在r中减去两个矩阵

Question

I have created two matrices as followings:

    A = c(1,2,3)
    B = c(2,4,6)
    c = as.matrix(c(3,6,9))

    z = as.matrix(cbind(A, B))

Now I want to take the matrix c and subtract it row by row eg 1-3 = -2 & 2-3 =-1 Do get a good understanding of R programming, I would like to create a for loop. PLEASE ALL YOUR ANSWERS SHOULD IMPROVE MY FOR LOOP.

 for (i in 1:nrow(z))# for the rows in matrix z
  for (j in 1:nrow(c)) # for the rows in matrix c 
     {
      sub = matrix(NA, 3,2) # make a placeholder 
      sub [i,]= z[i,]-c[j,] # i am not sure whether this right
      return((sub))
   }

I get the following error:

    Error: no function to return from, jumping to top level

I believe my for loop is wrong, can anyone help. The purpose is to learn more about R programming. Thanks

Answer 1

This is a good case for sweep :

sweep(z, 1, c)
#      A  B
#[1,] -2 -1
#[2,] -4 -2
#[3,] -6 -3

Answer 2

If you write your loop that way:

sub = matrix(NA, 3,2) # make a placeholder 
for (i in 1:nrow(z))# for the rows in matrix z
  for (j in 1:nrow(c)) # for the rows in matrix c 
  {
    sub [i,]= z[i,]-c[j,] # i am not sure whether this right
  }
sub

it will end without error but you'll get:

     [,1] [,2]
[1,]   -8   -7
[2,]   -7   -5
[3,]   -6   -3

which is not what you expected... because the last j was always 3 so you replaced sub[i,] by z[i,]-c[3,] which is z[i,]-9

Now if you replace the loops by :

for (i in 1:nrow(z)) #(nrow(z)==nrow(c))
  {
    sub [i,]= z[i,]-c[i,]
  }

then you'll have:

     [,1] [,2]
[1,]   -2   -1
[2,]   -4   -2
[3,]   -6   -3

Answer 3

although not recommended, this will work

 sub = matrix(NA, 3,2)
 for (i in 1:nrow(z)) {
     sub[i,]=z[i,]-c[i,1]
 }

you don't need the second loop, also c is a column matrix.

Answer 4

You could also just create a matrix where both columns are given by c :

z - cbind(c, c)
##       A  B
## [1,] -2 -1
## [2,] -4 -2
## [3,] -6 -3

This won't be a convenient solution for a matrix with many columns. You could use cbind together with do.call to have more flexibility:

z - do.call(cbind, rep(list(c), 2))

It is necessary to put c in a list. Otherwise, rep(...) will return a single vector instead of a list of column matrixes:

rep(c, 2)
## [1] 3 6 9 3 6 9
rep(list(c), 2)
## [[1]]
##      [,1]
## [1,]    3
## [2,]    6
## [3,]    9
## 
## [[2]]
##      [,1]
## [1,]    3
## [2,]    6
## [3,]    9