R: Matrix by Vector Multiplication

R: matrix by vector multiplication

The %*% operator in R does matrix multiplication:

> mymat %*% myvec
      [,1]
 [1,]  116
 [2,]  122
 ...
[10,]  170

Matrix and vector multiplication operation in R

See ?`%*%`:

Description:

 Multiplies two matrices, if they are conformable.  If one argument
 is a vector, it will be promoted to either a row or column matrix
 to make the two arguments conformable.  If both are vectors of the
 same length, it will return the inner product (as a matrix).

Multiply rows of matrix by vector?

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.

R: multiply vector by matrix according to names

You can use subsetting to create a prices vector in the correct order:

pricesm <- prices[rownames(M)]
pricesm
#banana       mango pomegranate       apple 
#  0.42        1.24        2.25        0.85 

rev <- pricesm %*% M
rev
#          M     Tu      W     Th      F
#[1,] 310.44 170.93 243.72 189.85 315.22

Looping over a matrix - vector multiplication, with elements of both changing with each round

a <- matrix(1:36, nrow = 6, byrow = TRUE)
b <- matrix(rep(1, times =  36), nrow = 6, byrow = TRUE)
b1 <- b
b1[1:3,1:3] <- 0
b2 <- b
b2[4:6,4:6] <- 0

my loop codes...

for (i in seq(6)){
temp = rep(0, 6)
temp[i] = 1
assign(paste('c', i, sep = ''), temp)
}

for (i in seq(6)){ 
cx = get(paste('c', i, sep = ''))
if(i<4){
  px = sum(colSums(b1 * diag(as.vector(a %*% cx), nrow = 6)))
  assign(paste('d', i, sep = ''), px)
} else{
  px = sum(colSums(b2 * diag(as.vector(a %*% cx), nrow = 6)))
  assign(paste('d', i, sep = ''), px)
}
}

then the results

results <- cbind(d1, d2, d3, d4, d5, d6)

Nice way to multiply each row of a matrix by a vector in r

I think this is what you are looking for...

t( t(mat) * vec )
     [,1] [,2] [,3] [,4]
[1,]    1    2    3    4
[2,]    1    2    3    4

* like most other operators in R is vectorised. The t is necessary because R will recycle column-wise. The apply solution is:

t( apply( mat , 1 , `*` , vec ) )
     [,1] [,2] [,3] [,4]
[1,]    1    2    3    4
[2,]    1    2    3    4

Fastest way to multiply matrix columns with vector elements in R

Use some linear algebra and perform matrix multiplication, which is quite fast in R.

eg

m %*% diag(v)

some benchmarking

m = matrix(rnorm(1200000), ncol=6)

 v=c(1.5, 3.5, 4.5, 5.5, 6.5, 7.5)
library(microbenchmark)
microbenchmark(m %*% diag(v), t(t(m) * v))
##    Unit: milliseconds
##            expr      min       lq   median       uq      max neval
##   m %*% diag(v) 16.57174 16.78104 16.86427 23.13121 109.9006   100
##     t(t(m) * v) 26.21470 26.59049 32.40829 35.38097 122.9351   100

Multiply elements of a matrix with vector values

First you need to coerce coords to a matrix for indexing, then reverse the column order. Then it's just a simple lapply() loop.

coords <- as.matrix(coords)[, 2:1]
lapply(multis, function(x) {
    M[coords] <- M[coords] * x
    M
})

resulting in

[[1]]
     [,1] [,2] [,3] [,4] [,5]
[1,]  0.1    5    9 13.0 17.0
[2,]  2.0    6   10  1.4 18.0
[3,]  3.0    7   11 15.0  1.9
[4,]  4.0    8   12 16.0 20.0

[[2]]
     [,1] [,2] [,3] [,4] [,5]
[1,]    2    5    9   13   17
[2,]    2    6   10   28   18
[3,]    3    7   11   15   38
[4,]    4    8   12   16   20

[[3]]
     [,1] [,2] [,3] [,4] [,5]
[1,]  100    5    9   13   17
[2,]    2    6   10 1400   18
[3,]    3    7   11   15 1900
[4,]    4    8   12   16   20

Fastest way for multiplying a matrix to a vector

sweep seems to run a bit faster on my machine

sweep(mat, 2, v, FUN = "*")

Some benchmarks:

> microbenchmark(mat %*% diag(v),sweep(mat, 2, v, FUN = "*"))

Unit: milliseconds
  expr       min        lq   median        uq      max neval
   %*% 214.66700 226.95551 231.2366 255.78493 349.1911   100
 sweep  42.42987  44.72254  62.9990  70.87403 127.2869   100

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