres = 0, log = FALSE) 
dweibull(x - thres, shape, scale, log) 
<environment: namespace:FAdist> 

，所以我们有这个

> 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) 

编辑：作为在评论中提到，当试图符合分配时会出现各种警告离子这样

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 

对我起初它只是NaNs produced，而不是当我看到它，所以我认为它不是那么有意义，因为估计是很好的第一次。经过一番搜索，似乎是相当流行的问题，我找不到原因和解决办法。一种替代方案可能是使用stats4包和mle()函数，但它似乎也有一些问题。但我可以为您提供通过danielmedic使用的code修改后的版本，我已经检查了几次：

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