Hi. I would use the function aggregate, but first you will have to tag each row with a special code so R can recognize the group of data and apply the function you desire. For example, with your data I would do this:
Date Pm FF LL KK HH NN Ww DD code 01/01/2012 00:00:00 349 120 10 8 1178 1292 2005 762 01_01_2012_1 01/01/01/2012 00:00:05 356 119 12 7 1167 1289 1992 778 01_01_2012_1 01/01/2012 00:00:10 360 115 15 8 189 1302 2010 770 01_01_2012_1 01/01/2012 00:00:15 349 120 10 8 1178 1292 2005 762 01_01_2012_1 01/01/01/2012 00:00:20 356 119 12 7 1167 1289 1992 778 01_01_2012_2 01/01/2012 00:00:25 360 115 15 8 189 1302 2010 770 01_01_2012_2 01/01/2012 00:00:30 360 115 15 8 189 1302 2010 770 01_01_2012_2 ##Then apply the function aggregate aggregate(name of the variable you want to obtain the mean, by=list(variable used for grouping),FUN=mean) For example, if you want to aggregate Pm by groups of 15 min, you write it like this aggregate(Pm,by=list(code),FUN=sum) and you'll obtain the mean of the rows that have the same code . In this example, you'll obtain the mean of two groups: the Pm measurments wich their labels are 01_01_2012_1 and 01_01_2012_2. Hope it works M.C. Luis Antonio Arias Medellín National Institute of Public Health Cuernavaca, Morelos, Mexico -- View this message in context: http://r.789695.n4.nabble.com/Average-for-Huge-file-tp4586926p4587482.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.
