[R-sig-Epi] Ns(..., detrend=T)

[Stephen Wade]

When detrend=T I would have expected that Epi::Ns(…)would more or less project the basis given by splines::ns(…) onto the orthogonal complement of the column space of [1 t] and finally extract the set of linearly independent columns (to give a basis).

However, this doesn’t appear to be the _exactly_ the case; I tried

x=seq(-0.75, 0.75, length.out=5)
Ns(x, knots=c(-0.5,0,0.5), Boundary.knots=c(-1,1), detrend=T)

and

detrend(ns(x, knots=c(-0.5,0,0.5), Boundary.knots=c(-1,1)), x)

The matrices produced by the above code are not the same, however, they do have the same column space in this example suggesting that if plugged in to a linear model, the fitted coefficients will be different but the fit (itself) will be the same. I haven’t yet checked/proved that the column spaces should be the same ‘in general’.

The call to Ns above gives a warning about fixsl being ignored. It turns out that the value of const in ns.ld() is not the same compared to the same variable in the call to splines::ns() above. This is because the fixsl argument in ns.ld() will have value c(NA,NA) when passed via Ns(…, detrend=T).

Does anyone have an explanation for why fixsl=c(NA,NA) is used instead of fixsl=c(F,F)? Or for why the two matrices in the above example should/would have the same column space?

Kind regards,

Stephen Wade | Post-Doc Research Fellow
Cancer Council NSW
153 Dowling Street
Woolloomooloo NSW 2021
☏ : (02) 9334 1431 | ✉ : Stephen.Wade at nswcc.org.au
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