Creating a given matrix

Question

How can I construct this matrix in R by for (loop)?

{[1000,100][1000,200][1000,300][2000,100][2000,200][2000,300][3000,100][3000,200][3000,300]};

Answer 1

x <- expand.grid((1:3)*100,(1:3)*1000)  ## construct data frame of all combinations
as.matrix(x[2:1])                       ## reverse column order, convert to matrix

Your use case really isn't clear to me, but I'll just point out that if you're doing this from scratch (i.e. you're going to be entering values manually), then here's the appropriate syntax (whitespace/newlines are optional, for clarity):

matrix(byrow=TRUE, ncol=2,
       c(1000,100,
         1000,200,
         1000,300,
         2000,100,
         2000,200,
         2000,300,
         3000,100,
         3000,200,
         3000,300))

Answer 2

If you are starting from the string itself (and cannot generate it as Ben and Rui have discussed), then you can try parsing it:

txt <- '{[1000,100][1000,200][1000,300][2000,100][2000,200][2000,300][3000,100][3000,200][3000,300]};'
m <- do.call(rbind, strsplit(strsplit(txt, "[^,0-9]+")[[1]], ","))
m
#       [,1]   [,2] 
#  [1,] "1000" "100"
#  [2,] "1000" "200"
#  [3,] "1000" "300"
#  [4,] "2000" "100"
#  [5,] "2000" "200"
#  [6,] "2000" "300"
#  [7,] "3000" "100"
#  [8,] "3000" "200"
#  [9,] "3000" "300"

Then convert to numeric with:

m <- apply(m, 2, as.numeric)
m
#       [,1] [,2]
#  [1,] 1000  100
#  [2,] 1000  200
#  [3,] 1000  300
#  [4,] 2000  100
#  [5,] 2000  200
#  [6,] 2000  300
#  [7,] 3000  100
#  [8,] 3000  200
#  [9,] 3000  300