Very interesting - thanks! Most of my problems are not limited by
compute speed, but its clear that for some sorts of compute-intensive
problems, sweep might be a limiting approach.
On 2/29/2024 6:12 PM, Richard M. Heiberger wrote:
> I decided to do a direct comparison of transpose and sweep.
>
I added two more rows
library(microbenchmark)
NN <- matrix(c(1, 2, 3, 4, 5, 6), nrow = 2, byrow = TRUE) # Example matrix
lambda <- c(2, 3, 4) # Example vector
colNN <- t(NN)
matlam <- matrix(lambda, byrow=TRUE, nrow=2, ncol=3)
microbenchmark(
sweep = sweep(NN, 2, lambda, "/"),
I decided to do a direct comparison of transpose and sweep.
library(microbenchmark)
NN <- matrix(c(1, 2, 3, 4, 5, 6), nrow = 2, byrow = TRUE) # Example matrix
lambda <- c(2, 3, 4) # Example vector
colNN <- t(NN)
microbenchmark(
sweep = sweep(NN, 2, lambda, "/"),
transpose =
Many thanks for the collective answers -- consider this a thank you to
the group. I had 'guessed' it had something to do with 'columns then
rows' or vice versa (MATLAB convention vs R convention), but had never
heard about 'sweep' before. Most of the time when I run into 'matrix
orientation'
On Tue, 27 Feb 2024 21:37:52 +
"Richard M. Heiberger" wrote:
> > t(t(NN)/lambda)
> [,1] [,2] [,3]
> [1,] 0.5 0.667 0.75
> [2,] 2.0 1.667 1.50
> >
>
> R matrices are column-based. MATLAB matrices are row-based.
It might depend on what you mean with this statement,
> t(t(NN)/lambda)
[,1] [,2] [,3]
[1,] 0.5 0.667 0.75
[2,] 2.0 1.667 1.50
>
R matrices are column-based. MATLAB matrices are row-based.
> On Feb 27, 2024, at 14:54, Evan Cooch wrote:
>
> So, trying to convert a very long, somewhat technical bit of lin alg
> MATLAB code to R.
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