Try this:
coef2 <- function(i) cov(x[i,"order"], x[i, "y"]) / var(x[i, "order"])
tapply(1:nrow(x), x$subject, coef2)
On Thu, Mar 19, 2009 at 5:41 PM, Adam D. I. Kramer <a...@ilovebacon.org> wrote:
> Dear colleagues,
>
> Consider the following data frame:
>
> x <- data.frame(y=rnorm(100),order=rep(1:10,10),subject=rep(1:10,each=10))
>
> ...it is my goal to aggregate x to compute a linear effect of order
> for each subject. So, ideally, result would be a vector containing a single
> number for each subject, representing the linear relationship between y and
> order.
>
> I first tried this:
>
> result <- aggregate(x[1:2,],list(subject=x$subject),
> function (z) { lm(y ~ order, data=z)$coefficients[2] }
> )
>
> ...because lm(y ~ order, data=x, subset=x$subject==1)$coefficients[2] would
> give me the correct term for subject 1 (i.e., that is the number I am
> actually looking for).
>
> However, when used on data frames, aggregate() aggregates every
> COLUMN in x _separately_ using FUN...while lm needs both columns *together.*
>
> ...I then turned to tapply, but that is useful only on "atomic
> objects," and not data frames.
>
> I have two solutions, which I find inelegant and slow:
>
> 1) result <- sapply(levels(factor(x$subject)),
> function(z) { lm(y ~ order, data=x,
> subset=subject==z)$coefficients[2]}
> )
>
> ...this gets the job done, but is very slow.
>
> 2) result <- c();
> for (z in 1:nlevels(x$s2)) { result[z] <- lm(y ~ order, data=x,
> subset=x$s2==levels(x$s2)[z])$coefficients[2] };
> result <- unlist(result);
>
> ...also does the job, but is also very slow.
>
> Is there a better solution? I miss the speed of tapply and aggregate; the
> example has only 100 rows and 10 subjects, but the actual data has many more
> of each.
>
> Cordially,
> Adam D. I. Kramer
> Ph.D. Candidate, Social and Personality Psychology
> University of Oregon
>
> ______________________________________________
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>
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