baptiste auguie wrote:
it's documented as "unexpected"
?diag
Note
Using diag(x) can have unexpected effects if x is a vector that could be of
length one. Use diag(x, nrow = length(x)) for consistent behaviour.
And the result follows from this part,
else if (length(x) == 1L && nargs() == 1L) {
n <- as.integer(x)
x <- 1
}
Looks like the man page is wrong: it says diag(0.9) should produce a
1x1 matrix, but as the code shows, it produces a 0x0 one.
I'll fix it.
Duncan Murdoch
baptiste
2009/9/4 Markku Karhunen <markku.karhu...@helsinki.fi>
Hi,
Does anybody know, what is going on here?
diag(sqrt(1))
[,1]
[1,] 1
diag(sqrt(0.3333))
<0 x 0 matrix>
sqrt(1)
[1] 1
sqrt(0.3333)
[1] 0.5773214
BR, Markku Karhunen
researcher
University of Helsinki
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