Fitted model of linear spline regression in R

Question

If the linear model is fitted as follows, what is the formula of the fitted model? I need to use the coefficients outside R.

fit <- lm( y ~ bs( x, degree=1, knots=c(6,12,18) ) );

Coefficients:
(Intercept)                    9.2679
ns(x, knots = c(6, 12, 18))1   5.3064
ns(x, knots = c(6, 12, 18))2   5.1313
ns(x, knots = c(6, 12, 18))3   6.1153
ns(x, knots = c(6, 12, 18))4   1.3724

see `Function` and `latex.rms` in package `rms` -- they may be of some help — Glen_b, Feb 17 '15 at 04:46
@Glen_b: They only convert natural splines generated with `rcs` to the truncated power basis, unfortunately. — Scortchi - Reinstate Monica, Feb 17 '15 at 14:41
@Scortchi Hmm. Okay, thanks. I must have been thinking of something else - I'm sure Frank had something in there to help with this. — Glen_b, Feb 18 '15 at 11:29
Try pmin(x - yourknot, 0)... The first knot will be 0, I suspect... — Chris, Jul 09 '17 at 21:41

score 5 · Answer 1 · edited Apr 13 '17 at 12:44

The coefficients have the usual interpretation, but for the B-spline basis functions; which you can generate for new data easily enough in R :

bs(x, degree=1, knots=c(6,12,18)) -> x.bspline.bff
new.x <- c(10.2, 11.8, 13, 30)
predict(x.bspline.bff, new.x)

Most software will have functions to generate these (e.g. SAS, Stata); should you need to do it yourself, a recursive procedure is given in Hastie et.al (2009), The Elements of Statistical Learning, Ch.5, "Appendix: Computational considerations for splines".

You could also use an equivalent reëxpression with truncated power functions, but in general that's not a good idea—there's a danger of numerical instability with higher order splines & interactions. See here for an example of exporting a spline function to Excel.

Fitted model of linear spline regression in R

1 Answers1