[R] creating 'all' sum contrasts
OK, my last question didn't get any replies so I am going to try and ask a different way. When I generate contrasts with contr.sum() for a 3 level categorical variable I get the 2 orthogonal contrasts: contr.sum( c(1,2,3) ) [,1] [,2] 110 201 3 -1 -1 This provides the contrasts 1-3 and 2-3 as expected. But I also want it to create 1-2 (i.e. 1-3 - 2-3). So in general I want all possible orthogonal contrasts - think of it as the contrasts for all pairwise comparisons between the levels. Are there are any options for contrast() or other functions/libraries that will allow me to do this automatically? I could go through and create new columns but I am using this for complex multi-factor experiments with varying levels per factor and fitting the models within other functions (e.g. regsubsets()) so an automatic solution using what is already available would be far preferable. Michael Hopkins Algorithm and Statistical Modelling Expert Upstream 23 Old Bond Street London W1S 4PZ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating 'all' sum contrasts
G'day Michael, On Fri, 15 Oct 2010 12:09:07 +0100 Michael Hopkins hopk...@upstreamsystems.com wrote: OK, my last question didn't get any replies so I am going to try and ask a different way. When I generate contrasts with contr.sum() for a 3 level categorical variable I get the 2 orthogonal contrasts: contr.sum( c(1,2,3) ) [,1] [,2] 110 201 3 -1 -1 These two contrasts are *not* orthogonal. This provides the contrasts 1-3 and 2-3 as expected. But I also want it to create 1-2 (i.e. 1-3 - 2-3). So in general I want all possible orthogonal contrasts - think of it as the contrasts for all pairwise comparisons between the levels. You have to decide what you want. The contrasts for all pairwise comparaisons between the levels or all possible orthogonal contrasts? The latter is actually not well defined. For a factor with p levels, there would be p orthogonal contrasts, which are only identifiable up to rotation, hence infinitely many such sets. But there are p(p-1) pairwise comparisons. So unless p=2, yo have to decide what you want Are there are any options for contrast() or other functions/libraries that will allow me to do this automatically? Look at package multcomp, in particular functions glht and mcp, these might help. Cheers, Berwin == Full address Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr) School of Maths and Stats (M019)+61 (8) 6488 3383 (self) The University of Western Australia FAX : +61 (8) 6488 1028 35 Stirling Highway Crawley WA 6009e-mail: ber...@maths.uwa.edu.au Australiahttp://www.maths.uwa.edu.au/~berwin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating 'all' sum contrasts
On 15 Oct 2010, at 13:55, Berwin A Turlach wrote: G'day Michael, Hi Berwin Thanks for the reply On Fri, 15 Oct 2010 12:09:07 +0100 Michael Hopkins hopk...@upstreamsystems.com wrote: OK, my last question didn't get any replies so I am going to try and ask a different way. When I generate contrasts with contr.sum() for a 3 level categorical variable I get the 2 orthogonal contrasts: contr.sum( c(1,2,3) ) [,1] [,2] 110 201 3 -1 -1 These two contrasts are *not* orthogonal. I'm surprised. Can you please tell me how you calculated that. This provides the contrasts 1-3 and 2-3 as expected. But I also want it to create 1-2 (i.e. 1-3 - 2-3). So in general I want all possible orthogonal contrasts - think of it as the contrasts for all pairwise comparisons between the levels. You have to decide what you want. The contrasts for all pairwise comparaisons between the levels or all possible orthogonal contrasts? Well the pairwise contrasts are the most important as I am looking for evidence of whether they are zero (i.e. no difference between levels) or not. But see my above comment about orthogonality. The latter is actually not well defined. For a factor with p levels, there would be p orthogonal contrasts, which are only identifiable up to rotation, hence infinitely many such sets. But there are p(p-1) pairwise comparisons. So unless p=2, yo have to decide what you want Well of course the pairwise comparisons are bi-directional so in fact only p(p-1)/2 are of interest to me. Are there are any options for contrast() or other functions/libraries that will allow me to do this automatically? Look at package multcomp, in particular functions glht and mcp, these might help. Thanks I will have a look. But I want to be able to do this transparently within lm() using regsubsets() etc as I am collecting large quantities of summary stats from all possible models to use with a model choice criterion based upon true Bayesian model probabilities. Cheers, Berwin == Full address Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr) School of Maths and Stats (M019)+61 (8) 6488 3383 (self) The University of Western Australia FAX : +61 (8) 6488 1028 35 Stirling Highway Crawley WA 6009e-mail: ber...