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 <b.h.mevik@usit.uio.no>
wrote:

> Gavin Simpson <gavin.simpson@ucl.ac.uk> 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
>
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