On Sat, 31 Oct 2009, parkbomee wrote:
Thank you both. However, using tapply() instead of a loop does not seem to improve my code much. I am using this inside of an optimization function, and it still takes more than it needs...
Well, you haven't given us much to work with. The optimization choices depend on the particulars of your problem, which you've not detailed. It does not take long to run the tapply() code once, so you need to do it many times. Right? If so, you need to say how the structure varies (rendered as DF in David's reply) from iteration to iteration in the optimization. If it turns out that only 'value' changes and that the number of different times is not too large, then precomputing suitable indicator matrics may help: mat1 <- model.matrix( ~ 0 + factor(time):as.numeric(choice==1),DF) mat2 <- model.matrix( ~ 0 + factor(time), DF ) Inside the optimization use something like with(DF,(value%*%mat1)/(value%*%mat2)) If the structure can change or the number of unique times is large, then with so simple a calculation you should probably just inline some C code. http://cran.r-project.org/web/packages/inline/index.html HTH, Chuck
CC: bbom419 at hotmail.com; r-help at r-project.org
From: dwinsemius at comcast.net
To: d.rizopoulos at erasmusmc.nl
Subject: Re: [R] avoiding loop
Date: Sat, 31 Oct 2009 22:26:17 -0400
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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David Winsemius, MD Heritage Laboratories West Hartford, CT
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