具有多个组和多个特征的未配对 t 检验的 R 代码

Question

假设我有以下数据：

Class,a,b,c,d,e,f,g,h,i,j,k,l,m A,0.1,0.4,0.4,0.7,0.1,0.1,0.9,0.5,0.5,0.9,0.6,0.1,0.9 A,0.0,0.3,0.7,0.1,0.2,0.1,0.3,0.2,1.0,0.9,0.6,0.6,0.1 B,0.6,0.3,0.4,0.6,0.3,0.8,0.3,0.0,0.8,0.9,0.6,0.7,0.5 B,0.8,0.3,0.8,0.8,0.3,0.7,0.3,0.7,0.7,0.4,0.9,0.3,0.8 C,0.8,0.1,0.0,0.0,0.2,0.0,0.7,0.3,0.1,0.6,0.8,0.0,0.3 C,0.3,0.0,0.7,1.0,0.8,0.3,0.6,0.4,0.9,0.9,0.1,0.9,1.0 D,0.9,0.7,1.0,0.1,0.2,0.5,0.9,0.9,0.6,0.8,0.6,0.4,0.2 D,0.6,0.2,0.8,0.8,0.1,0.3,0.0,0.0,0.6,0.3,0.6,0.9,0.3 E,1.0,0.5,0.5,0.0,0.7,0.9,1.0,0.1,0.5,0.5,0.1,0.9,0.6 E,0.5,0.8,0.5,0.1,0.7,0.6,0.1,0.8,0.2,0.1,0.7,0.6,0.4 F,0.6,0.4,0.6,0.2,0.5,0.4,0.2,0.2,0.3,0.4,0.9,0.5,0.7 F,0.9,0.6,0.8,0.6,0.6,0.3,0.8,0.5,0.4,0.4,0.5,0.7,0.4 G,0.2,0.1,0.6,0.4,0.3,0.8,0.4,0.1,0.9,0.7,0.0,0.5,0.8 G,0.2,0.3,0.2,0.8,0.3,0.6,0.2,0.6,0.9,0.1,0.2,0.1,0.3 H,0.7,0.4,0.7,0.2,0.5,0.5,0.5,0.9,0.8,0.4,0.9,0.7,0.5 H,0.1,0.6,0.5,0.7,0.5,0.4,0.5,0.4,0.1,0.6,0.1,0.3,0.5 I,0.6,0.8,0.7,0.6,0.3,0.5,0.2,0.6,0.5,0.3,0.9,0.8,0.5 I,0.1,0.0,0.5,0.2,0.3,0.2,0.4,0.9,0.9,0.8,0.5,0.3,0.2 J,0.6,0.1,0.2,0.3,0.3,0.6,0.5,0.9,0.0,0.2,0.4,0.8,0.9 J,0.3,0.7,0.4,0.1,0.4,0.1,0.7,0.5,0.1,0.0,0.1,0.2,0.6 K,0.1,0.9,0.5,0.0,0.0,0.3,0.0,0.7,0.2,0.0,0.0,0.6,0.6 K,0.9,0.6,0.1,0.9,0.3,0.2,0.1,0.2,0.2,0.1,0.7,0.8,0.0 L,1.0,0.6,0.0,0.0,0.8,0.5,0.7,0.9,0.3,0.5,0.4,0.9,0.4 L,0.6,0.2,0.5,0.2,0.7,0.6,0.4,0.5,0.8,0.6,0.8,0.3,0.9 M,0.7,0.1,0.1,0.8,0.4,1.0,0.9,0.1,0.1,0.5,0.5,0.6,0.3 M,0.8,0.4,0.2,0.4,0.2,0.2,0.3,0.5,0.5,0.3,0.3,1.0,0.3 N,0.0,0.4,0.5,0.1,0.6,0.5,0.3,0.4,0.8,0.7,0.1,0.2,0.8 N,0.9,0.4,1.0,0.9,0.7,0.5,0.7,0.4,0.8,0.8,0.1,0.7,0.2 O,0.2,0.1,1.0,0.2,0.0,0.8,0.4,0.9,0.1,1.0,0.8,0.3,0.5 O,0.6,1.0,1.0,0.8,1.0,0.6,0.4,0.3,0.2,0.8,0.5,0.0,0.1 P,0.2,0.1,0.7,0.8,0.3,0.2,0.4,0.4,1.0,0.2,0.7,0.1,0.0 P,0.3,0.6,0.6,0.5,0.9,0.1,0.9,0.4,0.5,0.2,0.7,0.8,0.2 Q,0.9,0.7,0.6,0.7,0.4,0.3,0.3,0.8,0.6,0.6,1.0,0.1,0.7 Q,0.6,0.9,0.9,0.1,0.8,0.7,0.0,0.1,0.7,0.0,0.9,0.7,0.8 R,0.1,0.7,0.5,0.9,0.9,0.1,1.0,0.1,0.6,0.2,0.5,0.5,0.3 R,0.1,0.6,0.9,0.6,0.6,0.8,1.0,0.3,0.7,0.9,0.1,0.8,0.2

