如何在R中的两个glm模型之间执行anova（）时提取p值

时间：2012-11-05 18:06:48

标签： r model regression glm anova

所以，我试图比较两个模型，fit1和fit2。

最初，我只是在做anova（fit1，fit2），这产生了我理解的输出（包括p值）。

但是，当我将模型从基于lm（）的模型切换到基于glm（）的模型时，anova（fit1，fit2）现在产生了剩余自由度，残差和Df Deviances，我遇到了麻烦解释（解释这些指标的资源似乎很少）。我希望为两个模型之间的比较提取一个p值，但由于某种原因anova（fit1，fit2，test ='Chisq'）不起作用。有什么建议吗？

我意识到，根据我的glms中的链接功能，卡方可能不是最合适的测试，但我在适当的上下文中使用了'F'以及类似的失望。

这个问题对其他人来说是否熟悉？建议？非常感谢！

示例：

make_and_compare_models <- function(fitness_trait_name, data_frame_name, vector_for_multiple_regression, predictor_for_single_regression, fam){
        fit1<-glm(formula=as.formula(paste(fitness_trait_name,"~", paste(vector_for_multiple_regression, sep="+"))), family=fam, data=data_frame_name)
        print ("summary fit 1")
        print(summary(fit1))
        fit2<- glm(data=data_frame_name, formula=as.formula(paste(fitness_trait_name,"~",predictor_for_single_regression)), family=fam)

        print("summary fit 2")
        print(summary(fit2))
        print("model comparison stats:")
        mod_test<-anova(fit2,fit1)

        ##suggestion #1
        print(anova(fit2,fit1, test="Chisq"))

        #suggestion #2
        print ("significance:")
    print (1-pchisq( abs(mod_test$Deviance[2]),df=abs(mod_test$Df[2])))

        }


data<-structure(list(ID = c(1L, 2L, 4L, 7L, 9L, 10L, 12L, 13L, 14L, 
15L, 16L, 17L, 18L, 20L, 21L, 22L, 23L, 24L, 25L, 27L, 28L, 29L, 
31L, 34L, 37L, 38L, 39L, 40L, 41L, 43L, 44L, 45L, 46L, 47L, 48L, 
49L, 52L, 55L, 56L, 59L, 60L, 61L, 62L, 63L, 65L, 66L, 67L, 68L, 
69L, 71L), QnWeight_initial = c(158L, 165L, 137L, 150L, 153L, 
137L, 158L, 163L, 159L, 151L, 145L, 144L, 157L, 144L, 133L, 148L, 
151L, 151L, 147L, 158L, 178L, 164L, 134L, 151L, 148L, 142L, 127L, 
179L, 162L, 150L, 151L, 153L, 163L, 155L, 163L, 170L, 149L, 165L, 
128L, 134L, 145L, 147L, 148L, 160L, 131L, 155L, 169L, 143L, 123L, 
151L), Survived_eclosion = c(0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 
1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Days_wrkr_eclosion_minus20 = c(NA, 
1L, NA, 3L, 0L, 2L, 0L, 1L, 0L, 0L, 0L, 1L, NA, 0L, 7L, 1L, 0L, 
1L, 0L, 1L, 2L, 2L, NA, 2L, 3L, 2L, 2L, NA, 0L, 1L, NA, NA, 0L, 
0L, 0L, 0L, 3L, 3L, 3L, 1L, 0L, 2L, NA, 1L, 0L, 1L, 1L, 3L, 1L, 
2L), MLH = c(0.5, 0.666666667, 0.555555556, 0.25, 1, 0.5, 0.333333333, 
0.7, 0.5, 0.7, 0.5, 0.666666667, 0.375, 0.4, 0.5, 0.333333333, 
0.4, 0.375, 0.3, 0.5, 0.3, 0.2, 0.4, 0.875, 0.6, 0.4, 0.222222222, 
0.222222222, 0.6, 0.6, 0.3, 0.4, 0.714285714, 0.4, 0.3, 0.6, 
0.4, 0.7, 0.625, 0.555555556, 0.25, 0.5, 0.5, 0.6, 0.25, 0.428571429, 
0.3, 0.25, 0.375, 0.555555556), Acon5 = c(0.35387674, 0.35387674, 
0.35387674, 0.35387674, 0.35387674, 0.35387674, 0.35387674, 0, 
0, 1, 0, 1, 0.35387674, 0, 0, 0.35387674, 1, 1, 0, 0, 0, 1, 0, 
0.35387674, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 
0, 0, 1, 0, 0, 0, 1, 0, 0.35387674), Baez = c(1, 1, 1, 0.467836257, 
1, 1, 0, 0, 1, 1, 0, 0.467836257, 1, 0, 0, 0, 0, 1, 0, 0, 0, 
0, 0, 0.467836257, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 
1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1), C294 = c(0, 1, 0, 0, 1, 
0.582542694, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 
0, 1, 1, 0, 0, 0.582542694, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1), C316 = c(1, 1, 0, 0, 0.519685039, 
0.519685039, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0.519685039, 0, 
1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0.519685039, 1, 0, 1, 
1, 0, 0.519685039, 1, 0.519685039, 1, 1, 1, 0.519685039, 0.519685039, 
0, 0.519685039, 0.519685039, 0), i_120_PigTail = c(1, 1, 0, 1, 
0.631236443, 0.631236443, 1, 1, 1, 1, 1, 0, 0.631236443, 1, 1, 
1, 0, 0.631236443, 1, 1, 1, 0, 0, 1, 1, 1, 0.631236443, 0, 1, 
1, 0, 1, 0.631236443, 1, 0, 1, 0, 0, 1, 0.631236443, 0.631236443, 
0, 1, 0, 0.631236443, 0.631236443, 1, 0.631236443, 0.631236443, 
1), i129 = c(0L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 0L, 
1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 
0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 0L), Jackstraw_PigTail = c(0L, 1L, 1L, 0L, 
1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 
1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 
0L, 1L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), Neil_Young = c(0.529636711, 
0, 1, 0, 0.529636711, 0.529636711, 1, 1, 0, 1, 1, 1, 0, 0, 1, 
1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 
1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1), Ramble = c(0, 0, 0, 
0, 0.215163934, 0.215163934, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 
0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0.215163934, 0, 
0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0.215163934, 0, 0, 0, 0), Sol_18 = c(1, 
0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 
0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0.404669261, 
1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1)), .Names = c("ID", "QnWeight_initial", 
"Survived_eclosion", "Days_wrkr_eclosion_minus20", "MLH", "Acon5", 
"Baez", "C294", "C316", "i_120_PigTail", "i129", "Jackstraw_PigTail", 
"Neil_Young", "Ramble", "Sol_18"), class = "data.frame", row.names = c(NA, 
-50L))

