jamaas wrote: > > Long story short, I have a big iterative procedure that produces a long > list of data.frames such as the one called > "results" here. Is there an easy way to produce a similar list of > data.frames comprised of the mean of each of the > columns in results, such that it ends up like the one I've shown in > "resultsmean" below? > > I've tried apply and lapply, still not got the correct arguments. As > usual, TIA. > > Jim > >> results > [[1]] > name LOR23 BIA23 MSE23 H0R23 > 1 0.2111122 -1.012228 -0.095937 0.035650 1.00 > 2 0.2111122 -0.836300 0.079991 0.042322 0.75 > 3 0.2111122 -0.518631 0.397659 0.214593 0.50 > > > [[2]] > name LOR23 BIA23 MSE23 H0R23 > 1 0.2211122 -0.724630 0.191660 0.051308 1 > 2 0.2211122 -0.781812 0.134478 0.033872 1 > 3 0.2211122 -0.522109 0.394181 0.164628 0.75 > > > would like > >> resultsmean > [[1]] > name LOR23 BIA23 MSE23 H0R23 > 1 0.2111122 -0.78333 0.12734 0.097160 0.75 > > [[2]] > name LOR23 BIA23 MSE23 H0R23 > 1 0.2211122 -0.67566 0.2400 0.08266 0.916 >
lapply(results, FUN=colMeans) Berend -- View this message in context: http://r.789695.n4.nabble.com/howto-calculate-column-means-in-data-frame-tp3439932p3439959.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.