[R] rounding

[Moshe Olshansky]

Actually, your result is strange, since if x = 0.55 then a hexadecimal representation of x*100 is 404B8000 00000001 which is above 55, meaning that x is above 0.55 and should have been rounded to 0.6, while if x = 2.55 then the hexadecimal representation of x*100 is 406FDFFF FFFFFFFF which is below 255, so that x is less than 2.55 and should have been rounded to 2.5.


--- On Fri, 11/7/08, Moshe Olshansky <m_olshansky at yahoo.com> wrote:

> From: Moshe Olshansky <m_olshansky at yahoo.com>
> Subject: Re: [R] rounding
> To: r-help at stat.math.ethz.ch, "Korn, Ed (NIH/NCI) [E]" <korne at ctep.nci.nih.gov>
> Received: Friday, 11 July, 2008, 10:39 AM
> The problem is that neither 0.55 nor 2.55 are exact machine
> numbers (the computer uses binary representation), so it
> may happen that the machine representation of 0.55 is
> slightly less than 0.55 while the machine representation of
> 2.55 is slightly above 2.55.
> 
> 
> --- On Fri, 11/7/08, Korn, Ed (NIH/NCI) [E]
> <korne at ctep.nci.nih.gov> wrote:
> 
> > From: Korn, Ed (NIH/NCI) [E]
> <korne at ctep.nci.nih.gov>
> > Subject: [R] rounding
> > To: r-help at stat.math.ethz.ch
> > Received: Friday, 11 July, 2008, 6:07 AM
> > Hi,
> >  
> > Round(0.55,1)=0.5
> >  
> > Round(2.55,1)=2.6
> >  
> > Can this be right?
> >  
> > Thanks,
> >  
> > Ed
> > 
> > 	[[alternative HTML version deleted]]
> > 
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