R 使用 post-hoc 運行多個獨立的單向方差分析

Question

我有以下示例數據：

structure(list(Class = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 
4L, 5L, 5L), .Label = c("A", "B", "C ", "D ", "E"), class = "factor"), 
    a = c(0.10881116, 0.526242737, 0.982902999, 0.320738663, 
    0.652972541, 0.039061302, 0.866235756, 0.319948863, 0.49249116, 
    0.387460274), b = c(0.962253789, 0.504883561, 0.958827249, 
    0.112715995, 0.481341694, 0.022454068, 0.365585675, 0.243682534, 
    0.540064663, 0.79933528), c = c(0.68491864, 0.170941001, 
    0.067239671, 0.350063079, 0.303616697, 0.811791432, 0.986189818, 
    0.261161444, 0.366817736, 0.393204464), d = c(0.171410187, 
    0.795272464, 0.127037962, 0.729957086, 0.783967392, 0.836820247, 
    0.39774571, 0.727385402, 0.191486044, 0.316815623), e = c(0.018072241, 
    0.360542881, 0.435783461, 0.557028064, 0.645997614, 0.631136435, 
    0.316623636, 0.871827327, 0.615828269, 0.956653665), f = c(0.152489388, 
    0.500431046, 0.249617685, 0.855327742, 0.578962117, 0.510960229, 
    0.910920471, 0.8616062, 0.301616817, 0.691359783), g = c(0.016796537, 
    0.597620997, 0.169782711, 0.190080222, 0.781218649, 0.323382447, 
    0.968615432, 0.287030348, 0.754648917, 0.720887331)), class = "data.frame", row.names = c(NA, 
-10L))

我正在尋找運行幾個單向方差分析。 我想獨立地為每一列在“類”（A、B、C、D 等）之間運行方差分析（即一個方差分析用於“a”，另一個用於“b”，另一個用於“c”，等等。總共 7 個方差分析）。 對於每個，我想運行 Scheffe 事后測試。

因此，例如，對於一個方差分析，代碼將是

res.aov <- aov(a ~ Class, data = df)
library(DescTools)
ScheffeTest(res.aov)

有沒有辦法一次運行所有方差分析？

Answer 1

我們可以在循環中做到這一點

library(DescTools)
out <- lapply(names(df)[-1], function(nm) 
       ScheffeTest(aov(reformulate('Class', nm), data = df)))
names(out) <- names(df)[-1]

在dput輸出中，找到“Class”的一些前導/滯后空格

df$Class <- trimws(df$Class)

---

以防萬一，公式也可以用paste構造

out <- lapply(names(df)[-1], function(nm) 
       ScheffeTest(aov(as.formula(paste(nm, "~", "Class")), data = df)))
names(out) <- names(df)[-1]

或者用sprintf

out <- lapply(names(df)[-1], function(nm) 
       ScheffeTest(aov(as.formula(sprintf("%s ~ Class", nm)), data = df)))
names(out) <- names(df)[-1]

---

或者如果我們決定在tidyverse這樣做

library(purrr)    
library(dplyr)
map(names(df)[-1], ~ aov(reformulate('Class', .x), data = df) %>%
                  ScheffeTest)