Export fitted regression splines (constructed by 'bs' or 'ns') as piecewise polynomials

Question

Take for instance the following one-knot, degree two, spline:

library(splines)
library(ISLR)
fit.spline <- lm(wage~bs(age, knots=c(42), degree=2), data=Wa         


        

Answer 1

I was constantly asked to wrap up the idea in my original answer into a user-friendly function, able to reparametrize a fitted linear or generalized linear model with a bs or ns term. Eventually I rolled out a small R package SplinesUtils at https://github.com/ZheyuanLi/SplinesUtils (with a PDF version package manual). You can install it via

## make you have `devtools` package avaiable
devtools::install_github("ZheyuanLi/SplinesUtils")

The function to be used here is RegBsplineAsPiecePoly.

library(SplinesUtils)

library(splines)
library(ISLR)
fit.spline <- lm(wage ~ bs(age, knots=c(42), degree=2), data = Wage)

ans1 <- RegBsplineAsPiecePoly(fit.spline, "bs(age, knots = c(42), degree = 2)")
ans1
#2 piecewise polynomials of degree 2 are constructed!
#Use 'summary' to export all of them.
#The first 2 are printed below.
#8.2e-15 + 4.96 * (x - 18) + 0.0991 * (x - 18) ^ 2
#61.9 + 0.2 * (x - 42) + 0.0224 * (x - 42) ^ 2

## coefficients as a matrix
ans1$PiecePoly$coef
#              [,1]        [,2]
#[1,]  8.204641e-15 61.91542748
#[2,]  4.959286e+00  0.20033307
#[3,] -9.914485e-02 -0.02240887

## knots
ans1$knots
#[1] 18 42 80

The function defaults to parametrize piecewise polynomials in shifted form (see ?PiecePoly). You can set shift = FALSE for a non-shifted version.

ans2 <- RegBsplineAsPiecePoly(fit.spline, "bs(age, knots = c(42), degree = 2)",
                              shift = FALSE)
ans2
#2 piecewise polynomials of degree 2 are constructed!
#Use 'summary' to export all of them.
#The first 2 are printed below.
#-121 + 8.53 * x + 0.0991 * x ^ 2
#14 + 2.08 * x + 0.0224 * x ^ 2

## coefficients as a matrix
ans2$PiecePoly$coef
#              [,1]        [,2]
#[1,] -121.39007747 13.97219046
#[2,]    8.52850050  2.08267822
#[3,]   -0.09914485 -0.02240887

You can predict the splines with predict.

xg <- 18:80
yg1 <- predict(ans1, xg)  ## use shifted form
yg2 <- predict(ans2, xg)  ## use non-shifted form
all.equal(yg1, yg2)
#[1] TRUE

But since there is an intercept in the model, the predicted values would differ from model prediction by the intercept.

yh <- predict(fit.spline, data.frame(age = xg))
intercept <- coef(fit.spline)[[1]]
all.equal(yh, yg1 + intercept, check.attributes = FALSE)
#[1] TRUE

The package has summary, print, plot, predict and solve methods for a "PiecePoly" class. Explore the package for more.