Almost there; you need the transpose of v, since Bill originally had columns changing faster: e.g. x <- pt$x[t(ver)]
-----Original Message----- From: Marc Schwartz <[EMAIL PROTECTED]> Sent: Apr 22, 2005 9:17 AM To: Bill Simpson <[EMAIL PROTECTED]> Cc: R-Help <[email protected]> Subject: Re: [R] ugly loop On Fri, 2005-04-22 at 08:58 -0400, Bill Simpson wrote: > The following code is slow and ugly: > > count<-0 > for(i in 1:nrow(ver)) > for(j in 1:ncol(ver)) > { > count<-count+1 > x[count]<-pt$x[ver[i,j]] > y[count]<-pt$y[ver[i,j]] > z[count]<-pt$z[ver[i,j]] > } > > Please help me make it better. > > Thanks! The following should work: > ver <- matrix(sample(1:16, 16), ncol = 4) > pt <- data.frame(x = sample(1:16, 16), + y = sample(1:16, 16), + z = sample(1:16, 16)) > ver [,1] [,2] [,3] [,4] [1,] 8 9 5 13 [2,] 14 16 1 10 [3,] 12 2 11 7 [4,] 6 3 4 15 > pt x y z 1 6 15 15 2 9 2 3 3 11 1 5 4 14 4 10 5 13 7 14 6 1 14 7 7 15 10 4 8 10 5 12 9 4 12 2 10 8 8 13 11 16 11 1 12 7 13 9 13 2 16 11 14 3 9 16 15 5 6 8 16 12 3 6 > x <- pt$x[ver] > y <- pt$y[ver] > z <- pt$z[ver] > x [1] 10 3 7 1 4 12 9 11 13 6 16 14 2 8 15 5 > y [1] 5 9 13 14 12 3 2 1 7 15 11 4 16 8 10 6 > z [1] 12 16 9 7 2 6 3 5 14 15 1 10 11 13 4 8 Keep in mind that a matrix is a vector with dims, so you can fill a vector from the matrix simply by doing the indexing with a single value, which will do the fill indexed column by column. HTH, Marc Schwartz ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
