Question

我一直在努力将分布拟合到我在R中的样本数据。我已经看过使用fitdist以及fitdistr函数，但我似乎遇到了两者的问题。

快速背景;我的代码输出应该是最合适的分布（从分发列表）到提供的数据，带参数。这需要在没有人工交互的情况下发生，因此比较图表不是一种选择。我想我可以将每个分布拟合到数据中，从卡方检验中绘制p值并找到具有最高p值的分布。我在样本数据的正态分布方面取得了一些成功，但是一旦我尝试拟合更复杂的东西（伽玛分布，如代码中所示），我就会遇到各种错误。我究竟做错了什么？

library(fitdistrplus) 
require(MASS) 
set.seed(1) 
testData <- rnorm(1000) 
distlist <- c("norm","unif","exp")

(z <- fitdist(testData,"gamma",start=list(rate=0.1),fix.arg=list(shape=4)))

我得到的错误示例如下：

[1]“在optim中出错（par = vstart，fn = fnobj，fix.arg = fix.arg，obs = 数据，：\ n'vmmin'中的初始值不是有限的\ n“attr（，”class“）

和

fitdist出错（testData，“gamma”，start = list（rate = 0.1），fix.arg = list（shape = 4））：函数mle无法估计参数，错误代码100

我知道我可能错误地实现了fitdist函数，但我似乎无法找到可以适应我的代码目标的简单示例。有人可以帮忙吗？

Answer 1

您正在寻找Kolmogorov-Smirnov测试。空假设是数据样本来自假设分布。

fitData <- function(data, fit="gamma", sample=0.5){
 distrib = list()
 numfit <- length(fit)
 results = matrix(0, ncol=5, nrow=numfit)

 for(i in 1:numfit){
if((fit[i] == "gamma") | 
     (fit[i] == "poisson") | 
     (fit[i] == "weibull") | 
     (fit[i] == "exponential") |
     (fit[i] == "logistic") |
     (fit[i] == "normal") | 
     (fit[i] == "geometric")
) 
  distrib[[i]] = fit[i]
else stop("Provide a valid distribution to fit data" )
 }

 # take a sample of dataset
 n = round(length(data)*sample)
 data = sample(data, size=n, replace=F)

 for(i in 1:numfit) {
  if(distrib[[i]] == "gamma") {
  gf_shape = "gamma"
  fd_g <- fitdistr(data, "gamma")
  est_shape = fd_g$estimate[[1]]
  est_rate = fd_g$estimate[[2]]

  ks = ks.test(data, "pgamma", shape=est_shape, rate=est_rate)

  # add to results
  results[i,] = c(gf_shape, est_shape, est_rate, ks$statistic, ks$p.value)
}

else if(distrib[[i]] == "poisson"){
  gf_shape = "poisson"
  fd_p <- fitdistr(data, "poisson")
  est_lambda = fd_p$estimate[[1]]

  ks = ks.test(data, "ppois", lambda=est_lambda)
  # add to results
  results[i,] = c(gf_shape, est_lambda, "NA", ks$statistic, ks$p.value)
}

else if(distrib[[i]] == "weibull"){
  gf_shape = "weibull"
  fd_w <- fitdistr(data,densfun=dweibull,start=list(scale=1,shape=2))
  est_shape = fd_w$estimate[[1]]
  est_scale = fd_w$estimate[[2]]

  ks = ks.test(data, "pweibull", shape=est_shape, scale=est_scale)
  # add to results
  results[i,] = c(gf_shape, est_shape, est_scale, ks$statistic, ks$p.value) 
}

else if(distrib[[i]] == "normal"){
  gf_shape = "normal"
  fd_n <- fitdistr(data, "normal")
  est_mean = fd_n$estimate[[1]]
  est_sd = fd_n$estimate[[2]]

  ks = ks.test(data, "pnorm", mean=est_mean, sd=est_sd)
  # add to results
  results[i,] = c(gf_shape, est_mean, est_sd, ks$statistic, ks$p.value)
}

else if(distrib[[i]] == "exponential"){
  gf_shape = "exponential"
  fd_e <- fitdistr(data, "exponential")
  est_rate = fd_e$estimate[[1]]
  ks = ks.test(data, "pexp", rate=est_rate)
  # add to results
  results[i,] = c(gf_shape, est_rate, "NA", ks$statistic, ks$p.value)
}

else if(distrib[[i]] == "logistic"){
  gf_shape = "logistic"
  fd_l <- fitdistr(data, "logistic")
  est_location = fd_l$estimate[[1]]
  est_scale = fd_l$estimate[[2]]
  ks = ks.test(data, "plogis", location=est_location, scale=est_scale)
  # add to results
  results[i,] = c(gf_shape, est_location, est_scale, ks$statistic,    ks$p.value) 
    }
  }
  results = rbind(c("distribution", "param1", "param2", "ks stat", "ks    pvalue"),   results)
  #print(results)
  return(results)
  }

应用于您的示例：

library(MASS)
set.seed(1) 
testData <- rnorm(1000) 
res = fitData(testData, fit=c("logistic","normal","exponential","poisson"),
    sample=1)
res

您不会拒绝正常的零假设。

参考：http://worldofpiggy.com/2014/02/25/automatic-distribution-fitting-r/

Answer 2

我认为错误主要是因为您的数据。如错误消息中所示，创建NaN以便该函数似乎无法获得分数（通过区分密度函数）。 [密度函数的范围是非负的，不是吗？]

使用更简单的

Method of moments代替最大似然估计，并且尽管发出警告，它仍会生成参数估计值。

library(fitdistrplus) 
require(MASS) 
set.seed(1) 
testData <- rnorm(1000) 
fitdist(testData, "gamma", method = "mme", start = list(shape = 0.1, rate = 0.1))

Fitting of the distribution ' gamma ' by matching moments 
Parameters:
           estimate
shape  0.0001268054
rate  -0.0108863200
Warning message:
In dgamma(c(-0.626453810742332, 0.183643324222082, -0.835628612410047,  :
  NaNs produced

如何将分布拟合到R中的样本数据？

2 个答案: