Help with contrasts

Or alternatively (though very similar to Peter's idea) you can do

ci <- contrasts
formals(ci)$contrasts <- FALSE
dd <- data.frame(a = gl(3,4), b = gl(4,1,12))
mm <- model.matrix(~ a + b, dd, contrasts = list(a=ci(dd$a),
b=ci(dd$b))) 

Best,
Leo.

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of peter dalgaard
Sent: 2011, May, 11 7:31 AM
To: Lars Bishop
Cc: R-help at r-project.org Help
Subject: Re: [R] Help with contrasts

Hi,

I need to build a function to generate one column for each level of a 
factor in the model matrix created on an arbitrary formula (instead of

using the available contrasts options such as contr.treatment,

contr.SAS, etc).

My approach to this was first to use the built-in function for 
contr.treatment but changing the default value of the contrasts 
argument to FALSE (I named this function "contr.identity" and it shown

at the bottom of the email for reference).

So this function works fine,

contr.identity(4)

 1 2 3 4
1 1 0 0 0
2 0 1 0 0
3 0 0 1 0
4 0 0 0 1

contr.treatment(4)

 2 3 4
1 0 0 0
2 1 0 0
3 0 1 0
4 0 0 1

However, when I try to create a model matrix using contr.identity 
specified in options(), it actually uses the contr.treatment option. 
Why is that? Any hint on how can I do this?

It's not actually using contr.treatment, it's just calling
contr.identity with contrasts=TRUE...

I don't think there's a painless way to avoid this. The closest I can
think of is

cI <- contr.treatment
formals(cI)$contrasts <- FALSE
dd <- data.frame(a = gl(3,4), b = gl(4,1,12)) model.matrix(~ C(a,cI,3) +
C(b,cI,4), dd)



--
Peter Dalgaard
Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000
Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

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