Hi Lydia,
I compared my ratio function with Dimitris and Phil's suggestions. Please do
NOT use my approach because it's painfully slow for a large vector (as Phil
told me). Here is why (using Win XP SP2, Intel Core- 2 Duo 2.4 GHz, R 2.7.0
Patched):
# Vector
x=rnorm(100000,0,1)
# Suggestion
new.ratio=function(x) x[2:length(x)]/x[1:(length(x)-1)]
# My horrible function
my.ratio=function(x){
temp=NULL
for (n in 1:length(x)) temp=c(temp,x[n]/x[n-1])
temp
}
# System time
t=system.time(my.ratio(x))
tnr=system.time(new.ratio(x))
t
user system elapsed
38.79 0.06 39.31
tnr
user system elapsed
0 0 0
Thanks to all,
Jorge
On Mon, May 12, 2008 at 11:15 AM, Phil Spector <[EMAIL PROTECTED]>
wrote:
> Another alternative would be to take advantage of R's vectorization:
>
> x=c(1,2,3,2,1,2,3)
> > x[2:length(x)]/x[1:(length(x)-1)]
> >
> [1] 2.0000000 1.5000000 0.6666667 0.5000000 2.0000000 1.5000000
>
> The solution using your ratio function will be painfully slow
> for a large vector.
>
> - Phil Spector
> Statistical Computing Facility
> Department of Statistics
> UC Berkeley
> [EMAIL PROTECTED]
>
>
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