R help - Sep 2009 - coefficients of aov results has less number of elements?

Hi,

I run the following commands. 'A' has 3 levels and 'B' has 4
levels.
Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3,
B4)?
> a=3
> b=4
> n=1000
> A = rep(sapply(1:a,function(x){rep(x,n)}),b)
> B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}),
function(x){rep(x,a)}))
> Y = A + B + rnorm(a*b*n)
>
> fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B))
> afit=aov(Y ~ A + B - 1,fr)
> afit$coefficients
      A1       A2       A3       B2       B3       B4
1.993067 2.969252 3.965888 1.005445 2.020717 3.023940

Regards,
Peng

On Tue, Sep 15, 2009 at 10:34 AM, Peng Yu <pengyu.ut at gmail.com>
wrote:
> Hi,
>
> I run the following commands. 'A' has 3 levels and 'B' has
4 levels.
> Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3,
> B4)?
If you were to define a model matrix for those coefficients it would
be rank deficient.  (the sum of the first three columns is 1 and the
sum of the last four columns is 1)  You only have 6 degrees of freedom
in total, not 7.  Because of the -1 in the formula the first factor is
expanded to the indicator columns but the second is represented with a
set of 3 "contrasts" which are the contr.treatment version, by
default.
>> a=3
>> b=4
>> n=1000
>> A = rep(sapply(1:a,function(x){rep(x,n)}),b)
>> B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}),
function(x){rep(x,a)}))
>> Y = A + B + rnorm(a*b*n)
>>
>> fr = data.frame(Y=Y,A=as.factor(A),B=as.factor(B))
>> afit=aov(Y ~ A + B - 1,fr)
>> afit$coefficients
> ? ? ?A1 ? ? ? A2 ? ? ? A3 ? ? ? B2 ? ? ? B3 ? ? ? B4
> 1.993067 2.969252 3.965888 1.005445 2.020717 3.023940
>
> Regards,
> Peng
>
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