R 中 Welch 校正的单向方差分析的事后检验 [英] Post-hoc tests for one-way ANOVA with Welch's correction in R
问题描述
我在 R 中使用 oneway.test() 运行了带有韦尔奇校正的单向方差分析测试，因为我的数据违反了等方差假设(转换没有解决问题).
I have run a one-way ANOVA test with welch's correction using oneway.test() in R, as I have data that violate the assumption of equal variance (transformations did not solve the problem).
一个简单的数据示例:
> dput(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")
library(car)
grp = as.factor(c(rep(1, 10), rep(2, 10),rep(3, 10)))
leveneTest(df$Count,grp) #unequal variances
#one-way ANOVA with welch's correction
oneway.test(Count ~ Group, data=df, na.action=na.omit, var.equal=FALSE)
我有多个组，所以我现在想运行成对的事后测试.有没有办法用 oneway.test() 函数中的对象来做到这一点?如果不是，如何对方差不等的组进行成对测试?我无法在网上找到这个问题的答案.任何建议将不胜感激.
I have multiple groups so I would now like to run pairwise post-hoc tests. Is there anyway to do this with an object from the oneway.test() function? If not, how would one go about running pair-wise tests on groups with unequal variances? I have not been able to find an answer to this question online. Any advice would be appreciated.
推荐答案
这里有两种方法:
library(car)
df <- structure(list(Count = c(13, 14, 14, 12, 11, 13, 14, 15, 13, 12, 20, 15, 9, 5, 13, 14, 7, 17, 18, 14, 12, 12, 13, 14, 11, 10, 15, 14, 14, 13),
Group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("a", "b", "c" ), class = "factor")),
.Names = c("Count", "Group"),
row.names = c(NA, -30L), class = "data.frame")
基础 R
首先，Group因子的唯一对集合:
allPairs <- expand.grid(levels(df$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$groups <- p
ret
})
length(allResults)
## [1] 3
allResults[[1]]
## One-way analysis of means (not assuming equal variances)
## data: Count and Group
## F = 0.004, num df = 1.000, denom df = 10.093, p-value = 0.9508
如果你想要这是一个矩阵，也许是这样的:
If you want this is a matrix, perhaps this:
mm <- diag(length(levels(df$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$p.value
})
mm
## a b c
## a 1.0000000 0.9507513 0.6342116
## b 0.9507513 1.0000000 0.8084057
## c 0.6342116 0.8084057 1.0000000
(对于 F 统计量，这可以很容易地完成.)
(This can be done just as easily for the F-statistic.)
首先，Group因子的唯一对集合:
First, the set of unique pairs of the Group factor:
library(dplyr)
## http://stackoverflow.com/questions/28574006/unique-combination-of-two-columns-in-r/28574136#28574136
allPairs <- expand.grid(levels(df$Group), levels(df$Group), stringsAsFactors = FALSE) %>%
filter(Var1 != Var2) %>%
mutate(key = paste0(pmin(Var1, Var2), pmax(Var1, Var2), sep='')) %>%
distinct(key) %>%
select(-key)
allPairs
## Var1 Var2
## 1 b a
## 2 c a
## 3 c b
如果顺序真的很重要，您可以将 dplyr::arrange(Var1, Var2) 添加到此管道的早期，可能是在调用 expand.grid 之后.
If the order really matters, you can add dplyr::arrange(Var1, Var2) early into this pipeline, perhaps after the call to expand.grid.
现在分析:
ret <- allPairs %>%
rowwise() %>%
do({
data.frame(.,
oneway.test(Count ~ Group, filter(df, Group %in% c(.$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
(我不会对其中任何一个的性能做出任何声明；通常一个会像这个例子一样用很少的数据发光，但另一个会以更大的集合出现.它们似乎都执行相同的统计对具有相同结果的明智比较.交给你了！)
(I'm making no claims as to the performance of either of these; often one will shine with few data like this example, but the other will come out ahead with larger sets. They both appear to perform the same statistical pair-wise comparisons with the same results. Over to you!)
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