[R] Contrasts for ordered factors
Professor Brian Ripley
ripley at stats.ox.ac.uk
Mon Jan 8 11:37:11 CET 2007
contr.poly is using orthoogonal polynomials: look at poly() for
further information. It seems to me that you did not realize that,
or did not realize what they are, or ... and that may be enough of
a hint for you or you may need more help, in which case please ask
again.
lorenz.gygax at art.admin.ch wrote:
> Dear all,
>
> I do not seem to grasp how contrasts are set for ordered factors. Perhaps someone can elighten me?
>
> When I work with ordered factors, I would often like to be able to reduce the used polynomial to a simpler one (where possible). Thus, I would like to explicetly code the polynomial but ideally, the intial model (thus, the full polynomial) would be identical to one with an ordered factor.
>
> Here is a toy example with an explanatory variable (EV) with three distinct values (1 to 3) and a continuous response variable (RV):
>
> options (contrasts= c ('contr.treatment', 'contr.poly'))
> example.df <- data.frame (EV= rep (1:3, 5))
> set.seed (298)
> example.df$RV <- 2 * example.df$EV + rnorm (15)
>
> I evaluate this data using either an ordered factor or a polynomial with a linear and a quadratic term:
>
> lm.ord <- lm (RV ~ ordered (EV), example.df)
> lm.pol <- lm (RV ~ EV + I(EV^2), example.df)
>
> I then see that the estimated coefficients differ (and in other examples that I have come across, it is often even more extreme):
>
> coef (lm.ord)
> (Intercept) ordered(EV).L ordered(EV).Q
> 3.9497767 2.9740535 -0.1580798
> coef (lm.pol)
> (Intercept) EV I(EV^2)
> -0.9015283 2.8774032 -0.1936074
>
> but the predictions are the same (except for some rounding):
>
> table (round (predict (lm.ord), 6) == round (predict (lm.pol), 6))
> TRUE
> 15
>
> I thus conclude that the two models are the same and are just using a different parametrisation. I can easily interprete the parameters of the explicit polynomial but I started to wonder about the parametrisation of the ordered factor. In search of an answer, I did check the contrasts:
>
> contr.poly (levels (ordered (example.df$EV)))
> .L .Q
> [1,] -7.071068e-01 0.4082483
> [2,] -9.073264e-17 -0.8164966
> [3,] 7.071068e-01 0.4082483
>
> The linear part basically seems to be -0.707, 0 (apart for numerical rounding) and 0.707. I can understand that any even-spaced parametrisation is possible for the linear part. But does someone know where the value of 0.707 comes from (it seems to be the square-root of 0.5, but why?) and why the middle term is not exactly 0?
>
> I do not understand the quadratic part at all. Wouldn't that need the be the linear part to the power of 2?
>
> Thank you for your thoughts! Lorenz
> -
> Lorenz Gygax
> Swiss Federal Veterinary Office
> Centre for proper housing of ruminants and pigs
> Tänikon, CH-8356 Ettenhausen / Switzerland
>
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