我应该什么时候使用 aov() 什么时候使用 anova()？
When should I use aov() and when anova()?
我参考了很多网络文献，但它增加了我的困惑。很多讨论都过于技术化，涉及术语不平衡设计和 I、II 或 III 因子方差分析等等。
我只知道aov()内部使用了lm()，对有因子的数据有用。而 anova() 可用于同一数据集上的不同模型。
我的理解正确吗？
anova 与 aov 有本质区别。为什么不阅读 R 的文档 ?aov 和 ?anova？简而言之：
aov 拟合模型（如您所知，它在内部调用 lm），因此它产生回归系数、拟合值、残差等；它产生一个主要 class "aov" 的对象，但也产生一个次要的 class "lm" 对象。因此，它是 "lm" 对象的扩充。anova 是一个通用函数。在您的场景中，您指的是 anova.lm 或 anova.lmlist（阅读 ?anova.lm 了解更多信息）。前者分析一个拟合模型（由 lm 或 aov 生成），而后者分析几个嵌套（越来越大）的拟合模型（由 lm 或 aov 生成）。他们都旨在生产 I 型（顺序）ANOVA table.
在实践中，您首先使用 lm / aov 来拟合模型，然后使用 anova 来分析结果。没有什么比尝试一个小例子更好的了：
fit <- aov(sr ~ ., data = LifeCycleSavings) ## can also use `lm`
z <- anova(fit)
现在，看看它们的结构。 aov returns 大对象：
str(fit)
#List of 12
# $ coefficients : Named num [1:5] 28.566087 -0.461193 -1.691498 -0.000337 0.409695
# ..- attr(*, "names")= chr [1:5] "(Intercept)" "pop15" "pop75" "dpi" ...
# $ residuals : Named num [1:50] 0.864 0.616 2.219 -0.698 3.553 ...
# ..- attr(*, "names")= chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# $ effects : Named num [1:50] -68.38 -14.29 7.3 -3.52 -7.94 ...
# ..- attr(*, "names")= chr [1:50] "(Intercept)" "pop15" "pop75" "dpi" ...
# $ rank : int 5
# $ fitted.values: Named num [1:50] 10.57 11.45 10.95 6.45 9.33 ...
# ..- attr(*, "names")= chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# $ assign : int [1:5] 0 1 2 3 4
# $ qr :List of 5
# ..$ qr : num [1:50, 1:5] -7.071 0.141 0.141 0.141 0.141 ...
# .. ..- attr(*, "dimnames")=List of 2
# .. .. ..$ : chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# .. .. ..$ : chr [1:5] "(Intercept)" "pop15" "pop75" "dpi" ...
# .. ..- attr(*, "assign")= int [1:5] 0 1 2 3 4
# ..$ qraux: num [1:5] 1.14 1.17 1.16 1.15 1.05
# ..$ pivot: int [1:5] 1 2 3 4 5
# ..$ tol : num 1e-07
# ..$ rank : int 5
# ..- attr(*, "class")= chr "qr"
# $ df.residual : int 45
# $ xlevels : Named list()
# $ call : language aov(formula = sr ~ ., data = LifeCycleSavings)
# $ terms :Classes 'terms', 'formula' language sr ~ pop15 + pop75 + dpi + ddpi
# .. ..- attr(*, "variables")= language list(sr, pop15, pop75, dpi, ddpi)
# .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
# .. .. ..- attr(*, "dimnames")=List of 2
# .. .. .. ..$ : chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# .. .. .. ..$ : chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. ..- attr(*, "term.labels")= chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. ..- attr(*, "order")= int [1:4] 1 1 1 1
# .. ..- attr(*, "intercept")= int 1
# .. ..- attr(*, "response")= int 1
# .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
# .. ..- attr(*, "predvars")= language list(sr, pop15, pop75, dpi, ddpi)
# .. ..- attr(*, "dataClasses")= Named chr [1:5] "numeric" "numeric" "numeric" "numeric" ...
# .. .. ..- attr(*, "names")= chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# $ model :'data.frame': 50 obs. of 5 variables:
