Thanks Rolf. very nice but "pretty easy" is ALWAYS a relative
statement.
On Thu, Feb 12, 2009 at 3:59 PM, Rolf Turner wrote:
On 13/02/2009, at 9:06 AM, markle...@verizon.net wrote:
Hi Jason: below seems to work. you have to take the transpose because
the apply
returns the rows transposed. i'm also not sure how to make the NAs be
the last
ones but maybe someone can show us how to do that.
Pretty easy:
na.at.end <- function(x){
i <- is.na(x)
c(x[!i],rep(NA,sum(i)))
}
mat <- matrix(c(2,7,2,7,9,10,10,6,8,6,1,9,7,2,0),byrow=TRUE,nrow=3)
print(mat)
t(apply(mat,1, function(.row) {
.row[duplicated(.row)] <- NA
.row
}))
Then just change to:
t(apply(mat,1, function(.row) {
.row[duplicated(.row)] <- NA
na.at.end(.row)
}))
cheers,
Rolf
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