My goodness!
> x %*% diag(y)
[,1] [,2]
[1,] 2 12
[2,] 4 15
[3,] 6 18
will do.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Thu, Nov 3, 2016 at 2:33 PM, Sarah Goslee <sarah.gos...@gmail.com> wrote:
> Like this?
>
>> sweep(x, 2, y, "*")
> [,1] [,2]
> [1,] 2 12
> [2,] 4 15
> [3,] 6 18
>>
>
>
> On Thu, Nov 3, 2016 at 5:05 PM, Dimitri Liakhovitski
> <dimitri.liakhovit...@gmail.com> wrote:
>> Hello!
>>
>> I have a matrix x and a vector y:
>>
>> x <- matrix(1:6, ncol = 2)
>> y <- c(2,3)
>>
>> I need to multiply the first column of x by 2 (y[1]) and the second
>> column of x by 3 (y[2]).
>>
>> Of course, I could do this - but it's column by column:
>>
>> x[,1] <- x[,1] * y[1]
>> x[,2] <- x[,2] * y[2]
>> x
>>
>> Or I could repeat each element of y and multiply two matrices - that's
>> better:
>>
>> rep.row<-function(x,n){
>> matrix(rep(x,each=n),nrow=n)
>> }
>> y <- rep.row(y, nrow(x))
>> x * y
>>
>> However, maybe there is a more elegant r-like way of doing it?
>> Thank you!
>>
>> --
>> Dimitri Liakhovitski
>>
>
> --
> Sarah Goslee
> http://www.functionaldiversity.org
>
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.
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