This is pretty much equivalent:
tapply(DF$value[DF$choice==1], DF$time[DF$choice==1], sum) /
tapply(DF$value, DF$time, sum)
And both will probably fail if the number of groups with choice==1 is
different than the number overall.
--
David.
On Oct 31, 2009, at 5:14 PM, Dimitris Rizopoulos wrote:
one approach is the following:
# say 'DF' is your data frame, then
with(DF, {
ind <- choice == 1
n <- tapply(value[ind], time[ind], sum)
d <- tapply(value, time, sum)
n / d
})
I hope it helps.
Best,
Dimitris
parkbomee wrote:
Hi all,
I am trying to figure out a way to improve my code's efficiency by
avoiding the use of loop.
I want to calculate a conditional mean(?) given time.
For example, from the data below, I want to calculate sum((value|
choice==1)/sum(value)) across time.
Is there a way to do it without using a loop?
time cum_time choice value
1 4 1 3
1 4 0 2
1 4 0 3
1 4 0 3
2 6 1 4
2 6 0 4
2 6 0 2
2 6 0 4
2 6 0 2
2 6 0 2 3 4
1 2 3 4 0 3 3
4 0 5 3 4 0 2
My code looks like
objective[1] = value[1] / sum(value[1:cum_time[1])
for (i in 2:max(time)){
objective[i] = value[cum_time[i-1]+1] /
sum(value[(cum_time[i-1]+1) : cum_time[i])])
}
sum(objective)
Anyone have an idea that I can do this without using a loop??
Thanks.
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
______________________________________________
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