R：一次運行多個事后測試，使用 emmeans package

Question

我正在處理一個以幾種不同類型的蛋白質作為列的數據集。 有點像這個這是簡化的，原始數據集包含 100 多種蛋白質。 我想看看在考慮隨機效應 (=id) 時，蛋白質的濃度是否因處理而不同。 我設法一次運行多個重復的方差分析。 但我也想根據治療對所有蛋白質進行成對比較。 我首先想到的是使用 emmeans package，但我在編碼時遇到了問題。

#install packages 
library(tidyverse)
library(emmeans)

#Create a data set
set.seed(1)
id <- rep(c("1","2","3","4","5","6"),3)
Treatment <- c(rep(c("A"), 6), rep(c("B"), 6),rep(c("C"), 6))
Protein1 <- c(rnorm(3, 1, 0.4), rnorm(3, 3, 0.5), rnorm(3, 6, 0.8), rnorm(3, 1.1, 0.4), rnorm(3, 0.8, 0.2), rnorm(3, 1, 0.6))
Protein2 <- c(rnorm(3, 1, 0.4), rnorm(3, 3, 0.5), rnorm(3, 6, 0.8), rnorm(3, 1.1, 0.4), rnorm(3, 0.8, 0.2), rnorm(3, 1, 0.6))
Protein3 <- c(rnorm(3, 1, 0.4), rnorm(3, 3, 0.5), rnorm(3, 6, 0.8), rnorm(3, 1.1, 0.4), rnorm(3, 0.8, 0.2), rnorm(3, 1, 0.6))

DF <- data.frame(id, Treatment, Protein1, Protein2, Protein3) %>%
      mutate(id = factor(id),
      Treatment = factor(Treatment, levels = c("A","B","C")))

#First, I tried to run multiple anova, by using lapply
responseList <- names(DF)[c(3:5)]

modelList    <- lapply(responseList, function(resp) {
mF <- formula(paste(resp, " ~ Treatment + Error(id/Treatment)"))
aov(mF, data = DF)
})

lapply(modelList, summary)

#Pairwise comparison using emmeans. This did not work
wt_emm <- emmeans(modelList, "Treatment") 

> wt_emm <- emmeans(modelList, "Treatment")
Error in ref_grid(object, ...) : Can't handle an object of class  “list” 
 Use help("models", package = "emmeans") for information on supported models.

所以我嘗試了一種不同的方法

anova2 <- aov(cbind(Protein1,Protein2,Protein3)~ Treatment +Error(id/Treatment), data = DF)
summary(anova2)

#Pairwise comparison using emmeans. 
#I got only result for the whole dataset, instead of by different types of protein.
wt_emm2 <- emmeans(anova2, "Treatment")
pairs(wt_emm2)

> pairs(wt_emm2)
 contrast estimate   SE df t.ratio p.value
 A - B      -1.704 1.05 10 -1.630  0.2782 
 A - C       0.865 1.05 10  0.827  0.6955 
 B - C       2.569 1.05 10  2.458  0.0793 

我不明白為什么即使我在 anova model 中使用了“cbind(Protein1, Protein2, Protein3)”。R 仍然只給我一個結果而不是像下面這樣的結果

this is what I was hoping to get
 > Protein1
     contrast 
     A - B      
     A - C      
     B - C       
> Protein2
     contrast 
     A - B      
     A - C      
     B - C
> Protein3
     contrast 
     A - B      
     A - C      
     B - C

我該如何編碼或者我應該嘗試不同的包/功能？

我一次運行一種蛋白質沒有問題。 但是，由於我要運行 100 多個蛋白質，因此將它們一個一個地編碼真的很耗時。

任何建議表示贊賞。 謝謝！

Answer 1

這里

#Pairwise comparison using emmeans. This did not work
wt_emm <- emmeans(modelList, "Treatment") 

您需要像使用lapply lapply(modelList, summary)一樣對列表進行 lapply

modelList  <- lapply(responseList, function(resp) {
  mF <- formula(paste(resp, " ~ Treatment + Error(id/Treatment)"))
  aov(mF, data = DF)
})