make_and_compare_models("QnWeight_initial", data, c("Acon5","Baez","C294","C316","i_120_PigTail","i129","Jackstraw_PigTail","Neil_Young","Ramble","Sol_18"), "MLH", "gaussian")

2 个答案:

答案 0 :(得分：8)

“更大”或更复杂的模型与嵌套或“缩小”模型之间的偏差差异被分布（渐近）为卡方变量，两个模型的自由度差异。因此，您将提取偏差估计值和自由度差异，并将其与pchisq（偏差，差异（df））进行比较。 “p值”只是1减去该值。

> 1-pchisq(3.84,1)
[1] 0.05004352

如果你在glm帮助页面中运行第一个例子，然后添加一个没有“treatment”变量的简化模型，你会得到：

glm.D93.o <- glm(counts ~ outcome, family=poisson())
 anova.res <-anova(glm.D93, glm.D93.o)
 anova.res
#------------
Analysis of Deviance Table

Model 1: counts ~ outcome + treatment
Model 2: counts ~ outcome
  Resid. Df Resid. Dev Df    Deviance
1         4     5.1291               
2         6     5.1291 -2 -2.6645e-15
#---------------
 str(anova.res)
Classes ‘anova’ and 'data.frame':   2 obs. of  4 variables:
 $ Resid. Df : num  4 6
 $ Resid. Dev: num  5.13 5.13
 $ Df        : num  NA -2
 $ Deviance  : num  NA -2.66e-15
 - attr(*, "heading")= chr  "Analysis of Deviance Table\n" "Model 1: counts ~ outcome + treatment\nModel 2: counts ~ outcome"

因此，在查看事物如何存储在对象本身之后，这给出了“结果”的p值：

 1-pchisq( abs(anova.res$Deviance[2]), abs(anova.res$Df[2]))
[1] 1

这将是治疗+结果模型与仅治疗模型的相应程序：

> glm.D93.t <- glm(counts ~ treatment, family=poisson())
> anova.res2 <-anova(glm.D93, glm.D93.t)
> 1-pchisq( abs(anova.res2$Deviance[2]), abs(anova.res2$Df[2]))
[1] 0.06547071

答案 1 :(得分：1)

如果您的2个模型嵌套，那么您可以使用2个模型的偏差变化来查看包含额外参数的模型是否会产生更好的拟合。如果模型1包含k个参数且模型2包含相同的k个参数以及其他m个参数，那么偏差的变化遵循（近似）卡方分布{{1 }} 自由程度。您可以使用此测试统计信息来查看模型2是否是模型1的改进。

如果您是这个领域的新手，我强烈建议您阅读有关GLM的介绍性文字

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