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 <dwinsem...@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
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