# ..$ sr : num [1:50] 11.43 12.07 13.17 5.75 12.88 ...
# ..$ pop15: num [1:50] 29.4 23.3 23.8 41.9 42.2 ...
# ..$ pop75: num [1:50] 2.87 4.41 4.43 1.67 0.83 2.85 1.34 0.67 1.06 1.14 ...
# ..$ dpi : num [1:50] 2330 1508 2108 189 728 ...
# ..$ ddpi : num [1:50] 2.87 3.93 3.82 0.22 4.56 2.43 2.67 6.51 3.08 2.8 ...
# ..- attr(*, "terms")=Classes 'terms', 'formula' language sr ~ pop15 + pop75 + dpi + ddpi
# .. .. ..- attr(*, "variables")= language list(sr, pop15, pop75, dpi, ddpi)
# .. .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
# .. .. .. ..- attr(*, "dimnames")=List of 2
# .. .. .. .. ..$ : chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# .. .. .. .. ..$ : chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. .. ..- attr(*, "term.labels")= chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. .. ..- attr(*, "order")= int [1:4] 1 1 1 1
# .. .. ..- attr(*, "intercept")= int 1
# .. .. ..- attr(*, "response")= int 1
# .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
# .. .. ..- attr(*, "predvars")= language list(sr, pop15, pop75, dpi, ddpi)
# .. .. ..- attr(*, "dataClasses")= Named chr [1:5] "numeric" "numeric" "numeric" "numeric" ...
# .. .. .. ..- attr(*, "names")= chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# - attr(*, "class")= chr [1:2] "aov" "lm"
同时 anova returns:
str(z)
#Classes ‘anova’ and 'data.frame': 5 obs. of 5 variables:
# $ Df : int 1 1 1 1 45
# $ Sum Sq : num 204.1 53.3 12.4 63.1 650.7
# $ Mean Sq: num 204.1 53.3 12.4 63.1 14.5
# $ F value: num 14.116 3.689 0.858 4.36 NA
# $ Pr(>F) : num 0.000492 0.061125 0.359355 0.042471 NA
# - attr(*, "heading")= chr "Analysis of Variance Table\n" "Response: sr"
我参考了很多网络文献，但它增加了我的困惑。很多讨论都过于技术化，涉及术语不平衡设计和 I、II 或 III 因子方差分析等等。
我只知道aov()内部使用了lm()，对有因子的数据有用。而 anova() 可用于同一数据集上的不同模型。
我的理解正确吗？
anova 与 aov 有本质区别。为什么不阅读 R 的文档 ?aov 和 ?anova？简而言之：
aov拟合模型（如您所知，它在内部调用lm），因此它产生回归系数、拟合值、残差等；它产生一个主要 class "aov" 的对象，但也产生一个次要的 class "lm" 对象。因此，它是 "lm" 对象的扩充。anova是一个通用函数。在您的场景中，您指的是anova.lm或anova.lmlist（阅读?anova.lm了解更多信息）。前者分析一个拟合模型（由lm或aov生成），而后者分析几个嵌套（越来越大）的拟合模型（由lm或aov生成）。他们都旨在生产 I 型（顺序）ANOVA table.
在实践中，您首先使用 lm / aov 来拟合模型，然后使用 anova 来分析结果。没有什么比尝试一个小例子更好的了：
fit <- aov(sr ~ ., data = LifeCycleSavings) ## can also use `lm`
z <- anova(fit)
现在，看看它们的结构。 aov returns 大对象：
str(fit)
#List of 12
# $ coefficients : Named num [1:5] 28.566087 -0.461193 -1.691498 -0.000337 0.409695
# ..- attr(*, "names")= chr [1:5] "(Intercept)" "pop15" "pop75" "dpi" ...
# $ residuals : Named num [1:50] 0.864 0.616 2.219 -0.698 3.553 ...
# ..- attr(*, "names")= chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# $ effects : Named num [1:50] -68.38 -14.29 7.3 -3.52 -7.94 ...
