Apologies - you're right. Missed it in the pdf. K.
On Sun, Sep 28, 2014 at 10:22 AM, Bert Gunter <gunter.ber...@gene.com> wrote: > Inline. > > Bert Gunter > Genentech Nonclinical Biostatistics > (650) 467-7374 > > "Data is not information. Information is not knowledge. And knowledge > is certainly not wisdom." > Clifford Stoll > > > > > On Sun, Sep 28, 2014 at 6:38 AM, Kate Ignatius <kate.ignat...@gmail.com> > wrote: >> Strange that, >> >> I did put everything with as.character but all I got was the same... >> >> class of dbpmn[,2]) = factor >> class of dbpmn[,21] = factor >> class of dbpmn[,20] = data.frame >> >> This has to be a problem ??? > > Indeed -- your failure to read documentation. > > I suggest you do your due diligence, read Pat Burns's link, and follow > the advice given you by posting a reproducible example. More than > likely the last will be unnecessary as you will figure it out in the > course of doing what you should do. > > Cheers, > Bert > >> >> I can put reproducible output here but not sure if this going to of >> help here. I think its all about factors and data frames and >> characters... >> >> K. >> >> On Sun, Sep 28, 2014 at 1:15 AM, Jim Lemon <j...@bitwrit.com.au> wrote: >>> On Sun, 28 Sep 2014 12:49:41 AM Kate Ignatius wrote: >>>> Quick question: >>>> >>>> I am running the following code on some variables that are factors: >>>> >>>> dbpmn$IID1new <- ifelse(as.character(dbpmn[,2]) == >>>> as.character(dbpmn[,(21)]), dbpmn[,20], '') >>>> >>>> Instead of returning some value it gives me this: >>>> >>>> c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)) >>>> >>>> Playing around with the code, gives me some kind of variation to it. >>>> Is there some way to get me what I want. The variable that its >>>> suppose to give back is a bunch of sampleIDs. >>>> >>> Hi Kate, >>> If I create a little example: >>> >>> dbpmn<-data.frame(V1=factor(sample(LETTERS[1:4],20,TRUE)), >>> V2=factor(sample(LETTERS[1:4],20,TRUE)), >>> V3=factor(sample(LETTERS[1:4],20,TRUE))) >>> dbpmn[4]<- >>> ifelse(as.character(dbpmn[,1]) == as.character(dbpmn[,(2)]), >>> dbpmn[,3],"") >>> dbpmn >>> V1 V2 V3 V4 >>> 1 B D C >>> 2 C A D >>> 3 C B A >>> 4 A B C >>> 5 B D B >>> 6 D D A 1 >>> 7 D D D 4 >>> 8 B C A >>> 9 B D B >>> 10 D C A >>> 11 A D C >>> 12 A C B >>> 13 A A A 1 >>> 14 D C A >>> 15 C D B >>> 16 A A B 2 >>> 17 A C C >>> 18 B B C 3 >>> 19 C C C 3 >>> 20 D D D 4 >>> >>> I get what I expect, the numeric value of the third element in dbpmn >>> where the first two elements are equal. I think what you want is: >>> >>> dbpmn[4]<- >>> ifelse(as.character(dbpmn[,1]) == as.character(dbpmn[,(2)]), >>> as.character(dbpmn[,3]),"") >>> dbpmn >>> V1 V2 V3 V4 >>> 1 B D C >>> 2 C A D >>> 3 C B A >>> 4 A B C >>> 5 B D B >>> 6 D D A A >>> 7 D D D D >>> 8 B C A >>> 9 B D B >>> 10 D C A >>> 11 A D C >>> 12 A C B >>> 13 A A A A >>> 14 D C A >>> 15 C D B >>> 16 A A B B >>> 17 A C C >>> 18 B B C C >>> 19 C C C C >>> 20 D D D D >>> >>> Jim >>> >> >> ______________________________________________ >> 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. ______________________________________________ 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.