在 R 中使用外部协变量融化矩阵

Melt a matrix using extern covariate in R

我有一个矩阵:

mat1 <- matrix(rnorm(8), ncol = 4;
  ,dimnames=list(c('R1','R2'),c('C1','C2','C3','C4')))

> mat1
          C1         C2         C3        C4
R1  1.226139 -1.0604743 -0.1803689 0.3852505
R2 -1.232622 -0.5567295 -0.4146919 0.2433812

和一个与矩阵列名称匹配的协变量

> covariate   <- factor(c('A','A','B','B'))
> t(data.frame(covariate, colnames(mat1)))
               [,1] [,2] [,3] [,4]
covariate      "A"  "A"  "B"  "B" 
colnames.mat1. "C1" "C2" "C3" "C4"

我想根据协变量将其融化以获得以下结果:

融化数据给出:

> melt( mat1 )
      Var1 Var2      value
    1   R1   C1  1.2261395
    2   R2   C1 -1.2326215
    3   R1   C2 -1.0604743
    4   R2   C2 -0.5567295
    5   R1   C3 -0.1803689
    6   R2   C3 -0.4146919
    7   R1   C4  0.3852505
    8   R2   C4  0.2433812

但是我想得到以下结果:

covariate_2 <- factor( c(rep('A',4) , rep('B',4) ))
> data.frame( covariate_2 , melted_data )
  covariate_2 Var1 Var2      value
1           A   R1   C1  1.2261395
2           A   R2   C1 -1.2326215
3           A   R1   C2 -1.0604743
4           A   R2   C2 -0.5567295
5           B   R1   C3 -0.1803689
6           B   R2   C3 -0.4146919
7           B   R1   C4  0.3852505
8           B   R2   C4  0.2433812

我认为一定有一种方法可以使用标准熔化函数来获得结果。如果有任何帮助,我将不胜感激。

也许最简单的方法是先重命名矩阵的列,然后 melt

这里有几个例子,第一个使用 "data.table",第二个使用 "tidyverse":

library(data.table)
setDT(melt(`colnames<-`(mat1, paste(c('A','A','B','B'), colnames(mat1), sep = "_"))))[
  , c("cov", "V1") := tstrsplit(Var2, "_")][, Var2 := NULL][]
#    Var1      value cov V1
# 1:   R1  1.2261390   A C1
# 2:   R2 -1.2326220   A C1
# 3:   R1 -1.0604743   A C2
# 4:   R2 -0.5567295   A C2
# 5:   R1 -0.1803689   B C3
# 6:   R2 -0.4146919   B C3
# 7:   R1  0.3852505   B C4
# 8:   R2  0.2433812   B C4


library(tidyverse)
`colnames<-`(mat1, paste(c('A','A','B','B'), colnames(mat1), sep = "_")) %>% 
  as.data.frame() %>%
  rownames_to_column() %>%
  gather(var, val, -rowname) %>%
  separate(var, into = c("cov", "var1"))
#   rowname cov var1        val
# 1      R1   A   C1  1.2261390
# 2      R2   A   C1 -1.2326220
# 3      R1   A   C2 -1.0604743
# 4      R2   A   C2 -0.5567295
# 5      R1   B   C3 -0.1803689
# 6      R2   B   C3 -0.4146919
# 7      R1   B   C4  0.3852505
# 8      R2   B   C4  0.2433812

示例数据:

mat1 <- structure(c(1.226139, -1.232622, -1.0604743, -0.5567295, -0.1803689, 
    -0.4146919, 0.3852505, 0.2433812), .Dim = c(2L, 4L), .Dimnames = list(
        c("R1", "R2"), c("C1", "C2", "C3", "C4")))