Re: which is faster?
- Posted by CChris <christian.cuvier at a?riculture.gouv.?r> Nov 09, 2007
- 602 views
Chris Bensler wrote:
>
> CChris wrote:
> >
> > Here is a benchmark to prove and disprove stuff....
>
> }}}
<eucode>
> Try adding this bit of test code to verify that the variations are accurate.
>
> sequence test
> test = {
> st_dev0(s[1],0)
> ,st_dev1(s[1],0)
> ,st_dev2(s[1],0)
> ,st_dev3(s[1],0)
> ,st_dev4(s[1],0)
> ,st_dev5(s[1],0)
> }
> test = (test = test[1])
> test[1] = 0
> for i = 2 to length(test) do
> if test[i] = 0 then
> test[1] += 1
> printf(1,"st_dev%d() != st_dev0()\n",{i-1})
> end if
> end for
> if test[1] then
> machine_proc(26,{})
> abort(1)
> end if
> </eucode>
{{{
>
> Chris Bensler
> Code is Alchemy
I had tried this, and all values differ, though by very tiny amounts. This is
known as roundoff error accumulating.
Adding N numbers with some fuzz in a ]-eps..+eps[ range will result in a total
error whose magnitude is in the order of eps*sqrt(N). With 53 bit precision on fp
numbers and N=10000, you can easily expect that the few last bits are different
when you change the method of computation.
But there is hardly a point in estimating a deviation estimate with many
significant digits, is it?
You can compute a mean with more accuracy by doing:
m += ((s[i] - m)/i) w/eucode> starting with m = s[1]. The roundoff error buildup is way slower. But because of the many integer divisions, the algorithm is way slower too.
CChris }}}
