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)
In R, one column is created by default for a matrix, therefore, to create a matrix without a column we can use ncol =0.
To create a matrix in R you need to use the function called matrix(). The arguments to this matrix() are the set of elements in the vector. You have to pass how many numbers of rows and how many numbers of columns you want to have in your matrix. Note: By default, matrices are in column-wise order.
A matrix having at least one dimension equal to zero is called an empty matrix. The simplest empty matrix is 0-by-0 in size. Examples of more complex matrices are those of dimension 0 -by- 5 or 10 -by- 0 -by- 20. To create a 0-by-0 matrix, use the square bracket operators with no value specified.
The nrow() function in R programming R provides us nrow() function to get the rows for an object. That is, with nrow() function, we can easily detect and extract the number of rows present in an object that can be matrix, data frame or even a dataset.
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),]
This is the generic form to do it. You can adapt it to your problem
matrix <- NULL
for(...){
...
matrix <- rbind(matriz,vector)
}
where vector contains the row elements
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