Here are two solutions. seq(length = ...) instead of
just seq(...) is so that v can possibly contain zeros.
# data
v <- 3:5
# solution 1 - rbind/lapply
f <- function(n) {
s = seq(length = n)
replace(rep(NA, max(v)), s, s)
}
do.call(rbind, lapply(v, f))
# solution 2 - loop
mat <- matrix(NA, length(v), max(v))
for(i in seq(v)) {
s <- seq(length = v[i])
mat[i, s] <- s
}
On 8/22/06, Sara-Jane Dunn <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm having trouble applying the matrix function. I'd like to be able to
> create a matrix of vectors filled in by rows, which are not all the same
> length, and so I need it to fill in NAs where applicable.
>
> It's easiest to explain with a simple example:
>
> Suppose vec = c(3,4,5). How can I form a matrix of the vectors 1:vec[j]
> for j=1:3?
> i.e. 1 2 3 NA NA
> 1 2 3 4 NA
> 1 2 3 4 5
> I've tried matrix(c(1:vec[j]),nrow=max(j),ncol=max(vec)) but it will
> only give me a matrix with repeated values for j=1, like 1 2 3 1
> 2
> 3 1 2 3 1
> 2 3 1 2 3
>
> Also using the list function hasn't got me anywhere either..
>
> Any help/ideas would be greatly appreciated!
>
> Many thanks,
> Sara-Jane Dunn
>
> --
> This message (and any attachments) is for the recipient only...{{dropped}}
>
> ______________________________________________
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