Question

我有一个这种形式的矩阵：

a b c
d e 0
f 0 0

我想把它变成这样的东西：

a b c
0 d e
0 0 f

转变模式是：

shift by 0 for row 1
shift by 1 for row 2
shift by 2 for row 3
...
shift by n-1 for row n

当然可以使用for循环来完成。我想知道是否有更好的方法吗？

Answer 1

假设您的示例具有代表性，即您始终使用字母和零的三角形结构：

mat <- structure(c("a", "d", "f", "b", "e", "0", "c", "0", "0"), 
                 .Dim = c(3L, 3L), .Dimnames = list(NULL, NULL))
res <- matrix(0, nrow(mat), ncol(mat))
res[lower.tri(res, diag=TRUE)] <- t(mat)[t(mat)!="0"]
t(res)
#     [,1] [,2] [,3]
# [1,] "a"  "b"  "c" 
# [2,] "0"  "d"  "e" 
# [3,] "0"  "0"  "f"

Answer 2

head和tail解决方案对我来说似乎不如for循环可读，甚至可能不那么快。尽管如此...

t( sapply( 0:(nrow(mat)-1) , function(x) c( tail( mat[x+1,] , x ) , head( mat[x+1,] , nrow(mat)-x ) ) ) )
#     [,1] [,2] [,3]
#[1,] "a"  "b"  "c" 
#[2,] "0"  "d"  "e" 
#[3,] "0"  "0"  "f"

此for循环版本可能是......

n <- nrow(mat)
for( i in 1:n ){
    mat[i,] <- c( tail( mat[i,] , i-1 ) , head( mat[i,] , n-(i-1)  ) )
}

Answer 3

我认为这就是你所需要的：

 mat<-matrix(1:25,5)
 mat
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    6   11   16   21
[2,]    2    7   12   17   22
[3,]    3    8   13   18   23
[4,]    4    9   14   19   24
[5,]    5   10   15   20   25
 for(j in 2:nrow(mat) ) mat[j,]<-mat[j, c(j:ncol(mat),1:(j-1))]
 mat
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    6   11   16   21
[2,]    7   12   17   22    2
[3,]   13   18   23    3    8
[4,]   19   24    4    9   14
[5,]   25    5   10   15   20

如何在R中移动矩阵的每一行

3 个答案: