Some timings of the different suggestions below
> library(microbenchmark)
> x <- matrix(1:20, nrow=10)
> s <- c(1,2)
> option1 <- sapply(1:2, function(i) x[,i]/s[i])
> option2 <- x %*% diag(1/s)
> option3 <- sweep(x, 2, s, '/')
> all.equal(option1,option2)
[1] TRUE
> all.equal(option2,option3)
[1] TRUE
> microbenchmark(sapply(1:2, function(i) x[,i]/s[i]), x %*% diag(1/s), sweep(x,
> 2, s, '/'))
Unit: microseconds
expr min lq mean median uq
max neval
sapply(1:2, function(i) x[, i]/s[i]) 18.425 19.8800 37.42147 21.5765 22.789
1583.518 100
x %*% diag(1/s) 1.456 2.1830 2.86150 2.9100 3.395
9.213 100
sweep(x, 2, s, "/") 32.486 34.1825 37.31964 35.3950 36.364
131.395 100
From: Eric Berger <ericjber...@gmail.com>
Sent: Wednesday, March 3, 2021 3:19 PM
To: Harold Doran <harold.do...@cambiumassessment.com>
Cc: Steven Yen <st...@ntu.edu.tw>; R-help Mailing List <r-help@r-project.org>
Subject: Re: [R] Column-by-column division
Why not use standard matrix multiplication which is straightforward here:
x %*% diag(1/s)
HTH,
Eric
On Wed, Mar 3, 2021 at 7:13 AM Harold Doran
<harold.do...@cambiumassessment.com<mailto:harold.do...@cambiumassessment.com>>
wrote:
To make sure the scalar is used instead of using the recycled vector s, maybe
like this
x <- matrix(1:20, nrow=10)
s <- c(1,2)
sapply(1:2, function(i) x[,i]/s[i])
-----Original Message-----
From: R-help
<r-help-boun...@r-project.org<mailto:r-help-boun...@r-project.org>> On Behalf
Of Steven Yen
Sent: Wednesday, March 3, 2021 6:00 AM
To: R-help Mailing List <r-help@r-project.org<mailto:r-help@r-project.org>>
Subject: [R] Column-by-column division
I have a 10 x 2 matrix x. Like to divide the first column by s[1] and second
column by s[2]. The following lines work but are clumsy. Any idea? Thanks.
> x
[,1] [,2]
[1,] 1 11
[2,] 2 12
[3,] 3 13
[4,] 4 14
[5,] 5 15
[6,] 6 16
[7,] 7 17
[8,] 8 18
[9,] 9 19
[10,] 10 20
> s
[1] 1 2
> t(t(x)/s)
[,1] [,2]
[1,] 1 5.5
[2,] 2 6.0
[3,] 3 6.5
[4,] 4 7.0
[5,] 5 7.5
[6,] 6 8.0
[7,] 7 8.5
[8,] 8 9.0
[9,] 9 9.5
[10,] 10 10.0
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