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:
Yes
PROTECTED]
Sent: Thursday, May 12, 2005 3:39 AM
To: r-devel@stat.math.ethz.ch
Subject: RE: [Rd] bug in modulus operator %% (PR#7852)
I've now found a Windows system that does this. This is also Windows
XP,
fully patched, and with the same rw2010. So it may be chip-specific:
the
one that works
=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
On Thu, 12 May 2005, Kjetil Brinchmann Halvorsen wrote:
Prof Brian Ripley wrote:
I've now found a Windows system that does this. This is also Windows XP,
fully patched, and with the same rw2010. So it may be chip-specific: the
one that works is a P4 and the one that does not is a latest
[EMAIL PROTECTED] writes:
The following can't be right,
first rw2010:
1 %% 0.001
[1] 0.001
Then rw2001:
1 %% 0.001
[1] -2.081668e-17
and the last seems about right.
A negative remainder? I don't think so. Presumably the result comes
from
o %% now warns if its
On 11-May-05 Peter Dalgaard wrote:
[EMAIL PROTECTED] writes:
The following can't be right,
first rw2010:
1 %% 0.001
[1] 0.001
Then rw2001:
1 %% 0.001
[1] -2.081668e-17
and the last seems about right.
A negative remainder? I don't think so. Presumably the result comes
to not be effected by rounding
errors or lack of precision.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Wednesday, May 11, 2005 3:25 PM
To: Peter Dalgaard
Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED];
r-devel@stat.math.ethz.ch
Subject: Re: [Rd] bug in modulus operator
McGehee, Robert [EMAIL PROTECTED] writes:
Yes, but from ?%%:
It is guaranteed that 'x == (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
McGehee, Robert [EMAIL PROTECTED] writes:
Yes, but from ?%%:
It is guaranteed that 'x == (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
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:
Yes, but from ?%%:
It is guaranteed that 'x =3D=3D (x
@stat.math.ethz.ch
Subject: Re: [Rd] bug in modulus operator %% (PR#7852)
McGehee, Robert [EMAIL PROTECTED] writes:
Yes, but from ?%%:
It is guaranteed that 'x =3D=3D (x %% y) + y * (x %/% y)' (up to =
rounding
error) ...
=20
(R 2.1.0)
x - 1
y - 0.2
x %% y
[1] 0.2
(x %% y) + y * (x %/% y)
[1] 1.2
=20
-devel@stat.math.ethz.ch
Subject: Re: [Rd] bug in modulus operator %% (PR#7852)
McGehee, Robert [EMAIL PROTECTED] writes:
Yes, but from ?%%:
It is guaranteed that 'x =3D=3D (x %% y) + y * (x %/% y)' (up to =
rounding
error) ...
=20
(R 2.1.0)
x - 1
y - 0.2
x %% y
[1] 0.2
(x %% y) + y
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