Hi Steven, One would be
with(yourdataset, aggregate(x, list(lc1, id), mean)) Group.1 Group.2 x 1 85 ga1 45.47261 2 95 ga1 53.38831 3 105 ga1 58.18282 4 115 ga1 63.77469 5 125 ga1 66.98222 6 85 ga2 47.55711 7 95 ga2 54.78450 8 105 ga2 60.67968 9 115 ga2 64.83192 10 125 ga2 66.98222 11 85 ga3 51.99211 12 95 ga3 56.70202 13 105 ga3 61.69574 14 115 ga3 65.90707 15 125 ga3 66.98222 HTH, Jorge On Fri, Mar 25, 2011 at 1:36 PM, Steven Ranney <> wrote: > All > > I have an example data frame > > x lc1 id > 43.38812035 85 ga1 > 47.55710661 85 ga1 > 47.55710661 85 ga2 > 47.55710661 85 ga2 > 51.99211429 85 ga3 > 51.99211429 85 ga3 > 51.99211429 95 ga1 > 54.78449958 95 ga1 > 54.78449958 95 ga2 > 54.78449958 95 ga2 > 56.70201864 95 ga3 > 56.70201864 95 ga3 > 56.70201864 105 ga1 > 59.66361903 105 ga1 > 59.66361903 105 ga2 > 61.69573564 105 ga2 > 61.69573564 105 ga3 > 61.69573564 105 ga3 > 63.77469479 115 ga1 > 63.77469479 115 ga1 > 64.83191994 115 ga2 > 64.83191994 115 ga2 > 64.83191994 115 ga3 > 66.98222118 115 ga3 > 66.98222118 125 ga1 > 66.98222118 125 ga1 > 66.98222118 125 ga2 > 66.98222118 125 ga2 > 66.98222118 125 ga3 > 66.98222118 125 ga3 > > and Im trying to extract means for every lc1 and id level so that I > would have a data frame that looks like > > x.m lc1 id > x.1 085 ga1 > x.2 085 ga2 > x.3 085 ga3 > x.4 095 ga1 > x.5 095 ga2 > x.6 095 ga3 > x.7 105 ga1 > x.8 105 ga2 > x.9 105 ga3 > x.10 115 ga1 > x.11 115 ga2 > x.12 115 ga3 > x.13 125 ga1 > x.14 125 ga2 > x.15 125 ga3 > > Now, I can use brute force to get the second data frame by > > write.table(tapply(x[lc1=="085"], id[lc1=="085"], mean), "file.xls", > sep=",") > write.table(tapply(x[lc1=="095"], id[lc1=="095"], mean), file.xls", > sep=",", append=T) > write.table(tapply(x[lc1=="105"], id[lc1=="105"], mean), file.xls", > sep=",", append=T) > etc > > and add the values for the lc1 column in the .xls file until Ive > worked my way through every level for lc1 then read the file back into > R, but this would require a great deal of my time. (I have 72 levels > for lc1 and 346 levels for id totalling over 20,000 lines.) > > I am confident that there is a simple, more elegant solution available > to me that I am overlooking. I am sure that I could use a for loop, > but as someone that is new to R programming, I am unsure of how to go > about creating the for loop to build the second data frame. I've > tried to modify existing for loops that I already have but have been > unsuccessful. > > Do you have any suggestions? > > Thank you > > Steven Ranney > > ______________________________________________ > 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. > [[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.
