[R] FAQ 7.31

Robin Hankin r.hankin at noc.soton.ac.uk
Fri May 6 14:14:11 CEST 2005


On May 6, 2005, at 09:00 am, Prof Brian Ripley wrote:

> On Fri, 6 May 2005, Robin Hankin wrote:
>>
[snip]

>> I worry about the use of identical() here, because I am comparing two 
>> floats for equality
>> as discussed in FAQ 7.31.    Perhaps all.equal() would be better:
>>
>> all.equal(1e99,1e99+1e83, tol=.Machine$double.eps)
>
> Well, probably 2*.Machine$double.eps since each of two numbers could 
> be one bit from the truth.
>

Yes, this makes a lot of sense, and will likely prevent me from 
flogging a dead horse by executing
dozens of pointless iterations in pursuit of the dregs of those last 
few binary digits at the tail end of
some float (not to mention infinite loops).

I would have thought that if x, y are scalars then identical(x,y) would 
be the same as
all.equal(x,y,tol=.Machine$double.eps).

And it is, most of the time.

But this caught me out just now:

R>  x <- 5352970674736366
R>  identical(x,x+1)
[1] FALSE

R> all.equal(x,x+1,tol=.Machine$double.eps)
[1] TRUE


Thus  x and x+1  are not identical but nevertheless agree to within my 
machine precision, and
evidently my understanding of "machine precision" doesn't tally with 
Rreality.

Any comments?


> The registers on ix86 machines have more precision than the 
> representation stored in RAM.  So it is possible that out.new has more 
> precision because it has been kept in registers and out.old has been 
> retrieved from RAM and so has less.  The result may well be continued 
> looping.
>
> I think the scenario in the previous para is quite unlikely (it would 
> need a very clever C compiler), but it is theoretically possible and 
> may become likely with semi-compiled versions of R.  The C-level 
> equivalent is quite common.
>

>
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
  tel  023-8059-7743




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