Hi Robert,

Try this:

reg2 <- lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) +
factor(x5) - 1, data = X  )
cof(ref2)

HTH,
Jorge


On Sun, Jun 12, 2011 at 4:40 PM, Robert Ruser <> wrote:

> Prof. Ripley, thank you very much for the answer but wanted to get
> something else. There is an example and an explanation:
>
> options(contrasts=c("contr.sum","contr.poly")) # contr.sum uses ‘sum
> to zero contrasts’
> Y <- c(6,3,5,2,3,1,1,6,6,6,7,4,1,6,6,6,6,1)
> X <- structure(list(x1 = c(2L, 3L, 1L, 3L, 3L, 2L, 1L, 1L, 3L, 2L,
> 3L, 2L, 1L, 1L, 2L, 1L, 2L, 3L), x2 = c(3L, 3L, 2L, 3L, 1L, 3L,
> 2L, 3L, 2L, 1L, 2L, 2L, 3L, 1L, 2L, 1L, 1L, 1L), x3 = c(1L, 1L,
> 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 1L
> ), x4 = c(1L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 2L, 2L, 1L, 2L), x5 = c(1L, 1L, 2L, 2L, 3L, 3L, 3L, 2L, 2L,
> 2L, 1L, 3L, 3L, 1L, 1L, 1L, 2L, 3L)), .Names = c("x1", "x2",
> "x3", "x4", "x5"), row.names = c(NA, 18L), class = "data.frame")
>
> reg <- lm( Y ~ factor(X$x1) + factor(X$x2) + factor(X$x3) +
> factor(X$x4) + factor(X$x5)   )
> coef(reg)
>
> and e.g. I get two coefficients for variable x1 (3-levels variable)
> but I would like to get the third. Of course I can calculate a3=
> -(a1+a2) where a1 and a2 are coefficients of the variable x1.
>
> I hope that I manage to explain my problem.
>
> Robert
>
> 2011/6/12 Prof Brian Ripley <>:
> > ?dummy.coef
> >
> > (NB: 'R' does as you tell it, and if you ask for the default contrasts
> you
> > get coefficients a2 and a3, not a1 and a2.  So perhaps you did something
> > else and failed to tell us?  And see the comment in ?dummy.coef about
> > treatment contrasts.)
> >
> >
> > On Sun, 12 Jun 2011, Robert Ruser wrote:
> >
> >> Dear R Users,
> >> Using lm() function with categorical variable R use contrasts. Let
> >> assume that I have one X independent variable with 3-levels. Because R
> >> estimate only 2 parameters ( e.g. a1, a2)  the coef function returns
> >> only 2 estimators. Is there any function or trick to get another a3
> >> values. I know that using contrast sum (?contr.sum) I could compute a3
> >> = -(a1+a2). But I have many independent categorical variables and I'm
> >> looking for a fast solution.
> >>
> >> Robert
> >>
> >> ______________________________________________
> >> 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.
> >>
> >
> > --
> > Brian D. Ripley,                  ripley@stats.ox.ac.uk
> > Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> > University of Oxford,             Tel:  +44 1865 272861 (self)
> > 1 South Parks Road,                     +44 1865 272866 (PA)
> > Oxford OX1 3TG, UK                Fax:  +44 1865 272595
> >
>
> ______________________________________________
> 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.
>

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