R: creating a matrix with an unknown number of rows

Question

I wrote the code below to generate a matrix containing something that is rather complicated for me. In this case, I determined that in the finished matrix there are 136 rows in the traffic jam and error.

I could write a function to calculate the number of rows of a matrix in advance, but the function would be a bit complicated. In this example, the number of rows in the matrix = ((4 * 3 + 1) + (3 * 3 + 1) + (2 * 3 + 1) + (1 * 3 + 1)) * 4.

Is there a simple and efficient way to create matrices in R without hard wiring the number of rows in a matrix expression? In other words, is there an easy way to let R just add a row to the matrix if necessary using for-loops?

I presented one solution that uses rbind on each pass through the loops, but this seems a bit confusing, and I was wondering if there could be a much easier solution.

Sorry if this question is redundant with an earlier question. I could not find a similar question using the search function on this site or using the Internet search engine today, although I think I found a similar question somewhere in the past.

Below are two sets of sample code, one of which uses rbind and the other where I used the trial version and error to set nrow = 136 in advance.

Thanks for any suggestions.

v1 <- 5 v2 <- 2 v3 <- 2 v4 <- (v1-1) my.matrix <- matrix(0, nrow=136, ncol=(v1+4) ) i = 1 for(a in 1:v2) { for(b in 1:v3) { for(c in 1:v4) { for(d in (c+1):v1) { if(d == (c+1)) ls = 4 else ls = 3 for(e in 1:ls) { my.matrix[i,c] = 1 if(d == (c+1)) my.matrix[i,d] = (e-1) else my.matrix[i,d] = e my.matrix[i,(v1+1)] = a my.matrix[i,(v1+2)] = b my.matrix[i,(v1+3)] = c my.matrix[i,(v1+4)] = d i <- i + 1 } } } } } my.matrix2 <- matrix(0, nrow=1, ncol=(v1+4) ) my.matrix3 <- matrix(0, nrow=1, ncol=(v1+4) ) i = 1 for(a in 1:v2) { for(b in 1:v3) { for(c in 1:v4) { for(d in (c+1):v1) { if(d == (c+1)) ls = 4 else ls = 3 for(e in 1:ls) { my.matrix2[1,c] = 1 if(d == (c+1)) my.matrix2[1,d] = (e-1) else my.matrix2[1,d] = e my.matrix2[1,(v1+1)] = a my.matrix2[1,(v1+2)] = b my.matrix2[1,(v1+3)] = c my.matrix2[1,(v1+4)] = d i <- i+1 if(i == 2) my.matrix3 <- my.matrix2 else my.matrix3 <- rbind(my.matrix3, my.matrix2) my.matrix2 <- matrix(0, nrow=1, ncol=(v1+4) ) } } } } } all.equal(my.matrix, my.matrix3) 

Answer 1

If you have an upper bound on the size of the matrix, you can create a matrix large enough to hold all the data

 my.matrix <- matrix(0, nrow=v1*v2*v3*v4*4, ncol=(v1+4) ) 

and truncate it at the end.

 my.matrix <- my.matrix[1:(i-1),] 

Answer 2

This is the general form for this. You can adapt it to your problem.

 matrix <- NULL for(...){ ... matrix <- rbind(matriz,vector) } 

where the vector contains string elements

Answer 3

I stumbled upon this solution today: convert matrix to data.frame . Since new lines are needed for-loop , these lines are automatically added to data.frame . You can then convert data.frame back to matrix at the end, if you want. I'm not sure if this seems like iterative use of rbind . Perhaps with big data.frames it becomes very slow. I dont know.

 my.data <- matrix(0, ncol = 3, nrow = 2) my.data <- as.data.frame(my.data) j <- 1 for(i1 in 0:2) { for(i2 in 0:2) { for(i3 in 0:2) { my.data[j,1] <- i1 my.data[j,2] <- i2 my.data[j,3] <- i3 j <- j + 1 } } } my.data my.data <- as.matrix(my.data) dim(my.data) class(my.data) 

EDIT: July 27, 2015

You can also delete the first matrix statement, create an empty data.frame , and then convert data.frame to matrix at the end:

 my.data <- data.frame(NULL,NULL,NULL) j <- 1 for(i1 in 0:2) { for(i2 in 0:2) { for(i3 in 0:2) { my.data[j,1] <- i1 my.data[j,2] <- i2 my.data[j,3] <- i3 j <- j + 1 } } } my.data my.data <- as.matrix(my.data) dim(my.data) class(my.data)