R中韦尔奇校正的单向方差分析的事后测试

这是两种方法：

数据

library(car) 
df <- structure(list(Count = c(13, 14, 14, 12, 11, 13, 14, 15, 13, 12, 20, 15, 9, 5, 13, 14, 7, 17, 18, 14, 12, 12, 13, 14, 11, 10, 15, 14, 14, 13),
                     Group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("a", "b", "c" ), class = "factor")),
                .Names = c("Count", "Group"),
                row.names = c(NA, -30L), class = "data.frame")

基数R

首先，一组唯一的Group因素对：

allPairs <- expand.grid(levels(df$Group), levels(df$Group))
## http://stackoverflow.com/questions/28574006/unique-combination-of-two-columns-in-r/28574136#28574136
allPairs <- unique(t(apply(allPairs, 1, sort)))
allPairs <- allPairs[ allPairs[,1] != allPairs[,2], ]
allPairs
##      [,1] [,2]
## [1,] "a"  "b" 
## [2,] "a"  "c" 
## [3,] "b"  "c" 

现在分析：

allResults <- apply(allPairs, 1, function(p) {
    dat <- df[ df$Group %in% p, ]
    ret <- oneway.test(Count ~ Group, data = dat, na.action = na.omit, var.equal = FALSE)
    ret$groups <- p
    ret
})
length(allResults)
## [1] 3
allResults[[1]]
##  One-way analysis of means (not assuming equal variances)
## data:  Count and Group
## F = 0.004, num df = 1.000, denom df = 10.093, p-value = 0.9508

如果您希望这是一个矩阵，则可能是这样：

mm <- diag(length(levels(df$Group)))
dimnames(mm) <- list(levels(df$Group), levels(df$Group))
pMatrix <- lapply(allResults, function(res) {
    ## not fond of out-of-scope assignment ...
    mm[res$groups[1], res$groups[2]] <<- mm[res$groups[2], res$groups[1]] <<- res$p.value
})
mm
##           a         b         c
## a 1.0000000 0.9507513 0.6342116
## b 0.9507513 1.0000000 0.8084057
## c 0.6342116 0.8084057 1.0000000

（对于F统计量，这可以轻松完成。）

使用 dplyr

首先，一组唯一的Group因素对：

library(dplyr)
## http://stackoverflow.com/questions/28574006/unique-combination-of-two-columns-in-r/28574136#28574136
allPairs <- expand.grid(levels(df$Group), levels(df$Group), stringsAsFactors = FALSE)  %>%
    filter(Var1 != Var2) %>%
    mutate(key = paste0(pmin(Var1, Var2), pmax(Var1, Var2), sep='')) %>%
    distinct(key) %>%
    select(-key)
allPairs
##   Var1 Var2
## 1    b    a
## 2    c    a
## 3    c    b

如果顺序确实很重要，则可以dplyr::arrange(Var1, Var2)在调用之后，尽早添加到此管道中expand.grid。

现在分析：

ret <- allPairs %>%
    rowwise() %>%
    do({
        data.frame(.,
                   oneway.test(Count ~ Group, filter(df, Group %in% c(.$Var1, .$Var2)),
                               na.action = na.omit, var.equal = FALSE)[c('statistic', 'p.value')],
                   stringsAsFactors = FALSE)
    })

ret
## Source: local data frame [3 x 4]
## Groups: <by row>
##   Var1 Var2   statistic   p.value
## 1    b    a 0.004008909 0.9507513
## 2    c    a 0.234782609 0.6342116
## 3    c    b 0.061749571 0.8084057

（我没有对这两种方法的性能做出任何声明；通常像这样的示例那样，其中的一种数据很少，但是另一种数据会以更大的数据集领先。它们似乎都执行相同的统计成对比较。结果相同。交给您！）