在 R 中拟合零膨胀泊松分布

发布于 2024-11-30 22:32:36 字数 373 浏览 0 评论 0原文

我有一个严重过度分散且零膨胀的计数数据向量。

该向量如下所示：

i.vec=c(0,63,1,4,1,44,2,2,1,0,1,0,0,0,0,1,0,0,3,0,0,2,0,0,0,0,0,2,0,0,0,0,
0,0,0,0,0,0,0,0,6,1,11,1,1,0,0,0,2)
m=mean(i.vec)
# 3.040816
sig=sd(i.vec)
# 10.86078

我想对此进行拟合，我强烈怀疑该分布将是零膨胀泊松（ZIP）。但我需要执行显着性检验来证明 ZIP 分布适合数据。

如果我有正态分布，我可以使用 vcd 包中的函数 goodfit() 进行卡方拟合优度检验，但我不知道可以对零膨胀数据执行任何测试。

原文

I have a vector of count data that is strongly over dispersed and zero inflated.

The vector looks like this:

i.vec=c(0,63,1,4,1,44,2,2,1,0,1,0,0,0,0,1,0,0,3,0,0,2,0,0,0,0,0,2,0,0,0,0,
0,0,0,0,0,0,0,0,6,1,11,1,1,0,0,0,2)
m=mean(i.vec)
# 3.040816
sig=sd(i.vec)
# 10.86078

I would like to fit a distribution to this, which I strongly suspect will be a zero inflated poisson (ZIP). But I need to perform a significance test to demonstrate that a ZIP distribution fits the data.

If I had a normal distribution, I could do a chi square goodness of fit test using the function goodfit() in the package vcd, but I don't know of any tests that I can perform for zero inflated data.

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记忆消瘦 2024-12-07 22:32:36

这是一种方法

# LOAD LIBRARIES
library(fitdistrplus)    # fits distributions using maximum likelihood
library(gamlss)          # defines pdf, cdf of ZIP


# FIT DISTRIBUTION (mu = mean of poisson, sigma = P(X = 0)
fit_zip = fitdist(i.vec, 'ZIP', start = list(mu = 2, sigma = 0.5))

# VISUALIZE TEST AND COMPUTE GOODNESS OF FIT    
plot(fit_zip)
gofstat(fit_zip, print.test = T)

基于此，ZIP 看起来并不合适。

Here is one approach

# LOAD LIBRARIES
library(fitdistrplus)    # fits distributions using maximum likelihood
library(gamlss)          # defines pdf, cdf of ZIP


# FIT DISTRIBUTION (mu = mean of poisson, sigma = P(X = 0)
fit_zip = fitdist(i.vec, 'ZIP', start = list(mu = 2, sigma = 0.5))

# VISUALIZE TEST AND COMPUTE GOODNESS OF FIT    
plot(fit_zip)
gofstat(fit_zip, print.test = T)

Based on this, it does not look like ZIP is a good fit.

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