[R] applying a function over a matrix 2N x M to produce a matrix N x M

[David Winsemius]

On Apr 22, 2011, at 2:38 PM, Richard M. Heiberger wrote:

> I like David's answer and it can be made much faster.
> I show three refinements, each faster than the preceding one.
>
> Rich
>
> > system.time(for (i in 1:1000)
> +   mat[seq(1, nrow(mat), by=2), ]+mat[seq(2, nrow(mat), by=2), ]
> + )
>    user  system elapsed
>    0.18    0.00    0.19
> >
> > system.time(for (i in 1:1000)
> +   mat[seqn <- seq(1, nrow(mat), by=2), ]+mat[seqn+1, ]
> + )
>    user  system elapsed
>    0.08    0.00    0.08

Yawn. Who cares about doubling speed?

> >
> > system.time(for (i in 1:1000)
> +   mat[seqn <- seq(1, length=nrow(mat)/2, by=2), ]+mat[seqn+1, ]
> + )
>    user  system elapsed
>    0.05    0.00    0.05
> >
> > system.time(for (i in 1:1000)
> +   {mat2 <- mat; dim(mat2) <- c(2,3,6); mat2[1,,]+mat2[2,,]}
> + )

Strong work, Richard. I like this answer much better than mine and I  
think it is truly novel. It does what I had thought should be  
possible, but I abandoned that effort and now see my brain was only  
working in two dimensions. Here you are now thinking "inside a  
box" ...  except the box is now 3-dimensional!

-- 
David.

>    user  system elapsed
>    0.01    0.00    0.02
> >
>
> On Fri, Apr 22, 2011 at 12:28 PM, David Winsemius <dwinsemius at comcast.net 
> > wrote:
>
> On Apr 22, 2011, at 12:13 PM, Christine SINOQUET wrote:
>
> Hello,
>
> mat1 only consists of 0s and 1s:
> 0 0 1 0 0 0
> 1 1 0 1 1 0
> 1 1 1 0 1 0
> 0 1 1 0 0 1
> 1 0 0 1 0 0
> 0 1 0 1 0 1
>
> N = 3
> M = 6
>
> I would like to "compress" mat1 every two rows, applying summation  
> over the two rows (per column), at each step, to yield:
>
> mat2
> 1 1 1 1 1 0
> 1 2 2 0 1 1
> 1 1 0 2 0 1
>
> > mat[seq(1, nrow(mat), by=2), ]+mat[seq(2, nrow(mat), by=2), ]
>     [,1] [,2] [,3] [,4] [,5] [,6]
> [1,]    1    1    1    1    1    0
> [2,]    1    2    2    0    1    1
> [3,]    1    1    0    2    0    1
>
>
>

David Winsemius, MD
West Hartford, CT