Question

Is there a way to do it?

I’m stuck with this:

m <- 10 # nof row
n <- 5  # nof column

# We will fill each cell with '0'

all <-c()
for (i in 1:m) {
   row_i <- c(rep(0,n))
   all <- c(all,row_i)
}

Which only create 1 row as output.

Answer 1

Why not use a matrix? data.frames are for storing columns of varying types.

So,

m = 10
n = 5
mat = matrix(0, nrow = m, ncol = n)

If you really want a data.frame, coerce to one – the column names will simply be a default:

dat = as.data.frame(mat)
names(dat)
[1] "V1" "V2" "V3" "V4" "V5"

The problem with your approach is that you simply append the values one after the other, ignoring the dimensions you want. You can do it like this, but it’s not a good idea to grow data, better to allocate it all upfront as above. Also, this results in a matrix anyway, which I think is what you should use.

WARNING: bad code ahead!

m <- 10 # nof row
n <- 5  # nof column

all <- NULL
for (i in 1:m) {
  row_i <- c(rep(0,n))
  all <- rbind(all,row_i)
}