Dear Uwe, Petr and Berend,
Thank's a lot !!
I just checked and it's
c(cumprod(a)[M-1],cumprod(a[-1])[M-1]))
that gives an identical result to the my initial loop (at 10x speed of
my initial loop ... )
Wolfgang
Berend Hasselman a écrit :
Petr Pikal wrote:
Hi
[EMAIL PROTECTED] napsal dne 03.09.2008 16:39:07:
....
In many cases I've noticed that using apply, sapply etc can help
speeding up processes, but in this case I don't see how I could do so.
a <- runif(10000000,0.5,1.6)
C <- 2
M <- 10000000
system.time( for (i in 1:(M-1)) {C <- C* c(a[i],a[i+1])} )
Maybe simple math? You want last two members of 2*cumprod(a).
So
system.time(2*cumprod(a)[9999999:10000000])
user system elapsed
1.97 0.04 2.00
shall be a little bit quicker then for cycle. But it is valid only for the
above calculation.
I think
2*c(cumprod(a)[M-1],cumprod(a[-1])[M-1])
is the answer.
Berend
--
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wolfgang Raffelsberger, PhD
Laboratoire de BioInformatique et Génomique Intégratives
CNRS UMR7104, IGBMC
1 rue Laurent Fries, 67404 Illkirch Strasbourg, France
Tel (+33) 388 65 3300 Fax (+33) 388 65 3276
http://www-bio3d-igbmc.u-strasbg.fr/~wraff
[EMAIL PROTECTED]
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