structure(list(Class = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 
4L, 5L, 5L, 6L, 6L, 7L, 7L, 8L, 8L, 9L, 9L, 10L, 10L, 11L, 11L, 
12L, 12L, 13L, 13L, 14L, 14L, 15L, 15L, 16L, 16L, 17L, 17L, 18L, 
18L), .Label = c("A", "B", "C", "D", "E", "F", "G", "H", "I", 
"J", "K", "L", "M", "N", "O", "P", "Q", "R"), class = "factor"), 
    a = c(0.7, 0.4, 0.9, 0.5, 0.4, 0.5, 0, 0.3, 0.7, 0.7, 0.6, 
    0.5, 0.4, 0.1, 0.3, 0.1, 0.3, 0.4, 0.5, 0.8, 0.8, 0.6, 0.2, 
    0.3, 0.2, 0.6, 0.6, 0.4, 0.9, 0.1, 0.6, 0.6, 0.7, 0.7, 1, 
    0.2), b = c(0.3, 0.1, 0.3, 1, 0.2, 0.8, 0.9, 0.5, 0.6, 0.3, 
    0.2, 0.9, 0.9, 0.8, 0, 0.2, 0.8, 0.7, 0.8, 0.8, 0.5, 0, 0.3, 
    0.9, 0.4, 0.9, 0.1, 0.4, 0.5, 0.7, 0.8, 0.9, 0.1, 0.8, 1, 
    0.7), c = c(0.2, 0.6, 0.7, 0.5, 0.4, 0.7, 0.9, 0.1, 0.2, 
    0.8, 0.8, 0.9, 0.2, 0.6, 0.4, 0.3, 0.1, 0.9, 0, 0.8, 0.5, 
    0.9, 0.4, 0.9, 0.8, 0.4, 0.9, 0.7, 0.9, 1, 0.2, 0.1, 1, 0.9, 
    0.8, 0.8), d = c(0.8, 0.1, 0.5, 0.8, 0.3, 0.2, 0.9, 0.9, 
    0, 0.3, 0.7, 0, 0.1, 0, 0, 0.8, 1, 0.8, 0.6, 1, 0.6, 0.6, 
    0.2, 0.1, 1, 0.1, 0.7, 0.3, 0.7, 0.3, 0.9, 0.8, 0.1, 0.2, 
    0.1, 0.4), e = c(0.6, 0.6, 0.3, 0.6, 0.7, 0.6, 0.9, 0.8, 
    0.6, 0.4, 0.6, 0.7, 0.7, 0.2, 1, 1, 1, 0.5, 0.4, 0.5, 0.6, 
    0.7, 0.1, 0.1, 0.6, 0.2, 0.4, 1, 0.1, 0.3, 0.7, 0, 0.4, 0.2, 
    0.7, 0.1), f = c(0.9, 0.2, 0.4, 0.8, 0, 0.2, 1, 0.2, 0.1, 
    0.1, 0.4, 1, 0.1, 0.9, 0.2, 0.2, 0.3, 0.6, 0.9, 0.2, 0.5, 
    0.4, 0.4, 0.2, 0.6, 0.1, 0.6, 0.1, 0.5, 0.1, 0.6, 0.9, 0.5, 
    0.4, 0.5, 0.8), g = c(0.6, 0.3, 0.9, 0.1, 0.9, 0.3, 0.1, 
    0.2, 0, 0.4, 1, 0.3, 0.7, 0.8, 0.5, 0.5, 0.6, 0.2, 0.2, 0, 
    1, 0.6, 0.2, 0.7, 0.6, 0.6, 0.1, 0.4, 0.6, 0.2, 0.6, 0.4, 
    0.6, 0.7, 0.3, 1), h = c(0.1, 0.8, 0.3, 0.3, 0.3, 0.8, 0.3, 
    0.7, 0.4, 0.8, 0.3, 0.1, 0.5, 0.7, 0.1, 0.3, 0.7, 0.4, 0.7, 
    0.5, 0.4, 0.1, 0.9, 0.8, 0.9, 0.7, 0.3, 0.1, 0.3, 0.1, 0.2, 
    0.1, 0.9, 0.3, 0.1, 0.6), i = c(0.8, 0.8, 0.4, 0.9, 0.5, 
    0.8, 0.6, 0.6, 0.2, 0.7, 0.1, 0.8, 0.1, 0.5, 0.7, 0.6, 0.8, 
    0.4, 0.2, 0.9, 0.7, 0.8, 0.5, 0.8, 0.3, 0.3, 1, 0.6, 0.9, 
    0.9, 0.7, 0.5, 0.1, 1, 0.5, 0.1), j = c(0.5, 0, 0.4, 0.7, 
    0.5, 0.6, 0.3, 0.9, 0.1, 0.4, 1, 0.5, 0.7, 0.4, 0.3, 0.9, 
    0, 0.5, 0.7, 0.4, 0, 0.1, 0.4, 0.3, 0.6, 0.9, 0.5, 0.5, 0.9, 
    0.5, 0.5, 0.3, 0.1, 0.1, 0.9, 0.7), k = c(0.8, 0.1, 0, 0.5, 
    0.7, 0.9, 0.4, 0.9, 0.3, 0.8, 0.7, 0.8, 0.2, 0.2, 0.8, 0.4, 
    0.8, 0.7, 0.2, 0.6, 0.5, 0.4, 1, 0.2, 0.7, 0.3, 0.1, 0, 0.4, 
    0.7, 0.8, 0.3, 0.3, 0.8, 0.5, 0.9), l = c(0.8, 0.6, 0.3, 
    0.6, 0.6, 0.1, 0.2, 0.3, 0.7, 0.6, 0.1, 0.8, 1, 1, 0.5, 0.3, 
    0.3, 0, 0.4, 0.1, 0.5, 0.6, 0.9, 0.4, 0.6, 0.4, 0.3, 0.4, 
    0.5, 0.5, 0.5, 0.5, 0.9, 0.2, 0.5, 0.9), m = c(0.4, 0, 0.7, 
    0.1, 0.9, 0.8, 0.6, 0.1, 0.2, 0.2, 0.9, 0.9, 0.4, 0.4, 0.9, 
    0.6, 0.2, 0.1, 0.7, 0.5, 0.2, 0.8, 0.4, 0.6, 0.9, 0.7, 0, 
    0.8, 0.7, 0.2, 0.1, 0.8, 0.2, 0.2, 0.3, 0.2)), row.names = c(NA, 
-36L), class = "data.frame")

