R help - May 2011 - displaying derived coefficients in lm

Hello R-help,

Is there a way to get R to tell you the coefficients in a lm that it 
wouldn't normally tell you because of identifiability constraints? For 
instance, if you use contr.sum() to generate contrasts for a factor, say

## y <- some data
## x <- a factor with levels 1:6
contrasts(x) <- contr.sum(levels(x))
lm.1 <- lm(y ~ x)

how would one persuade summary.lm to give the coefficient for x6 as well 
as x1 up to x5? I know in this case that x6 is minus the sum of x1 to 
x5, but I am working on a problem where the constraints are very 
complicated and it would be a real help to see all the coefficients.

Thanks in advance

James Lawrence.

Look at the dummy.coef function.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of James Lawrence
> Sent: Wednesday, May 11, 2011 5:43 AM
> To: r-help at r-project.org
> Subject: [R] displaying derived coefficients in lm
> 
> Hello R-help,
> 
> Is there a way to get R to tell you the coefficients in a lm that it
> wouldn't normally tell you because of identifiability constraints? For
> instance, if you use contr.sum() to generate contrasts for a factor,
> say
> 
> ## y <- some data
> ## x <- a factor with levels 1:6
> contrasts(x) <- contr.sum(levels(x))
> lm.1 <- lm(y ~ x)
> 
> how would one persuade summary.lm to give the coefficient for x6 as
> well
> as x1 up to x5? I know in this case that x6 is minus the sum of x1 to
> x5, but I am working on a problem where the constraints are very
> complicated and it would be a real help to see all the coefficients.
> 
> Thanks in advance
> 
> James Lawrence.
> 
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