I am not sure that the guarantee on the help page has been supported for a while. I've altered the code so it is more likely to be.
BTW,
options(digits=20) 1 %% 0.001
[1] 0.0009999999999999792
shows why the original is not a bug in R.
On Thu, 12 May 2005 [EMAIL PROTECTED] wrote:
On Wed, 11 May 2005 [EMAIL PROTECTED] wrote:
Yes, you are correct. I had only checked one of my platforms. Linux works as you suggest. But for me on Windows,
[1] 5 ## I get a 4 in Linuxx <- 1 y <- 0.2 x %/% y
I get 5 on Windows, but
(x %% y) + y * (x %/% y)[1] 1
so is there a problem particular to your Windows runtime?
version _ =20 platform i386-pc-mingw32 arch i386 =20 os mingw32 =20 system i386, mingw32 =20 status =20 major 2 =20 minor 1.0 =20 year 2005 =20 month 04 =20 day 18 =20 language R =20
-----Original Message----- From: Peter Dalgaard [mailto:[EMAIL PROTECTED] Sent: Wednesday, May 11, 2005 4:14 PM To: McGehee, Robert Cc: [EMAIL PROTECTED]; Peter Dalgaard; [EMAIL PROTECTED]; [EMAIL PROTECTED]; r-devel@stat.math.ethz.ch Subject: Re: [Rd] bug in modulus operator %% (PR#7852)
"McGehee, Robert" <[EMAIL PROTECTED]> writes:
roundingYes, but from ?"%%": "It is guaranteed that 'x =3D=3D (x %% y) + y * (x %/% y)' (up to =error) ..." =20 (R 2.1.0)[1] 0.2x <- 1 y <- 0.2 x %% y(x %% y) + y * (x %/% y)[1] 1.2 =20 Certainly 1 does not equal 1.2 as the documentation would suggest, and these seem like large enough numbers to not be effected by rounding errors or lack of precision.
Now that looks a bit odd, but it isn't universal:
[1] 0.2x <- 1 y <- 0.2 x %% yx %/% y[1] 4(x %% y) + y * (x %/% y)[1] 1
So what platform was that happening on?
-- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595
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