我想运行在列（即 A-B、A-C、B-C 等）内的行（即 A-B、A-C、B-C 等）之间进行非配对 t 检验 "a"，然后是 "b" 内的 A-B、A-C、B-C 等）。有没有一种简单的方法可以用 R 来做到这一点，而不是运行单独地每个基因？理想情况下，我想要每个基因中每个样本配对的 p 值输出列表。

非常感谢， L.

Answer 1

我们可以使用 combn 得到 'Class' 列值的成对组合并在 t.test

中使用

lstOut <- combn(unique(df1$Class), 2, FUN = function(x) {
           dat1 <- subset(df1, Class == x[1])
           dat2 <- subset(df1, Class == x[2])
           Map(function(x, y) tryCatch(t.test(x, y),
                    error = function(e) NA), dat1[-1], dat2[-1])
       }, simplify = FALSE)

names(lstOut) <- combn(as.character(unique(df1$Class)), 2, FUN = paste, collapse="_")

list 输出可以 tidy 并转换为单个 data.frame

library(broom)
library(purrr)
out <- map_depth(lstOut, .depth = 2, tidy)
out2 <- map_dfr(out, ~bind_rows(.x, .id = 'colname'), .id = 'classCompare')

具有多个组和多个特征的未配对 t 检验的 R 代码

R code for unpaired t-test with multiple groups & multiple traits

r

statistics

t-test