dividing first row of each matrix in list of matrices by the lagged sum of all rows in each matrix of a list matrices

Question

I have a list of 100 - 3 x 51 matrices in r and would like to divide the first row of each matrix in the list by the lagged (n=2) sum of all rows in each corresponding matrix iteratively corresponding with the lag. I know how to achieve this within the same row within a list of vectors with the following code

Example_List2 <- lapply(Example_List1,function(x){x/lag(x,n=2)})

My attempt at the 3 row list is coded below. I would ultimately like to make this the replacement first row of a new DB and repeat this process for each row with dif lags. My attempted code is

List2 <- List1
lapply(List2, `[`,1,) <- lapply(List1,function(x){lapply(x, `[`,1,)/lag(colSums(x),n=2)})
lapply(List2, `[`,2,) <- lapply(List1,function(x){lapply(x, `[`,2,)/lag(colSums(x),n=3)})
lapply(List2, `[`,3,) <- lapply(List1,function(x){lapply(x, `[`,3,)/lag(colSums(x),n=4)})

Answer 1

We may use

library(rbindlist)
List2 <- lapply(List1, \(x) x/do.call(rbind, shift(colSums(x), n = 2:4)))

---

For the second case

List3 <- lapply(List1, \(x) { 
      n1 <- 2:4
     x1 <- colSums(x)
     x2 <- x
   for(i in seq_along(n1)) {
      x2[i,] <- shift(x[i,], n = n1[i], type = "lead")}
        colSums(x2)/x1
})

data

set.seed(24)
List1 <- replicate(100, matrix(rnorm(3*51), nrow = 3), simplify = FALSE)