OK, just for fun, I added a couple of *obvious* (or should have been)
approaches and got results that exactly agreed with Peter D's
comments: all O(n^2) results were essentially the same, and my matrix
multiplication O(n^3) solution was **far** worse. So I'm not sure
quite where Jeff's result came
> On 4 Nov 2016, at 17:35, peter dalgaard wrote:
>
> Notice though, that Bert loses (or _should_ lose) for larger values of N,
> since that method involves O(N^3) operations whereas the other two are
> O(N^2). I am a bit surprised that sweep() is so inefficient even at N=1000.
I also got diff
If you want to get a comparable result, then run the same code. You may not
want a comparable result though... your version straightens out the issue Peter
was puzzled by.
--
Sent from my phone. Please excuse my brevity.
On November 4, 2016 11:35:02 AM PDT, Berend Hasselman wrote:
>
>> On 4 N
I love R for it: if you experiment enough with t() you can find a way
to multiply almost anything by almost anything!
:)
On Thu, Nov 3, 2016 at 5:32 PM, Rui Barradas wrote:
> Hello,
>
> Take advantage that in R '*' recycles its arguments (the shorter one) and
> that the operation is performed col
Berend, et. al.:
"I don't quite understand why I get a very different ranking."
Nor do I. But, as Peter noted, my matrix multiplication solution
"should" be worst in terms of efficiency, and it clearly is. The other
two *should* be "similar", and they are. So your results are
consistent with what
> On 4 Nov 2016, at 19:27, Berend Hasselman wrote:
>
>>
>> On 4 Nov 2016, at 16:41, Jeff Newmiller wrote:
>>
>> Sara wins on memory use.
>>
>> Rui wins on speed.
>>
>> Bert wins on clarity.
>>
>> library(microbenchmark)
>>
>> N <- 1000
>> x <- matrix( runif( N*N ), ncol=N )
>> y <- seq.in
> On 4 Nov 2016, at 16:41, Jeff Newmiller wrote:
>
> Sara wins on memory use.
>
> Rui wins on speed.
>
> Bert wins on clarity.
>
> library(microbenchmark)
>
> N <- 1000
> x <- matrix( runif( N*N ), ncol=N )
> y <- seq.int( N )
>
> microbenchmark( { t( y * t(x) ) }
> , { x %*% d
Notice though, that Bert loses (or _should_ lose) for larger values of N, since
that method involves O(N^3) operations whereas the other two are O(N^2). I am a
bit surprised that sweep() is so inefficient even at N=1000.
-pd
On 04 Nov 2016, at 16:41 , Jeff Newmiller wrote:
> Sara wins on mem
Sara wins on memory use.
Rui wins on speed.
Bert wins on clarity.
library(microbenchmark)
N <- 1000
x <- matrix( runif( N*N ), ncol=N )
y <- seq.int( N )
microbenchmark( { t( y * t(x) ) }
, { x %*% diag( y ) }
, { sweep( x, 2, y, `*` ) }
)
Unit: milli
Nice!
Thanks a lot, everybody!
Dimitri
On Fri, Nov 4, 2016 at 10:35 AM, Bert Gunter wrote:
> 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
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
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
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
Hello,
Take advantage that in R '*' recycles its arguments (the shorter one)
and that the operation is performed column-wise:
t(y * t(x))
Hope this helps,
Rui Barradas
Em 03-11-2016 21:05, Dimitri Liakhovitski escreveu:
Hello!
I have a matrix x and a vector y:
x <- matrix(1:6, ncol = 2)
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
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