On 2012-07-08 06:57, fabiano wrote:
Thanks Peter.We had a look at both Hab and habitat. These are integers representing habitat types. habitat <- read.csv("Ungulate_vegetation.csv") habitat <- habitat[,3] habitat [1] 3 3 4 3 3 3 4 4 3 3 3 3 3 4 2 3 2 3 2 3 Hab <- cbind(seq(1,20),habitat) Hab habitat [1,] 1 3 [2,] 2 3 [3,] 3 4 [4,] 4 3 [5,] 5 3 [6,] 6 3 [7,] 7 4 [8,] 8 4 [9,] 9 3 [10,] 10 3 [11,] 11 3 [12,] 12 3 [13,] 13 3 [14,] 14 4 [15,] 15 2 [16,] 16 3 [17,] 17 2 [18,] 18 3 [19,] 19 2 [20,] 20 3 hab [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] NA NA NA NA NA NA NA [2,] NA NA NA NA NA NA NA for(i in 1:2){ hab[i, ] <- Hab[habitat==i,1] } The idea is to assign habitat types to hab.
I don't understand how this relates to a 2-by-7 matrix, but it's easy to see where your problem with the loop is. Try this and see if it sheds light: hab[1,] <- Hab[habitat==1,1] #Error in hab[1, ] <- Hab[habitat == 1, 1] : replacement has length zero hab[2,] <- Hab[habitat==2,1] #Error in hab[2, ] <- Hab[habitat == 2, 1] : number of items to replace is not a multiple of replacement length Now do: Hab[habitat==1,1] ## how many 1s are in habitat? Hab[habitat==2,1] ## how many 2s are in habitat? I think that you may have to rethink your approach. Peter Ehlers
Thanks -- View this message in context: http://r.789695.n4.nabble.com/replacement-has-length-zero-tp4635700p4635777.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