但是當你這樣做時，會出現錯誤：

lapply(modelList, function(x) pairs(emmeans(x, "Treatment")))

注意：重新擬合 model 和與零的對比 Error in terms(formula, "Error", data = data): object 'mF' not found

attr(modelList[[1]], 'call')$formula
# mF

請注意， mF是formula object 的名稱，因此似乎emmeans出於某種原因需要原始公式。 您可以將公式添加到調用中：

modelList  <- lapply(responseList, function(resp) {
  mF <- formula(paste(resp, " ~ Treatment + Error(id/Treatment)"))
  av <- aov(mF, data = DF)
  attr(av, 'call')$formula <- mF
  av
})

lapply(modelList, function(x) pairs(emmeans(x, "Treatment")))

# [[1]]
# contrast estimate   SE df t.ratio p.value
#   A - B       -1.89 1.26 10 -1.501  0.3311 
#   A - C        1.08 1.26 10  0.854  0.6795 
#   B - C        2.97 1.26 10  2.356  0.0934 
# 
# P value adjustment: tukey method for comparing a family of 3 estimates 
# 
# [[2]]
# contrast estimate   SE df t.ratio p.value
#   A - B       -1.44 1.12 10 -1.282  0.4361 
#   A - C        1.29 1.12 10  1.148  0.5082 
#   B - C        2.73 1.12 10  2.430  0.0829 
# 
# P value adjustment: tukey method for comparing a family of 3 estimates 
# 
# [[3]]
# contrast estimate   SE df t.ratio p.value
#   A - B       -1.58 1.15 10 -1.374  0.3897 
#   A - C        1.27 1.15 10  1.106  0.5321 
#   B - C        2.85 1.15 10  2.480  0.0765 
# 
# P value adjustment: tukey method for comparing a family of 3 estimates 

Answer 2

按列名循環 function。

responseList <- names(DF)[c(3:5)]

for(n in responseList) {
  anova2 <- aov(get(n) ~ Treatment +Error(id/Treatment), data = DF)
  summary(anova2)
  wt_emm2 <- emmeans(anova2, "Treatment")
  print(pairs(wt_emm2))
}

這返回

Note: re-fitting model with sum-to-zero contrasts
Note: Use 'contrast(regrid(object), ...)' to obtain contrasts of back-transformed estimates
 contrast estimate   SE df t.ratio p.value
 A - B       -1.41 1.26 10 -1.122  0.5229 
 A - C        1.31 1.26 10  1.039  0.5705 
 B - C        2.72 1.26 10  2.161  0.1269 

Note: contrasts are still on the get scale 
P value adjustment: tukey method for comparing a family of 3 estimates 
Note: re-fitting model with sum-to-zero contrasts
Note: Use 'contrast(regrid(object), ...)' to obtain contrasts of back-transformed estimates
 contrast estimate   SE df t.ratio p.value
 A - B       -2.16 1.37 10 -1.577  0.2991 
 A - C        1.19 1.37 10  0.867  0.6720 
 B - C        3.35 1.37 10  2.444  0.0810 

Note: contrasts are still on the get scale 
P value adjustment: tukey method for comparing a family of 3 estimates 
Note: re-fitting model with sum-to-zero contrasts
Note: Use 'contrast(regrid(object), ...)' to obtain contrasts of back-transformed estimates
 contrast estimate   SE df t.ratio p.value
 A - B       -1.87 1.19 10 -1.578  0.2988 
 A - C        1.28 1.19 10  1.077  0.5485 
 B - C        3.15 1.19 10  2.655  0.0575 

Note: contrasts are still on the get scale 
P value adjustment: tukey method for comparing a family of 3 estimates

如果您想將 output 作為列表：

responseList <- names(DF)[c(3:5)]

output <- list()

for(n in responseList) {
  anova2 <- aov(get(n) ~ Treatment +Error(id/Treatment), data = DF)
  summary(anova2)
  wt_emm2 <- emmeans(anova2, "Treatment")
  output[[n]] <- pairs(wt_emm2)
  }