On May 17, 2012, at 6:45 AM, Axel Urbiz wrote:

Dear List,

Is there a way I can sort a sample based on a sort index constructed from the data from which the sample is taken? Basically, I need to take 'many'
samples from the same source data and sort them. This can be very time
consuming for long vectors. Is there any way I can sort the data only once
initially, and use that sort order for the samples?

I believe that idea is what is implemented in tree-based classifiers, so the data is sorted only once initially and that sort order is used for the
child nodes.


set.seed(12345)
x <- sample(0:100, 10)
x.order <- order(x)
x.sorted <- x[x.order]

sample.ind <- sample(1:length(x), 5, replace = TRUE) #sample 1/2 size with
replacement
x.sample <- x[sample.ind]

x.sample.sorted <-   #??? (without sorting again)

Thanks for any help on this.


If you had created a named vector or used a dataframe with rownames you would have a record of the original sort order.

Regards,
Axel.

        [[alternative HTML version deleted]]

______________________________________________
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.

David Winsemius, MD
West Hartford, CT

______________________________________________
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.

Reply via email to