From ?contrasts
Usage:
contrasts(x, how.many) <- value
...
how.many: How many contrasts should be made. Defaults to one less than
the number of levels of 'x'. This need not be the same as
the number of columns of 'ctr'.
so that is 2 in your example, and it takes the first 2 of the 3 you
supplied.
(The posting guide does ask you to read the help before posting.)
On Tue, 20 May 2008, [EMAIL PROTECTED] wrote:
Hi Folks,
I'm a bit puzzled by the following (example):
N<-factor(sample(c(1,2,3),1000,replace=TRUE))
unique(N)
# [1] 3 2 1
# Levels: 1 2 3
So far so good. Now:
contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE)
contrasts(N)
# 1 2
# 1 1 0
# 2 0 1
# 3 0 0
whereas:
contr.treatment(3, base=1, contrasts=FALSE)
# 1 2 3
# 1 1 0 0
# 2 0 1 0
# 3 0 0 1
contr.treatment(3, base=1, contrasts=TRUE)
# 2 3
# 1 0 0
# 2 1 0
# 3 0 1
I can follow the last two fine -- they are what is implied
by the code for contr.treatment().
Likewise:
contrasts(factor(Nlevs <-c(1,2,3)))
# 2 3
# 1 0 0
# 2 1 0
# 3 0 1
But why the different result when applied to N?
With thanks,
Ted.
--------------------------------------------------------------------
E-Mail: (Ted Harding) <[EMAIL PROTECTED]>
Fax-to-email: +44 (0)870 094 0861
Date: 20-May-08 Time: 01:12:30
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--
Brian D. Ripley, [EMAIL PROTECTED]
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
______________________________________________
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and provide commented, minimal, self-contained, reproducible code.
