Question

I have written the code below to generate a matrix containing what is, to me, a fairly complex pattern. In this case I determined that there are 136 rows in the finished matrix by trial and error.

I could write a function to calculate the number of matrix rows in advance, but the function would be a little complex. 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 an easy and efficient way to create matrices in R without hard-wiring the number of rows in the matrix statement? In other words, is there an easy way to let R simply add a row to a matrix as needed when using for-loops?

I have presented one solution that employs rbind at each pass through the loops, but that seems a little convoluted and I was wondering if there might be a much easier solution.

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

Below are 2 sets of example code, one using rbind and the other where I used trial 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)) l.s = 4 
        else           l.s = 3

        for(e in 1:l.s) {

          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)) l.s = 4 
        else           l.s = 3

        for(e  in 1:l.s) {

          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 some 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),]