subtract a constant vector from each row in a matrix in r

33
votes

I have a matrix with 5 columns and 4 rows. I also have a vector with 3 columns. I want to subtract the values in the vector from columns 3,4 and 5 respectively at each row of the matrix.

b <- matrix(rep(1:20), nrow=4, ncol=5)
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    5    9   13   17
[2,]    2    6   10   14   18
[3,]    3    7   11   15   19
[4,]    4    8   12   16   20

c <- c(5,6,7)

to get

     [,1] [,2] [,3] [,4] [,5]
[1,]    1    5    4    7   10
[2,]    2    6    5    8   11
[3,]    3    7    6    9   12
[4,]    4    8    7   10   13
5
rvectormatrix

5 Answers

68
votes

This is exactly what sweep was made for:

b <- matrix(rep(1:20), nrow=4, ncol=5)
x <- c(5,6,7)

b[,3:5] <- sweep(b[,3:5], 2, x)
b

#     [,1] [,2] [,3] [,4] [,5]
#[1,]    1    5    4    7   10
#[2,]    2    6    5    8   11
#[3,]    3    7    6    9   12
#[4,]    4    8    7   10   13

..or even without subsetting or reassignment:

sweep(b, 2, c(0,0,x))
10
votes

Perhaps not that elegant, but

b <- matrix(rep(1:20), nrow=4, ncol=5)
x <- c(5,6,7)

b[,3:5] <- t(t(b[,3:5])-x)

should do the trick. We subset the matrix to change only the part we need, and we use t() (transpose) to flip the matrix so simple vector recycling will take care of subtracting from the correct row.

If you want to avoid the transposed, you could do something like

b[,3:5] <- b[,3:5]-x[col(b[,3:5])]

as well. Here we subset twice, and we use the second to get the correct column for each value in x because both those matrices will index in the same order.

I think my favorite from the question that @thelatemail linked was

b[,3:5] <- sweep(b[,3:5], 2, x, `-`)
3
votes

Another way, with apply:

b[,3:5] <- t(apply(b[,3:5], 1, function(x) x-c))
2
votes

A simple solution:

b <- matrix(rep(1:20), nrow=4, ncol=5)
c <- c(5,6,7)

for(i in 1:nrow(b)) {
  b[i,3:5] <- b[i,3:5] - c
}
0
votes

This can be done with the rray package in a very satisfying way (using its (numpy-like) broadcasting - operator %b-%):

#install.packages("rray")
library(rray)

b <- matrix(rep(1:20), nrow=4, ncol=5)
x <- c(5, 6, 7)

b[, 3:5] <- b[, 3:5] %b-% matrix(x, 1)
b
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]    1    5    4    7   10
#> [2,]    2    6    5    8   11
#> [3,]    3    7    6    9   12
#> [4,]    4    8    7   10   13

For large matrices this is even faster than sweep:

#install.packages("bench")
res <- bench::press(
  size = c(10, 1000, 10000),
  frac_selected = c(0.1, 0.5, 1),
  {
  B <- matrix(sample(size*size), nrow=size, ncol=size)
  B2 <- B
  x <- sample(size, size=ceiling(size*frac_selected))
  idx <- sample(size, size=ceiling(size*frac_selected))

  bench::mark(rray = {B2[, idx] <- B[, idx, drop = FALSE] %b-% matrix(x, nrow = 1); B2}, 
              sweep = {B2[, idx] <- sweep(B[, idx, drop = FALSE], MARGIN = 2, x); B2}
  )
  }
)
plot(res)

benchmark results