On 11-10-2013, at 15:26, Steven Ranney <steven.ran...@gmail.com> wrote:
> Hello all - > > I have an example column in a dataFrame > > id.name > 123.45 > 123.45 > 123.45 > 123.45 > 234.56 > 234.56 > 234.56 > 234.56 > 234.56 > 234.56 > 234.56 > 345.67 > 345.67 > 345.67 > 456.78 > 456.78 > 456.78 > 456.78 > 456.78 > 456.78 > 456.78 > 456.78 > 456.78 > ... > [truncated] > > And I'd like to create a second vector of sequential values (i.e., 1:N) for > each unique id.name value. In other words, I need > > id.name x > 123.45 1 > 123.45 2 > 123.45 3 > 123.45 4 > 234.56 1 > 234.56 2 > 234.56 3 > 234.56 4 > 234.56 5 > 234.56 6 > 234.56 7 > 345.67 1 > 345.67 2 > 345.67 3 > 456.78 1 > 456.78 2 > 456.78 3 > 456.78 4 > 456.78 5 > 456.78 6 > 456.78 7 > 456.78 8 > 456.78 9 > > The number of unique id.name values is different; for some values, nrow() > may be 42 and for others it may be 36, etc. > > The only way I could think of to do this is with two nested for loops. I > tried it but because this data set is so large (nrow = 112,679 with 2,161 > unique values of id.name), it took several hours to run. > > Is there an easier way to create this vector? I'd appreciate your thoughts. I named your dataframe dat1. You can also do this unlist(sapply(rle(dat1$id.name)$lengths, function(k) 1:k )) And as Petr told you: beware of FAQ 7.31 Berend ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
