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: bbom...@hotmail.com; r-help@r-project.org
From: dwinsem...@comcast.net
To: d.rizopou...@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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Charles C. Berry (858) 534-2098
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