Yes, but from ?"%%": "It is guaranteed that 'x =3D=3D (x %% y) + y * (x %/% y)' (up to = rounding error) ..." (R 2.1.0)
x <- 1 y <- 0.2 x %% y
[1] 0.2
(x %% y) + y * (x %/% y)
[1] 1.2 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. -----Original Message----- From: Ted.Harding@nessie.mcc.ac.uk [mailto:Ted.Harding@nessie.mcc.ac.uk] Sent: Wednesday, May 11, 2005 3:25 PM To: Peter Dalgaard Cc: R-bugs@biostat.ku.dk; kjetil@acelerate.com; r-devel@stat.math.ethz.ch Subject: Re: [Rd] bug in modulus operator %% (PR#7852)
On 11-May-05 Peter Dalgaard wrote:
kjetil@acelerate.com writes: =20
The following can't be right, first rw2010: =20
> 1 %% 0.001
[1] 0.001 =20 Then rw2001: =20
> 1 %% 0.001
[1] -2.081668e-17
>
=20 and the last seems about right. =20
=20
A negative remainder? I don't think so. Presumably the result comes
from
=20
o %% now warns if its accuracy is likely to be affected by lack
of precision (as in 1e18 %% 11, the unrealistic expectation of
PR#7409), and tries harder to return a value in range when it
is.
=20
So, not a bug.
Agreed! One should always be aware of such fuzzy edges in any case where the mathematical result depends on mathematically exact representation, as here. In such cases what I often do is on the lines of ((1000*x)%%(1000*y))/1000 Using R-2.1.0beta, first of all I get the same as Kjetil: 1 %% 0.001 ## [1] 0.001 Secondly I get (1000*1)%%(1000*0.001) ## [1] 0 Although this trick does not guarantee the exact result, it makes it more likely; and also, any little inexactness will now show up more clearly since in the form ((BIG*x) %% (BIG*y))/BIG the possible discrepancy before division by BIG is at most y in magnitude. While the properties of this trick are not entirely transparent, one can for instance compare the result of using it with the result of not using it, and then come to a decision as to which to use. A further example on the same lines is 1 %% (0.00001) # [1] 1e-05 (Similar to Kjetil's first example), while ((1000*1) %% (1000*0.00001))/1000 ## [1] -2.081668e-17 Any comments would be interesting! Best wishes, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 094 0861 Date: 11-May-05 Time: 20:13:22 ------------------------------ XFMail ------------------------------ ______________________________________________ R-devel@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-devel