Re: [R] contr.treatments query

Apologies -- I have just observed my oversight below! Please ignore the message below. Ted. On 20-May-08 09:14:30, Ted Harding wrote: > On 20-May-08 06:10:48, Prof Brian Ripley wrote: >>>From ?contrasts >> Usage: >> contrasts(x, how.many) <- value >> ... >> how.many: How many contrasts shoul

Re: [R] contr.treatments query

On 20-May-08 06:10:48, Prof Brian Ripley wrote: >>From ?contrasts > Usage: > contrasts(x, how.many) <- value > ... > how.many: How many contrasts should be made. Defaults to one less than >the number of levels of 'x'. This need not be the same as >the number of column

Re: [R] contr.treatments query

From ?contrasts Usage: contrasts(x, how.many) <- value ... how.many: How many contrasts should be made. Defaults to one less than the number of levels of 'x'. This need not be the same as the number of columns of 'ctr'. so that is 2 in your example, and it takes the

[R] contr.treatments query

Hi Folks, I'm a bit puzzled by the following (example): N<-factor(sample(c(1,2,3),1000,replace=TRUE)) unique(N) # [1] 3 2 1 # Levels: 1 2 3 So far so good. Now: contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE) contrasts(N) # 1 2 # 1 1 0 # 2 0 1 # 3 0 0 whereas: contr.treatment(3, ba