在R中，如何使用DPLYR对每N列进行堆栈/RBIND？

Question

我想知道有没有人能帮我把一个较长的行分成几个较短的行，然后再把它们拆开？

1. 在本例中，我有12列长的行，我希望将其分成更多的行和4列（请参阅stack_df)。
2. 总体计划是按行合并（）所有列，并将（）突变为一列（9行x 1列，请参阅merge_df)。
3. 之后，我希望将它们解压回一个大小为3行3列的数据帧（请参阅simple_df)

> df <- matrix(c("A", "B", "C"),nrow=3,ncol=12,byrow=F)
> df
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] "A"  "A"  "A"  "A"  "A"  "A"  "A"  "A"  "A"  "A"   "A"   "A"  
[2,] "B"  "B"  "B"  "B"  "B"  "B"  "B"  "B"  "B"  "B"   "B"   "B"  
[3,] "C"  "C"  "C"  "C"  "C"  "C"  "C"  "C"  "C"  "C"   "C"   "C"

> stack_df <- matrix(c(rep("A",3), rep("B",3), rep("C",3)), nrow = 9, ncol = 4) 
> stack_df
      [,1] [,2] [,3] [,4]
 [1,] "A"  "A"  "A"  "A" 
 [2,] "A"  "A"  "A"  "A" 
 [3,] "A"  "A"  "A"  "A" 
 [4,] "B"  "B"  "B"  "B" 
 [5,] "B"  "B"  "B"  "B" 
 [6,] "B"  "B"  "B"  "B" 
 [7,] "C"  "C"  "C"  "C" 
 [8,] "C"  "C"  "C"  "C" 
 [9,] "C"  "C"  "C"  "C"

    > merge_df <- stack_df %>% 
+   as.data.frame(.) %>% 
+   unite(stack_df, 1:4, na.rm = T) %>% 
+   print()
  stack_df
1  A_A_A_A
2  A_A_A_A
3  A_A_A_A
4  B_B_B_B
5  B_B_B_B
6  B_B_B_B
7  C_C_C_C
8  C_C_C_C
9  C_C_C_C

> mutate_df <- cbind(sample(letters,9)) %>% 
+   print()
      [,1]
 [1,] "w" 
 [2,] "q" 
 [3,] "t" 
 [4,] "p" 
 [5,] "r" 
 [6,] "k" 
 [7,] "i" 
 [8,] "o" 
 [9,] "d"

第3部分：解叠行（3行，3列）。（所需输出）

> simple_df <- matrix(mutate_df, nrow = 3, ncol=3, byrow = T)
> simple_df
     [,1] [,2] [,3]
[1,] "w"  "q"  "t" 
[2,] "p"  "r"  "k" 
[3,] "i"  "o"  "d" 

Answer 1

该问题将输入命名为df，就好像它是一个数据帧，但它是一个矩阵，dplyr通常与数据帧而不是矩阵一起使用。dplyr可能不是这里使用的正确工具，而是使用这个base R one-liner，我们在末尾的注释中使用输入m，使用一个更精确的名称，并且还更改内容以使结果明确无误。

matrix(t(m), ncol = 4, byrow = TRUE)

给予：

      [,1] [,2]  [,3]  [,4] 
 [1,] "A1" "A2"  "A3"  "A4" 
 [2,] "A5" "A6"  "A7"  "A8" 
 [3,] "A9" "A10" "A11" "A12"
 [4,] "B1" "B2"  "B3"  "B4" 
 [5,] "B5" "B6"  "B7"  "B8" 
 [6,] "B9" "B10" "B11" "B12"
 [7,] "C1" "C2"  "C3"  "C4" 
 [8,] "C5" "C6"  "C7"  "C8" 
 [9,] "C9" "C10" "C11" "C12"

或者，如果你打算让A一栏一栏地填写前3行，同样地让B和C一栏一栏地填写前3行

matrix(aperm(array(t(m), c(3, 4, 3)), c(1, 3, 2)), ncol = 4)

      [,1] [,2] [,3] [,4] 
 [1,] "A1" "A4" "A7" "A10"
 [2,] "A2" "A5" "A8" "A11"
 [3,] "A3" "A6" "A9" "A12"
 [4,] "B1" "B4" "B7" "B10"
 [5,] "B2" "B5" "B8" "B11"
 [6,] "B3" "B6" "B9" "B12"
 [7,] "C1" "C4" "C7" "C10"
 [8,] "C2" "C5" "C8" "C11"
 [9,] "C3" "C6" "C9" "C12"

library(magrittr)

m %>% t %>% matrix(m, ncol = 4, byrow = TRUE)

m %>% t %>% array(c(3, 4, 3)) %>% aperm(c(1, 3, 2)) %>% matrix(ncol = 4)

m0 <- matrix(c("A", "B", "C"), 3, 12)
m <- replace(m0, TRUE, paste0(m0, col(m)))
m

给予：

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] "A1" "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12"
[2,] "B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8" "B9" "B10" "B11" "B12"
[3,] "C1" "C2" "C3" "C4" "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12"