将矩阵中的值复制到 r 中的另一个位置

Question

我有一个所有不同维度的矩阵列表。我想取一个存在于最后一行、倒数第二列的值，并将该值复制到矩阵的右下角。

矩阵列表如下所示：

> matrix.list

$`1`
      [,1] [,2] [,3] [,4]      [,5] [,6]     [,7]      [,8] [,9]
 [1,]    0    0    0    0 0.0000000    0 0.000000 0.0000000 4.52
 [2,]    1    0    0    0 0.0000000    0 0.000000 0.0000000 0.00
 [3,]    0    1    0    0 0.0000000    0 0.000000 0.0000000 0.00
 [4,]    0    0    1    0 0.0000000    0 0.000000 0.0000000 0.00
 [5,]    0    0    0    1 0.0000000    0 0.000000 0.0000000 0.00
 [6,]    0    0    0    0 0.9733333    0 0.000000 0.0000000 0.00
 [7,]    0    0    0    0 0.0000000    1 0.000000 0.0000000 0.00
 [8,]    0    0    0    0 0.0000000    0 0.902029 0.0000000 0.00
 [9,]    0    0    0    0 0.0000000    0 0.000000 0.9733333 0.00

$`2`
     [,1] [,2] [,3] [,4] [,5] [,6]  [,7] [,8]
[1,]    0    0    0    0 0.00    0 0.000  6.5
[2,]    1    0    0    0 0.00    0 0.000  0.0
[3,]    0    1    0    0 0.00    0 0.000  0.0
[4,]    0    0    1    0 0.00    0 0.000  0.0
[5,]    0    0    0    1 0.00    0 0.000  0.0
[6,]    0    0    0    0 0.96    0 0.000  0.0
[7,]    0    0    0    0 0.00    1 0.000  0.0
[8,]    0    0    0    0 0.00    0 0.905  0.0

$`3`
     [,1] [,2] [,3] [,4]      [,5] [,6] [,7]
[1,]    0    0    0    0 0.0000000    0 4.82
[2,]    1    0    0    0 0.0000000    0 0.00
[3,]    0    1    0    0 0.0000000    0 0.00
[4,]    0    0    1    0 0.0000000    0 0.00
[5,]    0    0    0    1 0.0000000    0 0.00
[6,]    0    0    0    0 0.9866667    0 0.00
[7,]    0    0    0    0 0.0000000    1 0.00

我想把它改成这样：

> matrix.list

$`1`
      [,1] [,2] [,3] [,4]      [,5] [,6]     [,7]      [,8]      [,9]
 [1,]    0    0    0    0 0.0000000    0 0.000000 0.0000000 4.5200000
 [2,]    1    0    0    0 0.0000000    0 0.000000 0.0000000 0.0000000
 [3,]    0    1    0    0 0.0000000    0 0.000000 0.0000000 0.0000000
 [4,]    0    0    1    0 0.0000000    0 0.000000 0.0000000 0.0000000
 [5,]    0    0    0    1 0.0000000    0 0.000000 0.0000000 0.0000000
 [6,]    0    0    0    0 0.9733333    0 0.000000 0.0000000 0.0000000
 [7,]    0    0    0    0 0.0000000    1 0.000000 0.0000000 0.0000000
 [8,]    0    0    0    0 0.0000000    0 0.902029 0.0000000 0.0000000
 [9,]    0    0    0    0 0.0000000    0 0.000000 0.9733333 0.9733333

$`2`
     [,1] [,2] [,3] [,4] [,5] [,6]  [,7]   [,8]
[1,]    0    0    0    0 0.00    0 0.000  6.500
[2,]    1    0    0    0 0.00    0 0.000  0.000
[3,]    0    1    0    0 0.00    0 0.000  0.000
[4,]    0    0    1    0 0.00    0 0.000  0.000
[5,]    0    0    0    1 0.00    0 0.000  0.000
[6,]    0    0    0    0 0.96    0 0.000  0.000
[7,]    0    0    0    0 0.00    1 0.000  0.000
[8,]    0    0    0    0 0.00    0 0.905  0.905

$`3`
     [,1] [,2] [,3] [,4]      [,5] [,6] [,7]
[1,]    0    0    0    0 0.0000000    0 4.82
[2,]    1    0    0    0 0.0000000    0 0.00
[3,]    0    1    0    0 0.0000000    0 0.00
[4,]    0    0    1    0 0.0000000    0 0.00
[5,]    0    0    0    1 0.0000000    0 0.00
[6,]    0    0    0    0 0.9866667    0 0.00
[7,]    0    0    0    0 0.0000000    1 1.00

我尝试在感兴趣的位置提取值：

data.list2 <- lapply(data.list, function(x) x[nrow(x),ncol(x)-1])

但我收到许多人遇到的关于尺寸不正确的错误：

Error in x[nrow(x), ncol(x) - 1] : incorrect number of dimensions

有没有办法索引某个位置，然后将其复制到矩阵列表中的相同位置？类似于dplyr::nth 命令但用于矩阵？

Answer 1

您明白了，但没有分配/返回新矩阵。 这应该可以完成工作

# for the sake of reproducibility
set.seed(123)

# create a reproducible example
matrix.list <- list(
  matrix(runif(12), 4, 3),
  matrix(runif(24), 6, 4)
)

# our original matrix
matrix.list


[[1]]
[,1]      [,2]      [,3]
[1,] 0.2875775 0.9404673 0.5514350
[2,] 0.7883051 0.0455565 0.4566147
[3,] 0.4089769 0.5281055 0.9568333
[4,] 0.8830174 0.8924190 0.4533342

[[2]]
[,1]      [,2]      [,3]       [,4]
[1,] 0.67757064 0.3279207 0.6557058 0.96302423
[2,] 0.57263340 0.9545036 0.7085305 0.90229905
[3,] 0.10292468 0.8895393 0.5440660 0.69070528
[4,] 0.89982497 0.6928034 0.5941420 0.79546742
[5,] 0.24608773 0.6405068 0.2891597 0.02461368
[6,] 0.04205953 0.9942698 0.1471136 0.47779597

lapply(matrix.list, function(x) {
  x[nrow(x), ncol(x)] <- x[nrow(x), ncol(x)-1] # we assign
  x                                            # and return
  }
)

[[1]]
[,1]      [,2]      [,3]
[1,] 0.2875775 0.9404673 0.5514350
[2,] 0.7883051 0.0455565 0.4566147
[3,] 0.4089769 0.5281055 0.9568333
[4,] 0.8830174 0.8924190 0.8924190

[[2]]
[,1]      [,2]      [,3]       [,4]
[1,] 0.67757064 0.3279207 0.6557058 0.96302423
[2,] 0.57263340 0.9545036 0.7085305 0.90229905
[3,] 0.10292468 0.8895393 0.5440660 0.69070528
[4,] 0.89982497 0.6928034 0.5941420 0.79546742
[5,] 0.24608773 0.6405068 0.2891597 0.02461368
[6,] 0.04205953 0.9942698 0.1471136 0.14711365

这就是你要找的吗？

Answer 2

加点purrr包：

require(purrr)

map(matrix.list, function(x) {
x[nrow(x), ncol(x)] <- x[nrow(x), ncol(x) - 1] 
x
})

关于错误，可能是某些矩阵只有一列吗？