Multiply rows of matrix by vector?
RVectorMatrixMultiplicationR Problem Overview
I have a numeric matrix with 25 columns and 23 rows, and a vector of length 25. How can I multiply each row of the matrix by the vector without using a for loop?
The result should be a 25x23 matrix (the same size as the input), but each row has been multiplied by the vector.
Added reproducible example from @hatmatrix's answer:
matrix <- matrix(rep(1:3,each=5),nrow=3,ncol=5,byrow=TRUE)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 2 2 2 2 2
[3,] 3 3 3 3 3
vector <- 1:5
Desired output:
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[2,] 2 4 6 8 10
[3,] 3 6 9 12 15
R Solutions
Solution 1 - R
I think you're looking for sweep().
# Create example data and vector
mat <- matrix(rep(1:3,each=5),nrow=3,ncol=5,byrow=TRUE)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 2 2 2 2 2
[3,] 3 3 3 3 3
vec <- 1:5
# Use sweep to apply the vector with the multiply (`*`) function
# across columns (See ?apply for an explanation of MARGIN)
sweep(mat, MARGIN=2, vec, `*`)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 5
[2,] 2 4 6 8 10
[3,] 3 6 9 12 15
It's been one of R's core functions, though improvements have been made on it over the years.
Solution 2 - R
> MyMatrix <- matrix(c(1,2,3, 11,12,13), nrow = 2, ncol=3, byrow=TRUE)
> MyMatrix
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 11 12 13
> MyVector <- c(1:3)
> MyVector
[1] 1 2 3
You could use either:
> t(t(MyMatrix) * MyVector)
[,1] [,2] [,3]
[1,] 1 4 9
[2,] 11 24 39
or:
> MyMatrix %*% diag(MyVector)
[,1] [,2] [,3]
[1,] 1 4 9
[2,] 11 24 39
Solution 3 - R
Actually, sweep is not the fastest option on my computer:
MyMatrix <- matrix(c(1:1e6), ncol=1e4, byrow=TRUE)
MyVector <- c(1:1e4)
Rprof(tmp <- tempfile(),interval = 0.001)
t(t(MyMatrix) * MyVector) # first option
Rprof()
MyTimerTranspose=summaryRprof(tmp)$sampling.time
unlink(tmp)
Rprof(tmp <- tempfile(),interval = 0.001)
MyMatrix %*% diag(MyVector) # second option
Rprof()
MyTimerDiag=summaryRprof(tmp)$sampling.time
unlink(tmp)
Rprof(tmp <- tempfile(),interval = 0.001)
sweep(MyMatrix ,MARGIN=2,MyVector,`*`) # third option
Rprof()
MyTimerSweep=summaryRprof(tmp)$sampling.time
unlink(tmp)
Rprof(tmp <- tempfile(),interval = 0.001)
t(t(MyMatrix) * MyVector) # first option again, to check order
Rprof()
MyTimerTransposeAgain=summaryRprof(tmp)$sampling.time
unlink(tmp)
MyTimerTranspose
MyTimerDiag
MyTimerSweep
MyTimerTransposeAgain
This yields:
> MyTimerTranspose
[1] 0.04
> MyTimerDiag
[1] 40.722
> MyTimerSweep
[1] 33.774
> MyTimerTransposeAgain
[1] 0.043
On top of being the slowest option, the second option reaches the memory limit (2046 MB). However, considering the remaining options, the double transposition seems a lot better than sweep in my opinion.
Edit
Just trying smaller data a repeated number of times:
MyMatrix <- matrix(c(1:1e3), ncol=1e1, byrow=TRUE)
MyVector <- c(1:1e1)
n=100000
[...]
for(i in 1:n){
# your option
}
[...]
> MyTimerTranspose
[1] 5.383
> MyTimerDiag
[1] 6.404
> MyTimerSweep
[1] 12.843
> MyTimerTransposeAgain
[1] 5.428
Solution 4 - R
For speed one may create matrix from the vector before multiplying
mat <- matrix(rnorm(1e6), ncol=1e4)
vec <- c(1:1e4)
mat * matrix(vec, dim(mat)[1], length(vec))
library(microbenchmark)
microbenchmark(
transpose = t(t(mat) * vec),
make_matrix = mat * matrix(vec, dim(mat)[1], length(vec), byrow = TRUE),
sweep = sweep(mat,MARGIN=2,vec,`*`))
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# transpose 9.940555 10.480306 14.39822 11.210735 16.19555 77.67995 100 b
#make_matrix 5.556848 6.053933 9.48699 6.662592 10.74121 74.14429 100 a
# sweep 8.033019 8.500464 13.45724 12.331015 14.14869 77.00371 100 b
Solution 5 - R
These solutions using outer() or collapse::TRA() are is significantly faster than anything suggested here.
Solution 6 - R
If you want speed, you can use Rfast::eachrow. It is the fastest from all...