replacing specific positional value in each matrix within a list, with sequential values from a vect-CodePudding

I am attempting to replace a specific value (by position) for each matrix in my list with each sequential value in a vector. From the code below I would like to replace the value 2 (by position, so no simple replace all 2's with...) in each of my matrices within the list, with each value from one.to.two.s which is a vector of numbers. As an extension, I would like to be able repeat the one.to.two.s sequence if the vector had say length 50 and the list was say length 100. I have a forloop here which doesn't work, but believe this could handled with lapply somehow as well. Thanks in advance for the help.

 A <- lapply(1:50, function(x)  # construct list of matrices
    matrix(c(0, 0, 0, 0,
             2, 0, 0, 0,
             0, 0, 0, 0,
             0, 0, 0, 1), nrow = 4,ncol=4, byrow = TRUE, ))
  Anew <-A

  one.to.two.s <- c(seq(from = 0.40, to = 0.89,by=0.01))
  
  for(t in 1:length(Anew)) {
    Anew[[t]][2,1] <- one.to.two.s
  }

CodePudding user response：

You can try the following for loop if you have longer list than the vector

for(t in 1:length(Anew)) {
    Anew[[t]][2,1] <- one.to.two.s[(t-1)%%length(one.to.two.s) 1]
  }

CodePudding user response：

I forgot to add [t] to the end of my replacement as well. Also can repeat a vector ahead of time.

for(t in 1:length(Anew)) {
    Anew[[t]][2,1] <- one.to.two.s
  }

instead becomes

for(t in 1:length(Anew)) {
    Anew[[t]][2,1] <- one.to.two.s[t]
  }

CodePudding user response：

Using an example one.to.two.s which is shorter than length(A), you could use rep with length.out to make it the correct length, and then Map over that vector and A to create Anew

one.to.two.s <- seq(from = 0.4, to = 0.8, by = 0.01)

Anew <- Map(function(A, x) {
  A[2, 1] <- x
  A
}, A, rep(one.to.two.s, length.out = length(A)))

^{Created on 2022-01-27 by the reprex package (v2.0.1)}

CodePudding user response：

I believe this is what you are looking for.

# use replicate() instead of lapply()
B <- 50L
A <- replicate(B*2.1,
               matrix(c(0, 0, 0, 0,
                        2, 0, 0, 0,
                        0, 0, 0, 0,
                        0, 0, 0, 1), nrow = 4,ncol=4, byrow = TRUE),
               simplify = FALSE)
Anew <- A
one.to.two.s <- seq(from = 0.40, to = 0.89, by = 0.01)

# loop over all elements in Anew
for (t in seq_along(Anew)) {
  Anew[[t]][2,1] <- one.to.two.s[
    seq_len(length(Anew)   2L) %% (length(one.to.two.s)   1L)
  ][t]
}

# > head(sapply(Anew, '[', 2))
# [1] 0.40 0.41 0.42 0.43 0.44 0.45

# > tail(sapply(Anew, '[', 2))
# [1] 0.89 0.40 0.41 0.42 0.43 0.44