Thanks. This is what I was referring to:
x <- rep(NA, 3)
is.na(x)
[1] TRUE TRUE TRUE
if (is.na(x)) {print("True")}
[1] "True"
Warning message:
In if (is.na(x)) { :
the condition has length > 1 and only the first element will be used
You are of course right - the warning is generated by if(), not by is.na() and
the reason for the warning is that is.na() returns a vector if applied to a
vector. I should have been more clear.
Cheers,
B.
On Dec 20, 2014, at 3:29 PM, Duncan Murdoch <murdoch.dun...@gmail.com> wrote:
> This may be out of context, but on the face of it, this claim is wrong:
>
> On 20/12/2014, 1:57 PM, Boris Steipe wrote:
> "Moreover is.na() behaves differently when evaluated on its own, or as
> the condition of an if() statement."
>
> The conditions in an if() statement are not evaluated in special
> conditions at all. The only way you'll get a different value is if the
> argument to is.na() does tricky stuff like looking at the evaluation stack.
>
> Duncan Murdoch
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