Multiply rows of matrix by vector?

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
86338 次浏览
小开 
> 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
小开
最佳答案

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.

小开

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
小开

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
小开

These solutions using outer() or collapse::TRA() are is significantly faster than anything suggested here.

小开

If you want speed, you can use Rfast::eachrow. It is the fastest from all...


提交答案