Hi! I would like to suggest to make it possible, in one way or another, to get meaningful contrast names when using contr.sum(). Currently, when using contr.treatment(), one gets factor levels as contrast names; but when using contr.sum(), contrasts are merely numbered, which is not practical and can lead to mistakes (see code at the end of this message).
This issue was discussed quickly in 2005 by Brian Ripley in a reply to a message on R-help [1]. He rightly stressed that treatment and sum contrasts are not equivalent to levels of a factor, because one needs to know the reference (here, level or sum) to interpret them. But when one knows the type of contrasts that are being used, useful labels are still of high value. I don't think anybody does serious work with sum contrasts named myfactor1, myfactor2, myfactor3. (This reasoning does not so much apply to contr.helmert() since ordered factors can quite naturally be reported using numbers.) Thus, would it be possible to add an option to contr.sum() so that it returns a matrix whose column names are the levels of the input factor? Such an option could also be added to other contrasts with default to FALSE. Another solution, which could be even more practical, would be to add a new function, called for example contr.sum2(), which would do the same thing - after all, we already have contr.SAS() to implement a slightly different behavior while being essentially the same as contr.treatment(). This contr.sum() issue really sounds like a detail, but it's sad one given that factors work really great in R in all other situations. The only reason I can think of to explain this behavior is that people rarely use it. When fitting log-linear models with glm(), for example, this contrast is the most natural one, but currently gives poorly named coefficients when everything could be so easy to interpret if factor levels were used. This means people have to implement a replacement for contr.sum() by hand, which is not the end of the world but is definitely not optimal given how simple the solution is. Thanks for your attention! Illustration of the current difference between contr.sum() and contr.treatment(): > z <- factor(LETTERS[1:3]) > contr.treatment(z) B C A 0 0 B 1 0 C 0 1 > contr.sum(z) [,1] [,2] A 1 0 B 0 1 C -1 -1 1: https://stat.ethz.ch/pipermail/r-help/2005-July/075430.html ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
