Hi, Thanks for the reply.
Basically I dont have any label for my data except column 1 which is labeled with Sample1, Sample2...etc...(vertical). My cancerv1 data is data.frame. So, I used df <- data.frame( x=I(coef(cancerv1(,2:407))), y=cancerv1[,408])before feeding to PCR. However, I get the below error. Error in coef(cancerv1(, 2:407)) : could not find function "cancerv1". I wonder what mistakes did I made in thiscase. My response variable is on column 408 and my predictors are from column 2 to 407. Please advise. Thanks. On Tue, Jul 1, 2008 at 6:41 PM, Bjørn-Helge Mevik <[EMAIL PROTECTED]> wrote: > Gavin Simpson <[EMAIL PROTECTED]> writes: > > > You can do this another way though, that I feel is more natural. So lets > > assume that your data frame contains columns that are named, and that > > one of these is the response variable, the remaining columns are the > > predictors. Further assume that this response is called 'myresp', then > > you can proceed by the following: > > > > cancerv1.pcr <- pcr(myresp ~ . , ncomp = 6, data = cancerv1, > > validation = "CV") > > This works fine as long as the number of (predictor) variables is not > too large. With many variables (>> 1000), R will spend a very long time > dealing with the formula. > > -- > Bjørn-Helge Mevik > > ______________________________________________ > 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.
