Not tested but should work:
sums = tapply(x, group, sum);
sums.ext = sums[ match(group, names(sums))]
normalized = x/sums.ext
It may be that the tapply is just as slow as your loop though, I'm not sure.
HTH,
Peter
On Thu, Nov 29, 2012 at 10:55 AM, Noah Silverman <noahsilver...@ucla.edu> wrote:
> Hi,
>
> I have a very large data set (aprox. 100,000 rows.)
>
> The data comes from around 10,000 "groups" with about 10 entered per group.
>
> The values are in one column, the group ID is an integer in the second column.
>
> I want to normalize the values by group:
>
> for(g in unique(groups){
> x[group==g] / sum(x[group==g])
> }
>
> This works find in a loop, but is slow. Is there a faster way to do this?
>
> Thanks!
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