适合列表的gofstat问题

Question

当我使用fitdist列表执行gofstat() R函数时，出现错误。我不知道这个错误的含义和导致它的原因。

> gofstat(list(fitw, fitg, fitln), fitnames=c("weibull", "gamma", "lnorm"))
Error in names(Chi2$chisqpvalue) <- names(Chi2$chisqdf) <- fitnames[1] : 
  attempt to set an attribute on NULL

我使用的整个代码如下所示。我认为它与prob向量有关，因为相同的脚本和其他数据集提供了结果。

prob <- c(0.004926108, 0.510983890, 0.306590334, 0.048409000, 0.032272667, 0.005378778, 0.005378778, 0.037651445, 0.005378778, 0.037651445, 0.005378778)

fitw <- fitdist(prob, distr = "weibull", method = "mle")
fitg<- fitdist(prob, distr = "gamma", method = "mle")
fitln <- fitdist(prob, distr = "lnorm", method = "mle")
gofstat(list(fitw, fitg, fitln), fitnames=c("weibull", "gamma", "lnorm"))

对于我来说，werid部分是gofstat单独工作的部分：

> gofstat(fitw)
Goodness-of-fit statistics
                             1-mle-weibull
Kolmogorov-Smirnov statistic     0.2354052
Cramer-von Mises statistic       0.1426774
Anderson-Darling statistic       0.8750089

Goodness-of-fit criteria
                               1-mle-weibull
Akaike's Information Criterion     -33.74893
Bayesian Information Criterion     -32.95314
> gofstat(fitg)
Goodness-of-fit statistics
                             1-mle-gamma
Kolmogorov-Smirnov statistic   0.2760907
Cramer-von Mises statistic     0.1983721
Anderson-Darling statistic     1.1044287

Goodness-of-fit criteria
                               1-mle-gamma
Akaike's Information Criterion   -32.26493
Bayesian Information Criterion   -31.46914
> gofstat(fitln)
Goodness-of-fit statistics
                             1-mle-lnorm
Kolmogorov-Smirnov statistic   0.2781579
Cramer-von Mises statistic     0.1279150
Anderson-Darling statistic     0.8426833

Goodness-of-fit criteria
                               1-mle-lnorm
Akaike's Information Criterion   -36.57233
Bayesian Information Criterion   -35.77654

顺便说一句，你认为哪个分布更适合？

Answer 1

这个错误是因为它在定义chisq中断时有问题，如果你看一下你提供的例子，在0.005378778处有一个意外的峰值，这使得它非常有问题，特别是当n很小时：

table(prob)
prob
0.004926108 0.005378778 0.032272667 0.037651445    0.048409 0.306590334 
          1           4           1           2           1           1 
 0.51098389 
          1

你可以尝试下面这样的东西，但由于你最可能专注于mle，它应该是可以的：

library(fitdistrplus)

fits = lapply(fitnames,function(i){
fitdist(prob, distr = i, method = "mle")
})

 gofstat(fits,chisqbreaks=c(0,0.01,0.1,0.6))

Goodness-of-fit statistics
                             1-mle-weibull 2-mle-gamma 3-mle-lnorm
Kolmogorov-Smirnov statistic     0.2354052   0.2760907   0.2781579
Cramer-von Mises statistic       0.1426774   0.1983721   0.1279150
Anderson-Darling statistic       0.8750089   1.1044287   0.8426833

Goodness-of-fit criteria
                               1-mle-weibull 2-mle-gamma 3-mle-lnorm
Akaike's Information Criterion     -33.74893   -32.26493   -36.57233
Bayesian Information Criterion     -32.95314   -31.46914   -35.77654

再说一次，你的观察是有限的，所以chisq估计将是棘手的。你的第二个关于哪个分布更好的问题，很难分辨，因为n很小，有4个重复值。