applying a function over a matrix 2N x M to produce a matrix N x M
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