# ..- attr(*, "names")= chr [1:50] "(Intercept)" "pop15" "pop75" "dpi" ...
# $ rank : int 5
# $ fitted.values: Named num [1:50] 10.57 11.45 10.95 6.45 9.33 ...
# ..- attr(*, "names")= chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# $ assign : int [1:5] 0 1 2 3 4
# $ qr :List of 5
# ..$ qr : num [1:50, 1:5] -7.071 0.141 0.141 0.141 0.141 ...
# .. ..- attr(*, "dimnames")=List of 2
# .. .. ..$ : chr [1:50] "Australia" "Austria" "Belgium" "Bolivia" ...
# .. .. ..$ : chr [1:5] "(Intercept)" "pop15" "pop75" "dpi" ...
# .. ..- attr(*, "assign")= int [1:5] 0 1 2 3 4
# ..$ qraux: num [1:5] 1.14 1.17 1.16 1.15 1.05
# ..$ pivot: int [1:5] 1 2 3 4 5
# ..$ tol : num 1e-07
# ..$ rank : int 5
# ..- attr(*, "class")= chr "qr"
# $ df.residual : int 45
# $ xlevels : Named list()
# $ call : language aov(formula = sr ~ ., data = LifeCycleSavings)
# $ terms :Classes 'terms', 'formula' language sr ~ pop15 + pop75 + dpi + ddpi
# .. ..- attr(*, "variables")= language list(sr, pop15, pop75, dpi, ddpi)
# .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
# .. .. ..- attr(*, "dimnames")=List of 2
# .. .. .. ..$ : chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# .. .. .. ..$ : chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. ..- attr(*, "term.labels")= chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. ..- attr(*, "order")= int [1:4] 1 1 1 1
# .. ..- attr(*, "intercept")= int 1
# .. ..- attr(*, "response")= int 1
# .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
# .. ..- attr(*, "predvars")= language list(sr, pop15, pop75, dpi, ddpi)
# .. ..- attr(*, "dataClasses")= Named chr [1:5] "numeric" "numeric" "numeric" "numeric" ...
# .. .. ..- attr(*, "names")= chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# $ model :'data.frame': 50 obs. of 5 variables:
# ..$ sr : num [1:50] 11.43 12.07 13.17 5.75 12.88 ...
# ..$ pop15: num [1:50] 29.4 23.3 23.8 41.9 42.2 ...
# ..$ pop75: num [1:50] 2.87 4.41 4.43 1.67 0.83 2.85 1.34 0.67 1.06 1.14 ...
# ..$ dpi : num [1:50] 2330 1508 2108 189 728 ...
# ..$ ddpi : num [1:50] 2.87 3.93 3.82 0.22 4.56 2.43 2.67 6.51 3.08 2.8 ...
# ..- attr(*, "terms")=Classes 'terms', 'formula' language sr ~ pop15 + pop75 + dpi + ddpi
# .. .. ..- attr(*, "variables")= language list(sr, pop15, pop75, dpi, ddpi)
# .. .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
# .. .. .. ..- attr(*, "dimnames")=List of 2
# .. .. .. .. ..$ : chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# .. .. .. .. ..$ : chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. .. ..- attr(*, "term.labels")= chr [1:4] "pop15" "pop75" "dpi" "ddpi"
# .. .. ..- attr(*, "order")= int [1:4] 1 1 1 1
# .. .. ..- attr(*, "intercept")= int 1
# .. .. ..- attr(*, "response")= int 1
# .. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
# .. .. ..- attr(*, "predvars")= language list(sr, pop15, pop75, dpi, ddpi)
# .. .. ..- attr(*, "dataClasses")= Named chr [1:5] "numeric" "numeric" "numeric" "numeric" ...
# .. .. .. ..- attr(*, "names")= chr [1:5] "sr" "pop15" "pop75" "dpi" ...
# - attr(*, "class")= chr [1:2] "aov" "lm"
同时 anova returns:
str(z)
#Classes ‘anova’ and 'data.frame': 5 obs. of 5 variables:
# $ Df : int 1 1 1 1 45
# $ Sum Sq : num 204.1 53.3 12.4 63.1 650.7
# $ Mean Sq: num 204.1 53.3 12.4 63.1 14.5
# $ F value: num 14.116 3.689 0.858 4.36 NA
# $ Pr(>F) : num 0.000492 0.061125 0.359355 0.042471 NA
# - attr(*, "heading")= chr "Analysis of Variance Table\n" "Response: sr"