我在R中进行了一系列泊松回归,然后根据AIC对我的模型进行排名。但是我得到了这个结果:
> aictab(cand.set = Cand.models, sort = TRUE)
Model selection based on AICc :
K AICc Delta_AICc AICcWt Cum.Wt LL
Mod7 4 Inf NaN NaN NA -Inf
Mod6 3 Inf NaN NaN NA -Inf
Mod5 3 Inf NaN NaN NA -Inf
Mod4 3 Inf NaN NaN NA -Inf
Mod3 2 Inf NaN NaN NA -Inf
Mod2 2 Inf NaN NaN NA -Inf
Mod1 2 Inf NaN NaN NA -Inf
每个模型分别给出拦截结果,但没有给出AIC ...
> Cand.models[[1]]
Call: glm(formula = D ~ A, family = poisson(), data = d)
Coefficients:
(Intercept) Slope
-0.17356 0.07058
Degrees of Freedom: 251 Total (i.e. Null); 250 Residual
Null Deviance: 55.35
Residual Deviance: 54.99 AIC: Inf
当我用family = gaussian(identity)做同样的事情时,我得到了结果。当我进行泊松回归时,AIC怎么会不工作?
任何帮助将不胜感激。
答案 0 :(得分:1)
很难理解为什么在没有看到您的数据或代码的情况下获得结果(下次提示)。但AIC(c)模型选择肯定可以与泊松回归一起使用 - 下面是一个例子:
library(AICcmodavg)
# make some dummy data (taken from: http://stats.stackexchange.com/questions/11096/how-to-interpret-coefficients-in-a-poisson-regression)
treatment <- factor(rep(c(1, 2), c(43, 41)),
levels = c(1, 2),
labels = c("placebo", "treated"))
improved <- factor(rep(c(1, 2, 3, 1, 2, 3), c(29, 7, 7, 13, 7, 21)),
levels = c(1, 2, 3),
labels = c("none", "some", "marked"))
numberofdrugs <- rpois(84, 10) + 1
healthvalue <- rpois(84, 5)
y <- data.frame(healthvalue, numberofdrugs, treatment, improved)
# Model selection using AICc
# setup a list of candidate models
Cand.models <- list( )
Cand.models[[1]] <- glm(healthvalue~numberofdrugs+treatment+improved, data=y, family=poisson)
Cand.models[[2]] <- glm(healthvalue~treatment, data=y, family=poisson)
# create a vector of names to trace back models in set
Modnames <- paste("mod", 1:length(Cand.models), sep = " ")
# generate AICc table
aictab(cand.set = Cand.models, modnames = Modnames, sort = TRUE)
答案 1 :(得分:0)
确保公式中的D由整数非0值组成,否则Poisson glm LL倾向于爆炸。