@maths.uwa.edu.au Australiahttp://www.maths.uwa.edu.au/~berwin Michael Hopkins Algorithm and Statistical Modelling Expert Upstream 23 Old Bond Street London W1S 4PZ Mob +44 0782 578 7220 DL +44 0207 290 1326 Fax +44 0207 290 1321 hopk...@upstreamsystems.com www.upstreamsystems.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating 'all' sum contrasts
Michael, Let c_1 and c_2 be vectors representing contrasts. Then c_1 and c_2 are orthogonal if and only if the inner product is 0. In your example, you have vectors (1,0,-1) and (0,1,-1). The inner product is 1, so they are not orthogonal. It's impossible to have more orthogonal contrasts than you have levels in your factor, a result from basic linear algebra. You can get all possible pairwise contrasts, which is different from orthogonal contrasts (in fact, it's only possible to have floor(n/2) orthogonal pairwise contrasts). This is probably not the easiest way, but it works: n - 10 M - matrix(0,nrow=n,ncol=n*(n-1)/2) comb - combn(n,2) M[cbind(comb[1,],1:(n*(n-1)/2))] - 1 M[cbind(comb[2,],1:(n*(n-1)/2))] - -1 M is then a matrix containing all pairwise contrasts for n levels of a factor. Hope that helps, Jonathan On Fri, Oct 15, 2010 at 10:30 AM, Michael Hopkins hopk...@upstreamsystems.com wrote: On 15 Oct 2010, at 13:55, Berwin A Turlach wrote: G'day Michael, Hi Berwin Thanks for the reply On Fri, 15 Oct 2010 12:09:07 +0100 Michael Hopkins hopk...@upstreamsystems.com wrote: OK, my last question didn't get any replies so I am going to try and ask a different way. When I generate contrasts with contr.sum() for a 3 level categorical variable I get the 2 orthogonal contrasts: contr.sum( c(1,2,3) ) [,1] [,2] 1 1 0 2 0 1 3 -1 -1 These two contrasts are *not* orthogonal. I'm surprised. Can you please tell me how you calculated that. This provides the contrasts 1-3 and 2-3 as expected. But I also want it to create 1-2 (i.e. 1-3 - 2-3). So in general I want all possible orthogonal contrasts - think of it as the contrasts for all pairwise comparisons between the levels. You have to decide what you want. The contrasts for all pairwise comparaisons between the levels or all possible orthogonal contrasts? Well the pairwise contrasts are the most important as I am looking for evidence of whether they are zero (i.e. no difference between levels) or not. But see my above comment about orthogonality. The latter is actually not well defined. For a factor with p levels, there would be p orthogonal contrasts, which are only identifiable up to rotation, hence infinitely many such sets. But there are p(p-1) pairwise comparisons. So unless p=2, yo have to decide what you want Well of course the pairwise comparisons are bi-directional so in fact only p(p-1)/2 are of interest to me. Are there are any options for contrast() or other functions/libraries that will allow me to do this automatically? Look at package multcomp, in particular functions glht and mcp, these might help. Thanks I will have a look. But I want to be able to do this transparently within lm() using regsubsets() etc as I am collecting large quantities of summary stats from all possible models to use with a model choice criterion based upon true Bayesian model probabilities. Cheers, Berwin == Full address Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr) School of Maths and Stats (M019) +61 (8) 6488 3383 (self) The University of Western Australia FAX : +61 (8) 6488 1028 35 Stirling Highway Crawley WA 6009 e-mail: ber...@maths.uwa.edu.au Australia http://www.maths.uwa.edu.au/~berwin Michael Hopkins Algorithm and Statistical Modelling Expert Upstream 23 Old Bond Street London W1S 4PZ Mob +44 0782 578 7220 DL +44 0207 290 1326 Fax +44 0207 290 1321 hopk...@upstreamsystems.com www.upstreamsystems.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
