R help - Aug 2009 - Infinite != NaN?

Greetings.  I somehow had the impression that an infinite number, as obtained by
dividing by zero, for instance, would be flagged as both missing
("NA") and not a number ("NaN").  It appears that I was
wrong on both counts, although the is.finite function correctly returns FALSE in
such a case.  Please see the appended for some details.  I guess that the bottom
line is that R works the way it works, but if you can add anything that will
further instruct me, I'd appreciate it.

Thanks.

-- Mike

> y <- 2/0
> y
[1] Inf
> is.na(y)
[1] FALSE
> is.nan(y)
[1] FALSE
> is.finite(y)
[1] FALSE
> z <- log(-1)
Warning message:
In log(-1) : NaNs produced
> z
[1] NaN
> is.nan(z)
[1] TRUE
> is.na(z)
[1] TRUE

-----

R version 2.9.2 (2009-08-24)
Copyright (C) 2009 The R Foundation for Statistical Computing
ISBN 3-900051-07-0

-----

$ uname -a
Linux xxxxxxx.localdomain 2.6.29.6-217.2.16.fc11.x86_64 #1 SMP Mon Aug 24
17:17:40 EDT 2009 x86_64 x86_64 x86_64 GNU/Linux

It is not clear that you have read the help page that arise with:

  ?Inf

"Note:

In R, basically all mathematical functions (including basic  
Arithmetic), are supposed to work properly with +/- Inf and NaN as  
input or output.

The basic rule should be that calls and relations with Infs really are  
statements with a proper mathematical limit."

Also:

 > Inf+Inf
[1] Inf
 > Inf-Inf
[1] NaN
 > -2/0
[1] -Inf
 > Inf*Inf
[1] Inf
 > Inf*(-Inf)
[1] -Inf

On Aug 29, 2009, at 9:58 PM, Michael Hannon wrote:
> Greetings.  I somehow had the impression that an infinite number, as  
> obtained by dividing by zero, for instance, would be flagged as both  
> missing ("NA") and not a number ("NaN").  It appears
that I was
> wrong on both counts, although the is.finite function correctly  
> returns FALSE in such a case.  Please see the appended for some  
> details.  I guess that the bottom line is that R works the way it  
> works, but if you can add anything that will further instruct me,  
> I'd appreciate it.
>
So I suppose the obvious further instruction is to read the help pages.
> Thanks.
>
> -- Mike
>
>
>> y <- 2/0
>
>> y
> [1] Inf
>
>> is.na(y)
> [1] FALSE
>
>> is.nan(y)
> [1] FALSE
>
>> is.finite(y)
> [1] FALSE
>
>> z <- log(-1)
> Warning message:
> In log(-1) : NaNs produced
>
>> z
> [1] NaN
>
>> is.nan(z)
> [1] TRUE
>
>> is.na(z)
> [1] TRUE
>
> -----


David Winsemius, MD
Heritage Laboratories
West Hartford, CT