ddply多因素全成对t-检验

Question

如何用最少的编码线对所有可能的组对进行多因素t检验。

我的例子：

3x功能:1、2、3

4x组：：A、B、C、D

目的:对于每个功能，测试所有组对：

1(A-B，A-C，A-D，B-C，B-D，C-D)

2(A-B，A-C，A-D，B-C，B-D，C-D)

3(A-B，A-C，A-D，B-C，B-D，C-D) = 18个T检验

目前，我正在使用ddply和lapply：

library(plyr)

groupVector <- c(rep("A",10),rep("B",10),rep("C",10),rep("D",10))
featureVector <- rep(1:3,each=40)

mydata <- data.frame(feature=factorVector,group=groupVector,value=rnorm(120,0,1))

ddply(mydata,.(feature),function(x){
  grid <- combn(unique(x$group),2, simplify = FALSE)
  df <- lapply(grid,function(p){
    sub <- subset(x,group %in% p)
    pval <- t.test(sub$value ~ sub$group)$p.value
    data.frame(groupA=p[1],groupB=p[2],pval=pval)
  })
  res <- do.call("rbind",df)
  return(res)
})

Answer 1

以下是我的观点，尽管它是否“更好”尚有争议

split.data <- split(mydata, mydata$feature)
pairs <- as.data.frame(matrix(combn(unique(mydata$group), 2), nrow=2))
library(tidyverse)
map_df(split.data, function(x) map_df(pairs, function(y) tibble(groupA = y[1], groupB = y[2], 
                                        pval = t.test(value ~ group, data = x, subset = which(x$group %in% y))$p.value)), .id="feature") 

输出

# # A tibble: 18 x 4
   # feature groupA groupB       pval
     # <chr>  <chr>  <chr>      <dbl>
 # 1       1      A      B 0.28452419
 # 2       1      A      C 0.65114472
 # 3       1      A      D 0.77746420
 # 4       1      B      C 0.42546791
 # 5       1      B      D 0.39876582
 # 6       1      C      D 0.88079645
 # 7       2      A      B 0.57843592
 # 8       2      A      C 0.30726571
 # 9       2      A      D 0.55457986
# 10       2      B      C 0.74871464
# 11       2      B      D 0.24017130
# 12       2      C      D 0.04252878
# 13       3      A      B 0.01355117
# 14       3      A      C 0.08746756
# 15       3      A      D 0.24527519
# 16       3      B      C 0.15130684
# 17       3      B      D 0.09172577
# 18       3      C      D 0.64206517