Duncan, Berwin, Martin,
Thanks for your thoughtful explanations, which make perfect sense.
May I simply suggest that the non-identity between last(cumsum) and
sum might be worth mentioning in the cumsum doc page?
-s
__
R-help@r-project.org
G'day all,
On Wed, 18 Feb 2009 12:41:27 -0500
Stavros Macrakis wrote:
> Hmm. Why not use the same method to guarantee the same result?
Hmm, I did not look at the source code but, potentially, sum() could
use some tricks to reduce rounding errors further that would not be
available to cumsum(
On 18/02/2009 12:41 PM, Stavros Macrakis wrote:
Hmm. Why not use the same method to guarantee the same result? Or at
least document the possibility that cumsum(x)[length(x)] != sum(x)...
that seems like an easy trap to fall into.
Assuming equality of floating point numbers computed by two dif
Hmm. Why not use the same method to guarantee the same result? Or at
least document the possibility that cumsum(x)[length(x)] != sum(x)...
that seems like an easy trap to fall into.
-s
On Wed, Feb 18, 2009 at 11:39 AM, Martin Maechler
wrote:
>> "SM" == Stavros Macrakis
>>
> "SM" == Stavros Macrakis
> on Wed, 18 Feb 2009 10:00:40 -0500 writes:
SM> Nice! Glad to hear it. It sounds as though it is still possible for
SM> cumsum(x)[length(x)] to not be exactly equal to sum, though?
Well, possible, probably yes, platform-dependently;
However I vag
> "GaGr" == Gabor Grothendieck
> on Tue, 17 Feb 2009 20:53:18 -0500 writes:
GaGr> Check out sum.exact and cumsum.exact in the caTools package.
>> library(caTools)
GaGr> Loading required package: bitops
>> x <- 1/(12:14)
>> sum(x) - cumsum(x)[3]
GaGr> [1] 2.7755
Check out sum.exact and cumsum.exact in the caTools package.
> library(caTools)
Loading required package: bitops
> x <- 1/(12:14)
> sum(x) - cumsum(x)[3]
[1] 2.775558e-17
> sum.exact(x) - cumsum.exact(x)[3]
[1] 0
On Tue, Feb 17, 2009 at 5:12 PM, Stavros Macrakis wrote:
> I recently traced a bug
I recently traced a bug of mine to the fact that cumsum(s)[length(s)]
is not always exactly equal to sum(s).
For example,
x<-1/(12:14)
sum(x) - cumsum(x)[3] => 2.8e-17
Floating-point addition is of course not exact, and in particular is
not associative, so there are various possible r
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