On Tue, Mar 6, 2012 at 1:32 PM, O'Hanlon, Simon J <simon.ohan...@imperial.ac.uk> wrote: > Ah! > > Thanks. > > I had already made vector x2 previously and then went and changed it for some > reason, which was why I didn't notice the error (because the subsequent code > was able to run regardless). Sorry about that. > > so x2 should have read x2=c(rep(1,12)) which is what I originally had and > what I was basing my plea for help on.
That would explain the difference in results. Regardless, the method I suggested should work. x1=c(rep(0:1,6)) x2=c(rep(1,12)) x3=c(rep(1,6),rep(0,6)) df=data.frame(x1,x2,x3) tabledf=as.data.frame(with(df, table(x1,x2,x3))) tabledf <- cbind(tabledf, res=1:nrow(tabledf)) newdf <- merge(df, tabledf) Note that row order is not preserved; if you need that you can add an id column to df before merging and sort on it after. Please notice also that I've been copying the R-help list on my replies, so that other people who either have similar questions or might be moved to help can see what we've been discussing. Sarah > ________________________________________ > From: Sarah Goslee [sarah.gos...@gmail.com] > Sent: 06 March 2012 18:27 > To: O'Hanlon, Simon J; r-help > Subject: Re: [R] Label rows of table by factor level for groups of factors > > Well, if you can get this to run your version of R is markedly\ > different than mine. > >> #Start of code >> >> x1=c(rep(0:1,6)) >> x2=c(rep(c(1,1,0,0)6)) > Error: unexpected numeric constant in "x2=c(rep(c(1,1,0,0)6" >> x3=c(rep(1,6),rep(0,6)) > > > > On Tue, Mar 6, 2012 at 1:23 PM, O'Hanlon, Simon J > <simon.ohan...@imperial.ac.uk> wrote: >> Hi Sarah, >> Thanks a lot for your suggestion. I'll give it a go if I can (I just spent >> the last 3 hours using unique record filtering and vlookups in Excel to >> achieve what I'm sure can be accomplished in 3 or 4 lines of R code!). >> >> I think you might want to run the sample code again though. I just tried it >> (and there was no missing comma) and I get: >> >> x1 x2 x3 >> 1 0 1 1 >> 2 1 1 1 >> 3 0 1 1 >> 4 1 1 1 >> 5 0 1 1 >> 6 1 1 1 >> 7 0 1 0 >> 8 1 1 0 >> 9 0 1 0 >> 10 1 1 0 >> 11 0 1 0 >> 12 1 1 0 >>> tabledf >> x1 x2 x3 Freq >> 1 0 1 0 3 >> 2 1 1 0 3 >> 3 0 1 1 3 >> 4 1 1 1 3 >>> desired >> x1 x2 x3 res >> 1 0 1 1 3 >> 2 1 1 1 4 >> 3 0 1 1 3 >> 4 1 1 1 4 >> 5 0 1 1 3 >> 6 1 1 1 4 >> 7 0 1 0 1 >> 8 1 1 0 2 >> 9 0 1 0 1 >> 10 1 1 0 2 >> 11 0 1 0 1 >> 12 1 1 0 2 >>> nrow(tabledf) >> [1] 4 >>> dim(tabledf) >> [1] 4 4 >> >> #Start of code >> >> x1=c(rep(0:1,6)) >> x2=c(rep(c(1,1,0,0)6)) >> x3=c(rep(1,6),rep(0,6)) >> df=data.frame(x1,x2,x3) >> tabledf=as.data.frame(with(df, table(x1,x2,x3))) >> res=c(3,4,3,4,3,4,1,2,1,2,1,2) >> desired=data.frame(x1,x2,x3,res) >> df >> tabledf >> desired >> >> #End of code >> >> Cheers! >> >> Simon >> >> -------------------------------- >> Simon O'Hanlon, BSc MSc >> Department of Infectious Disease Epidemiology >> Imperial College London >> St. Mary's Hospital >> London >> W2 1PG >> ________________________________________ >> From: Sarah Goslee [sarah.gos...@gmail.com] >> Sent: 06 March 2012 18:16 >> To: O'Hanlon, Simon J >> Cc: r-help@R-project.org >> Subject: Re: [R] Label rows of table by factor level for groups of factors >> >> One possible approach is to use unique() to get the list of distinct >> combinations, cbind() an identifying variable to that list, then use >> merge() to join it to your existing data frame. >> >> But I'm not seeing how you are getting four unique combinations. >> Given your sample data (with the missing comma replaced): >>> dim(tabledf) >> [1] 8 4 >>> head(desired) >> x1 x2 x3 res >> 1 0 1 1 3 >> 2 1 1 1 4 >> 3 0 0 1 3 >> 4 1 0 1 4 >> 5 0 1 1 3 >> 6 1 1 1 4 >> >> tabledf has 8 rows, not 4, and I don't see how rows 1 and 3 >> or rows 2 and 4 of your desired df should get the same >> classification. >> >> Regardless, if you can make a data frame like tabledf with >> an additional column for your desired res variable, you can >> merge() it with your original data frame. >> >> Sarah >> >> On Tue, Mar 6, 2012 at 11:06 AM, O'Hanlon, Simon J >> <simon.ohan...@imperial.ac.uk> wrote: >>> Dear useRs, >>> I am sure this is a fairly simple problem, but I just cannot get my head >>> around it. >>> >>> >>> I have a dataframe which contains several factor variables. I can use >>> table() to tell me how many different combinations there are of these >>> variables. What I should like to do is to add a column to my original >>> dataframe which labels each row according to the unique combination of >>> factors. >>> >>> >>> E.g. in the simple example below I create a dataframe 'df' with 3 columns, >>> the values of which take 0 or 1. I can then classify each row in the table >>> and I find that I have 4 unique combinations of factors. I would now like >>> to add a fourth column to df which labels each row according to whether it >>> was unique combination 1,2,3 or 4: >>> >>> x1=c(rep(0:1,6)) >>> x2=c(rep(c(1,1,0,0)6)) >>> x3=c(rep(1,6),rep(0,6)) >>> df=data.frame(x1,x2,x3) >>> tabledf=as.data.frame(with(df, table(x1,x2,x3))) >>> res=c(3,4,3,4,3,4,1,2,1,2,1,2) >>> desired=data.frame(x1,x2,x3,res) >>> df >>> tabledf >>> desired >>> >>> >>> I realise that this is probably quite simple to do, I am just struggling to >>> get my head around it! Help much appreciated in advance. >>> ______________________________________________ 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.
