I have a preallocated matrix in R and need to fill it with data from a dataframe generated from a file having following format:
1-1-7
1-3-2
2-2-6
1-4-8
....
where the first column contains a row index, the second a column index, and the 3rd contains the values.
Is there a faster/better way then the following loop?
for(i in 1:nrow(file)){
matrix[file[i,1],file[i,2]]=file[i,3]
}
Use row/col index as the first and second columns of file and assign the file[,3] to m1 using those index.
m1[as.matrix(file[1:2])] <- file[,3]
m1
# [,1] [,2] [,3] [,4]
#[1,] 7 0 2 8
#[2,] 0 6 0 0
As I mentioned above, we assign the values based on the row/column index. If we take the first two columns and convert to matrix, the first column acts as the row index and second as the column index
as.matrix(file[1:2])
# v1 v2
#[1,] 1 1
#[2,] 1 3
#[3,] 2 2
#[4,] 1 4
If we subset the m1 based on the first row i.e. 1 1, we are getting the element of m1 at row=1 and column=1, similarly for the 2nd row, it would be the element at row=1 and column=3, and so on.. Here, we created a matrix of 0 with the specified dimensions. So, the 4 values based on the row/columns positions will be 0
m1[as.matrix(file[1:2])]
#[1] 0 0 0 0
Now, we are changing/replacing the values in 'm1' at the specified row/column positions by assigning the value to the 3rd column of 'file'
m1[as.matrix(file[1:2])] <- file[,3]
If we check the values at the specified position, it will be replaced
m1[as.matrix(file[1:2])]
#[1] 7 2 6 8
Or we can use another approach using sparseMatrix from library(Matrix)
library(Matrix)
sM <- sparseMatrix(file[,1], file[,2], x=file[,3])
as.matrix(sM)
# [,1] [,2] [,3] [,4]
#[1,] 7 0 2 8
#[2,] 0 6 0 0
file <- structure(list(v1 = c(1L, 1L, 2L, 1L), v2 = c(1L, 3L, 2L, 4L),
v3 = c(7, 2, 6, 8)), .Names = c("v1", "v2", "v3"),
class = "data.frame", row.names = c(NA, -4L))
m1 <- matrix(0, nrow=max(file[,1]), ncol=max(file[,2]))