重复测量ANOVA：ezANOVA与aov vs. lme语法

Question

This question is both about syntax and semantics, thus please find a (yet unanswered) duplicate on Cross-Validated: https://stats.stackexchange.com/questions/113324/repeated-measures-anova-ezanova-vs-aov-vs-lme-syntax

In the machine-learning domain, I evaluated 4 classifiers on the same 5 datasets, ie each classifier returned a performance measure for dataset 1, 2, 3, ... and 5. Now I want to know whether the classifiers differ significantly in their performance. Here's some toy data:

Performance<-c(2,3,3,2,3,1,2,2,1,1,3,1,3,2,3,2,1,2,1,2)
Dataset<-factor(c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5))
Classifier<-factor(c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4))
data<-data.frame(Classifier,Dataset,Performance)

Following a textbook, I conducted a repeated-measures one-way ANOVA. I interpreted my performance as dependent variable, the classifiers as subjects and the datasets as within-subjects factor. Using aov, I got:

model <- aov(Performance ~ Classifier + Error(factor(Dataset)), data=data)
summary(model)

Yielding the following output:

Error: factor(Dataset)
           Df Sum Sq Mean Sq F value Pr(>F)
Residuals  4    2.5   0.625               

Error: Within
            Df Sum Sq Mean Sq F value Pr(>F)  
Classifier  3    5.2  1.7333   4.837 0.0197 *
Residuals  12    4.3  0.3583 

I get similar results when using a linear mixed-effects model:

model <- lme(Performance ~ Classifier, random = ~1|Dataset/Classifier,data=data)
result<-anova(model)

I then tried to reproduce the results with ezANOVA in order to perform Mauchlys test for Sphericity:

 ezANOVA(data=data, dv=.(Performance), wid=.(Classifier), within=.(Dataset), detailed=TRUE, type=3)

Yielding the following output:

        Effect DFn DFd  SSn SSd         F          p p<.05       ges
 1 (Intercept)   1   3 80.0 5.2 46.153846 0.00652049     * 0.8938547
 2     Dataset   4  12  2.5 4.3  1.744186 0.20497686       0.2083333

This clearly doesn't correspond to the prior output with aov/lme. Nevertheless, when I exchange "Performance" with "Classifier" in the ezANOVA definition, I get the expected results.

I now wonder whether my textbook is wrong (aov definition) or if I misunderstood the ezANOVA syntax. Furthermore, why do I only get Mauchly's test results when rewriting the ezANOVA statement, but not in the first case?

Answer 1

Since you want to compare classifiers and not datasets, the within factor is classifier and the within ID is dataset. So the correct syntax for your ezANOVA example would be:

ezANOVA(data=data, dv=.(Performance), within=.(Classifier), wid=.(Dataset), detailed=TRUE)

Btw, there is no need to specifiy the type of sums of squares. Since you have only one factor all types of sums of squares will produce the same results anyway.