Vectorizing an easy loop in R - r

Question

I am trying to improve my code, but there is a task where I am not able to vectorize a loop. I have 5 sparse matrices, where I have to loop through and add a 1x3-vector resulting from the non-null elements to another matrix. In this example I work with only 1 sparse matrix to keep it simple.
m <- matrix(0, nrow = 3, ncol = 3)
m[1, 2] <- 5
m[3, 3] <- 8
n <- matrix(0, nrow=3, ncol=5)
for (i in 1:nrow(m)){
for (j in 1:ncol(m)){
if (!m[i,j]==0){
n[i, j:(j+2)] <- n[i, j:(j+2)] + rep(m[i, j], 3)
}
}
}
The code works, but I have the feeling that there are much better solutions with vectorized functions. I tried and failed with apply functions as the new matrix n has other dimensions as m.
Would be great to get some ideas here.

Answer 1

Here's a vectorized base R solution:
Create the sequence of indices where you want your values to be inserted. Check ?sequence to understand how this works under the hood.
Insert the values
idx <- sequence(nvec = rep(3, length(m[m != 0])), from = which(m != 0), by = nrow(m))
n[idx] <- rep(m[m != 0], each = 3)
n
[,1] [,2] [,3] [,4] [,5]
[1,] 0 5 5 5 0
[2,] 0 0 0 0 0
[3,] 0 0 8 8 8

Answer 2

I'm not sure this is all that much better, but you could use which() to identify the row and column indices of the non-zero values in m, then use a bit of dplyr code to expand the list of column indices by the required length. Then use those indices and the values to replace the appropriate cells in the matrix.
library(dplyr)
library(tidyr)
m <- matrix(0, nrow = 3, ncol = 3)
m[1, 2] <- 5
m[3, 3] <- 8
n <- matrix(0, nrow=3, ncol=5)
w <- which(m != 0, arr.ind=TRUE)
w <- cbind(w, value=m[w])
w <- as_tibble(w)
w <- w %>%
rowwise() %>%
mutate(col = list(col:(col+2))) %>%
unnest(col)
n[cbind(w$row, w$col)] <- w$value
n
#> [,1] [,2] [,3] [,4] [,5]
#> [1,] 0 5 5 5 0
#> [2,] 0 0 0 0 0
#> [3,] 0 0 8 8 8
Created on 2023-02-10 by the reprex package (v2.0.1)

Answer 3

Nested loops can often be solved with mapply() or outer().
#Sample Data
m <- matrix(0, nrow = 3, ncol = 3)
m[1, 2] <- 5
m[3, 3] <- 8
n <- matrix(0, nrow=3, ncol=5)
#Build the function using the <<- operators, which are normally only used in functions, and cause a search to be made through parent environments for an existing definition of the variable being assigned
FN <- function (i, j){
if (m[i, j] != 0){
n[i, j:(j+2)] <<- n[i, j:(j+2)] + rep(m[i, j], 3)
}
}
#Outer() function to iterate, with mapply() to vectorize the if statement
outer(1:nrow(m), 1:ncol(m), FUN=function(x, y) mapply(FN, x, y))
n
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 5 5 5 0
#[2,] 0 0 0 0 0
#[3,] 0 0 8 8 8