每两行生成一个升序矩阵，R中没有for循环

Question

I'm trying to generate this matrix without using a for loop (to speed up the processing time):

       Module 1 Module 2 Module 3
y0       20       20       20
y1       20       20       20
y0       40       40       40
y1       40       40       40
y0       60       60       60
y1       60       60       60
y0       80       80       80
y1       80       80       80
y0      100      100      100
y1      100      100      100
y0      120      120      120
y1      120      120      120
y0      140      140      140
y1      140      140      140
y0      160      160      160
y1      160      160      160
y0      180      180      180
y1      180      180      180
y0      200      200      200
y1      200      200      200

I have been attempting the apply functions, replicate() and do.call(). I've ended up with this code:

a = 10          # Number of increments
n1 = 100       # Sample size
m.fn <- function(n1, a) {
  b <- matrix(rep(n1), nrow = 2, ncol = 3)
  rownames(b) <- c("y0", "y1")
  c <- do.call(rbind, replicate(a, b, simplify=FALSE))

  colnames(c) <- c("Module 1", "Module 2", "Module 3")
  return(c)
}

Which produces a matrix similar to above, but with all entries as 20.

I've tried replacing the values with little success, as per below:

 a = 10          # Number of increments
 n1 = 100       # Sample size
    m.fn <- function(n1, a) {
      b <- matrix(rep(n1), nrow = 2, ncol = 3)
      rownames(b) <- c("y0", "y1")
      c <- do.call(rbind, replicate(a, b, simplify=FALSE))
      c[3:(2 * a),] <- (n1 * seq(3, 2 * a))
      colnames(c) <- c("Module 1", "Module 2", "Module 3")
      return(c)
    }

This is what results:

       Module 1 Module 2 Module 3
y0       20       20       20
y1       20       20       20
y0       60       60       60
y1       80       80       80
y0      100      100      100
y1      120      120      120
y0      140      140      140
y1      160      160      160
y0      180      180      180
y1      200      200      200
y0      220      220      220
y1      240      240      240
y0      260      260      260
y1      280      280      280
y0      300      300      300
y1      320      320      320
y0      340      340      340
y1      360      360      360
y0      380      380      380
y1      400      400      400

For reference, this is the code that I used to create my desired matrix, involving a for loop:

    n.fn <- function(n1, a) {
  b <- matrix(rep(1, 3), nrow = 1, ncol = 3)
  for (no in 1:a) {
    c <- matrix(rep(n1 * no, 6), nrow = 2, ncol = 3)
    rownames(c) <- c("y0", "y1")
    b <- rbind(b, c)
  }
  b <- as.matrix(b[-1, 1:3])
  colnames(b) <- c("Module 1", "Module 2", "Module 3")
  return(b)
}
n <- n.fn(n1, a)

Any help with creating the first matrix, without the use of a for loop, would be gratefully appreciated!

Answer 1

You can just generate the sequence from 20 to 400, repeat it as many times as there are columns times and arrange in a matrix:

 df <- matrix(rep(rep(seq(from = 20, to = 400, by = 20), each = 2), times = 3), ncol = 3)

Another way is per Adam Quek suggestion using replicate:

df <- replicate(3, unlist(lapply(seq(20,200,20), rep, 2) )) 

Now just give it the desired column and row names:

colnames(df) <- paste("Module", 1:ncol(df))
rownames(df) <- rep(paste0("y", 0:1), nrow(df)/2)