我有6个矩阵，我必须对6个矩阵的单元格值进行t检验，然后将p值存储在新矩阵中

Question

输入： 有6个具有相同维度的输入矩阵 3来自正常组织的输入矩阵：

              GeneA             GeneB          
GeneA          31                  4           
GeneB           5                  8 

              GeneA             GeneB          
GeneA           5                 14           
GeneB           5                  8 


              GeneA             GeneB          
GeneA          30                 14           
GeneB           45                 7 

3癌组织的输入矩阵：

              GeneA             GeneB          
GeneA          11                  4           
GeneB           5                  18 

              GeneA             GeneB          
GeneA           7                 14           
GeneB           15                 4 


              GeneA             GeneB          
GeneA          30                 14           
GeneB           45                 7 

输出：

                  GeneA                            GeneB          
GeneA        t-test({31,5,30},{11,7,13})    t-test({4,14,14},{4,14,14})        
GeneB        t-test({5,5,45},{5,15,45})     t-test({8,8,7},{18,4,7})

输出矩阵将具有测试中的p值

Answer 1

遵循的代码以3-DIM阵列形式对数据进行t检验。这使得循环遍历数据集，提取所需的向量并运行测试变得更加容易。

从表格数据到数组，

normal <- mget(ls(pattern = "^normal"))
cancer <- mget(ls(pattern = "^cancer"))

anorm <- array(dim = c(dim(normal[[1]]), length(normal)))
acanc <- array(dim = c(dim(cancer[[1]]), length(cancer)))
for(i in seq_along(normal)) {
  anorm[, , i] <- unlist(normal[[i]])
  acanc[, , i] <- unlist(cancer[[i]])
}

---

t检验

首先创建结果列表，然后运行测试，然后将p值提取到data.frame中。

t_test_list <- vector("list", length = prod(dim(anorm)[1:2]))
for(j in seq(dim(anorm)[2])) {
  for(i in seq(dim(anorm)[1])) {
    x <- anorm[i, j, ]
    y <- acanc[i, j, ]
    k <- i + dim(anorm)[1]*(j - 1)
    t_test_list[[k]] <- t.test(x, y)
  }
}

t_test_pval <- normal[[1]]
t_test_pval[] <- sapply(t_test_list, `[[`, 'p.value')

t_test_pval
#>           GeneA     GeneB
#> GeneA 0.6176386 1.0000000
#> GeneB 0.8618120 0.6850214

^由

---

创建

x<-'              GeneA             GeneB          
GeneA          31                  4           
GeneB           5                  8'
normal1 <- read.table(textConnection(x), header = TRUE)

x<-'              GeneA             GeneB          
GeneA           5                 14           
GeneB           5                  8 '
normal2 <- read.table(textConnection(x), header = TRUE)


x<-'              GeneA             GeneB          
GeneA          30                 14           
GeneB           45                 7 '
normal3 <- read.table(textConnection(x), header = TRUE)

x<-'              GeneA             GeneB          
GeneA          11                  4           
GeneB           5                  18 '
cancer1 <- read.table(textConnection(x), header = TRUE)

x<-'              GeneA             GeneB          
GeneA           7                 14           
GeneB           15                 4 '
cancer2 <- read.table(textConnection(x), header = TRUE)


x<-'              GeneA             GeneB          
GeneA          30                 14           
GeneB           45                 7 '
cancer3 <- read.table(textConnection(x), header = TRUE)

^由

Answer 2

也许最简单的方法是将两组矩阵绑定到2 x 2 x 3数组中，然后使用map在四个组合中的每一个中获取t测试。

假设您的矩阵被称为norm1，norm2，norm3的常规组织和ca1，ca2和CA3癌组织。然后我们可以做：

library(abind)

norm <- abind::abind(norm1, norm2, norm3, along = 3)
canc <- abind::abind(ca1, ca2, ca3, along = 3)

pvals <- Map(function(i, j) {
  t.test(norm[i, j, ], canc[i, j, ])$p.val
  }, i = c(1:2, 1:2), j = c(1, 1, 2, 2))

matrix(unlist(pvals), 2, dimnames = dimnames(norm1))
#>           GeneA     GeneB
#> GeneA 0.6176386 1.0000000
#> GeneB 0.8618120 0.6850214

^{在2022-06-01创建的 reprex package （v2.0.1）}

---

可重复的数据

norm1 <- structure(c(31L, 5L, 4L, 8L), dim = c(2L, 2L), dimnames = list(
    c("GeneA", "GeneB"), c("GeneA", "GeneB")))

norm2 <- structure(c(5L, 5L, 14L, 8L), dim = c(2L, 2L), dimnames = list(
    c("GeneA", "GeneB"), c("GeneA", "GeneB")))

norm3 <- structure(c(30L, 45L, 14L, 7L), dim = c(2L, 2L), dimnames = list(
    c("GeneA", "GeneB"), c("GeneA", "GeneB")))

ca1 <- structure(c(11L, 5L, 4L, 18L), dim = c(2L, 2L), dimnames = list(
    c("GeneA", "GeneB"), c("GeneA", "GeneB")))

ca2 <- structure(c(7L, 15L, 14L, 4L), dim = c(2L, 2L), dimnames = list(
    c("GeneA", "GeneB"), c("GeneA", "GeneB")))

ca3 <- structure(c(30L, 45L, 14L, 7L), dim = c(2L, 2L), dimnames = list(
    c("GeneA", "GeneB"), c("GeneA", "GeneB")))