Try this: lapply(mydf[-6], ccf, y = mydf[6])
On Tue, Jun 1, 2010 at 5:50 PM, Bunny, lautloscrew.com < bu...@lautloscrew.com> wrote: > Dear all, > > I am trying to avoid a for loop here and wonder if the following is > possible: > > I have a data.frame with 6 columns and i want to get a cross-correlogram > (by using ccf) . Obivously ccf only accepts two columns at once and then > returms a list. > In fact, with a for loop i´d do the following > > > for (i in 1:6) { > > x[[i]]=ccf(mydf[,i],mydf[,6]) > > > } > > Is there any chance to the same with lapply? e.g. lapply(mydf,"ccf", .... ) > with ... respresenting the changing arguments for ccf functions (note only > the first argument does actually change) > > thx for any suggestions in advance > > best > > matt > ______________________________________________ > 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. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[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.