Dear R-users,
I'm struggling in R in order to "squeeze" a matrix without using a
for-loop.
Although my case is a bit more complex, the following example should
help you to understand what I would like to do, but without the slow
for-loop.
Thanks in advance,
Carlo Giovanni Camarda
A <- matrix(1:54, ncol=6) # my original matrix
A.new <- matrix(nrow=3, ncol=6) # a new matrix which I'll fill
# for-loop
for(i in 1:nrow(A.new)){
B <- A[i:(i+2), ] # selecting the rows
C <- apply(B,2,sum) # summing by columns
A.new[i,] <- C # inserting in the new matrix
}
+++++
This mail has been sent through the MPI for Demographic Rese...{{dropped}}
The result is linear in A so its a matter of finding the matrix to multiply it by: matrix(c(rep(1,3), rep(0,7)), 3, 9, byrow = TRUE) %*% A On 1/30/06, Camarda, Carlo Giovanni <Camarda at demogr.mpg.de> wrote:> Dear R-users, > I'm struggling in R in order to "squeeze" a matrix without using a > for-loop. > Although my case is a bit more complex, the following example should > help you to understand what I would like to do, but without the slow > for-loop. > Thanks in advance, > Carlo Giovanni Camarda > > > A <- matrix(1:54, ncol=6) # my original matrix > A.new <- matrix(nrow=3, ncol=6) # a new matrix which I'll fill > # for-loop > for(i in 1:nrow(A.new)){ > B <- A[i:(i+2), ] # selecting the rows > C <- apply(B,2,sum) # summing by columns > A.new[i,] <- C # inserting in the new matrix > } > > > +++++ > This mail has been sent through the MPI for Demographic Rese...{{dropped}} > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html >
use 'filter':> x <- matrix(1:100,10) > x[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 1 11 21 31 41 51 61 71 81 91 [2,] 2 12 22 32 42 52 62 72 82 92 [3,] 3 13 23 33 43 53 63 73 83 93 [4,] 4 14 24 34 44 54 64 74 84 94 [5,] 5 15 25 35 45 55 65 75 85 95 [6,] 6 16 26 36 46 56 66 76 86 96 [7,] 7 17 27 37 47 57 67 77 87 97 [8,] 8 18 28 38 48 58 68 78 88 98 [9,] 9 19 29 39 49 59 69 79 89 99 [10,] 10 20 30 40 50 60 70 80 90 100> (y <- apply(x, 2, filter, c(1,1,1)))[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] NA NA NA NA NA NA NA NA NA NA [2,] 6 36 66 96 126 156 186 216 246 276 [3,] 9 39 69 99 129 159 189 219 249 279 [4,] 12 42 72 102 132 162 192 222 252 282 [5,] 15 45 75 105 135 165 195 225 255 285 [6,] 18 48 78 108 138 168 198 228 258 288 [7,] 21 51 81 111 141 171 201 231 261 291 [8,] 24 54 84 114 144 174 204 234 264 294 [9,] 27 57 87 117 147 177 207 237 267 297 [10,] NA NA NA NA NA NA NA NA NA NA> y[-c(1, nrow(y)),][,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 6 36 66 96 126 156 186 216 246 276 [2,] 9 39 69 99 129 159 189 219 249 279 [3,] 12 42 72 102 132 162 192 222 252 282 [4,] 15 45 75 105 135 165 195 225 255 285 [5,] 18 48 78 108 138 168 198 228 258 288 [6,] 21 51 81 111 141 171 201 231 261 291 [7,] 24 54 84 114 144 174 204 234 264 294 [8,] 27 57 87 117 147 177 207 237 267 297>On 1/30/06, Camarda, Carlo Giovanni <Camarda@demogr.mpg.de> wrote:> > Dear R-users, > I'm struggling in R in order to "squeeze" a matrix without using a > for-loop. > Although my case is a bit more complex, the following example should > help you to understand what I would like to do, but without the slow > for-loop. > Thanks in advance, > Carlo Giovanni Camarda > > > A <- matrix(1:54, ncol=6) # my original matrix > A.new <- matrix(nrow=3, ncol=6) # a new matrix which I'll fill > # for-loop > for(i in 1:nrow(A.new)){ > B <- A[i:(i+2), ] # selecting the rows > C <- apply(B,2,sum) # summing by columns > A.new[i,] <- C # inserting in the new matrix > } > > > +++++ > This mail has been sent through the MPI for Demographic Rese...{{dropped}} > > ______________________________________________ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html >-- Jim Holtman Cincinnati, OH +1 513 247 0281 What the problem you are trying to solve? [[alternative HTML version deleted]]
Hi> matrix(colSums(embed(A,3)[1:3,]),3,6, byrow=T)[3:1,][,1] [,2] [,3] [,4] [,5] [,6] [1,] 6 33 60 87 114 141 [2,] 9 36 63 90 117 144 [3,] 12 39 66 93 120 147 will do it. However I am not sure if it is quicker than your for loop. HTH Petr On 30 Jan 2006 at 16:29, Camarda, Carlo Giovanni wrote: Date sent: Mon, 30 Jan 2006 16:29:38 +0100 From: "Camarda, Carlo Giovanni" <Camarda at demogr.mpg.de> To: <r-help at stat.math.ethz.ch> Subject: [R] Subsetting a matrix without for-loop> Dear R-users, > I'm struggling in R in order to "squeeze" a matrix without using a > for-loop. Although my case is a bit more complex, the following > example should help you to understand what I would like to do, but > without the slow for-loop. Thanks in advance, Carlo Giovanni Camarda > > > A <- matrix(1:54, ncol=6) # my original matrix > A.new <- matrix(nrow=3, ncol=6) # a new matrix which I'll fill > # for-loop > for(i in 1:nrow(A.new)){ > B <- A[i:(i+2), ] # selecting the rows > C <- apply(B,2,sum) # summing by columns > A.new[i,] <- C # inserting in the new matrix > } > > > +++++ > This mail has been sent through the MPI for Demographic > Rese...{{dropped}} > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.htmlPetr Pikal petr.pikal at precheza.cz