Re: [Rd] is.numeric (NA + NA) is TRUE, should be FALSE (PR#8745)

From: Duncan Murdoch <murdoch_at_stats.uwo.ca>
Date: Tue 04 Apr 2006 - 19:52:06 GMT

On 4/4/2006 3:38 PM, cfillekes@gmail.com wrote:
> Full_Name: c fillekes
> Version: Version 2.2.1 (2005-12-20 r36812)
> OS: Gentoo Linux kernel 2.6.12
> Submission from: (NULL) (129.116.71.233)
>
>
>
>
> "Not Available" is of course not a numeric:
> R
>

>> is.numeric (NA)

> [1] FALSE
>
> But for some reason, all arithmetic operations on NA's are
> in fact numeric, even if it's with other NA's.
>
>
>> is.numeric (NA+NA)

> [1] TRUE
>> is.numeric (NA^2)

> [1] TRUE
>> is.numeric (NA-NA)

> [1] TRUE
>> is.numeric (NA*NA)

> [1] TRUE
>> is.numeric (NA/NA)

> [1] TRUE
>>

>
> This is not the expected thing.

But it is the documented thing, so it is not a bug:

"'is.numeric' returns 'TRUE' if its argument is of mode '"numeric"'
      (type '"double"' or type '"integer"') and not a factor, and
      'FALSE' otherwise."

The value is irrelevant. You probably want to use "is.numeric(x) && all(is.finite(x))" to get the result you're expecting.

Duncan Murdoch



R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel Received on Wed Apr 05 06:16:15 2006

This archive was generated by hypermail 2.1.8 : Tue 04 Apr 2006 - 22:16:53 GMT