Linear model - coefficients

this may work.
X<-data.frame(sapply(X,function(x) as.factor(x)))
reg3=lm(Y~.,data=X)
dummy.coef(reg3)

Weidong Gu

On Sun, Jun 12, 2011 at 4:55 PM, Robert Ruser <robert.ruser at gmail.com> wrote:

Hi,
but I want to get the coefficients for every variables from x1 to x5.
(x1 was an example)

Robert

2011/6/12 Jorge Ivan Velez <jorgeivanvelez at gmail.com>:

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

--
Brian D. Ripley, ? ? ? ? ? ? ? ? ?ripley at 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