Question

如何在R中执行简单的矩阵运算以找到矩阵C？

A = (a11 a12 a13)   B = (b1)   C = (a11-b1 a12-b1 a13-b1)
    (a21 a22 a23)       (b2)       (a21-b2 a22-b2 a23-b2)
    (a31 a32 a33)       (b3)       (a31-b3 a32-b3 a33-b3)

感谢您的时间！非常感激！：）

Answer 1

只需从B

中减去A即可

A = matrix(c(1:9),3,3)
A
#      [,1] [,2] [,3]
#[1,]    1    4    7
#[2,]    2    5    8
#[3,]    3    6    9
B = c(1:3)
B
#      [,1]
#[1,]    1
#[2,]    2
#[3,]    3
C= A - B
C
#      [,1] [,2] [,3]
#[1,]    0    3    6
#[2,]    0    3    6
#[3,]    0    3    6

Answer 2

以下是解决方案：

a=matrix(c(1:9), nrow = 3, ncol = 3)
b=matrix(c(1:3),nrow = 3,ncol = 1)
ans=c()
for(i in 1:ncol(a)){
ans=c(ans,a[,i]-b[,1])
}
final=matrix(ans,nrow = 3,ncol = 3)

上面的代码产生以下输出：

> a
      [,1] [,2] [,3]
[1,]    1    4    7
[2,]    2    5    8
[3,]    3    6    9
> b
      [,1]
[1,]    1
[2,]    2
[3,]    3
> final
      [,1] [,2] [,3]
[1,]    0    3    6
[2,]    0    3    6
[3,]    0    3    6

希望这适合你：）

R：从每行B [i，]中减去A [i，]

2 个答案: