将拟合回归样条（由 'bs' 或 'ns' 构造）导出为分段多项式

Question

以下面的一结二阶样条为例：

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

我看到了出乎我意料的估计值。

Coefficients:
                                  Estimate  Std. Error   t value    Pr(>|t|)   
(Intercept)                           57.349   3.950   14.518   < 2e-16 ***
bs(age, knots = c(42), degree = 2)1   59.511   5.786   10.285   < 2e-16 ***
bs(age, knots = c(42), degree = 2)2   65.722   4.076   16.122   < 2e-16 ***
bs(age, knots = c(42), degree = 2)3   37.170   9.722    3.823  0.000134 ***

有没有办法提取结前后的二次模型（及其系数）？即如何提取age = 42切点前后的两个二次模型？

使用 summary(fit.spline) 产生系数，但（据我所知）它们对解释没有意义。

Answer 1

我经常被要求将原始答案中的想法包装成一个用户友好的函数，能够用 bs 或 ns 项重新参数化拟合线性或广义线性模型。最终我在 https://github.com/ZheyuanLi/SplinesUtils 推出了一个小的 R 包 SplinesUtils（带有 PDF 版本的包手册）。您可以通过

安装它

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

这里要用到的函数是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

该函数默认以移位形式参数化分段多项式（参见 ?PiecePoly）。您可以为非移动版本设置 shift = FALSE。

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

您可以使用 predict.

预测样条曲线

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

但是由于模型中存在截距，因此预测值与截距的模型预测值会有所不同。

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

该包具有“PiecePoly”class 的 summary、print、plot、predict 和 solve 方法。浏览套餐了解更多。

将拟合回归样条（由 'bs' 或 'ns' 构造）导出为分段多项式

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

regression

r

spline

lm

polynomials