R中带有mgcv的粗薄板样条拟合（薄板样条插值）

Question

背景

我试图在《 统计学习入门 》一书中复制图2.6 ：

粗糙的薄板样条曲线适合图2.3中的收入数据。 这种拟合使训练数据的错误为零。

到目前为止，我尝试了什么？

我试图复制前面的图2.5，它是平滑的薄板样条曲线拟合，不确定是否成功。

income_2 <- read.csv("http://www-bcf.usc.edu/~gareth/ISL/Income2.csv")
library(mgcv)
model1 <- gam(Income ~ te(Education, Seniority, bs=c("tp", "tp")), data = income_2) 
x <- range(income_2$Education)
x <- seq(x[1], x[2], length.out=30)
y <- range(income_2$Seniority)
y <- seq(y[1], y[2], length.out=30)
z <- outer(x,y,
           function(Education,Seniority)
                     predict(model1, data.frame(Education,Seniority)))
p <- persp(x,y,z, theta=30, phi=30,
           col="yellow",expand = 0.5,shade = 0.2,
           xlab="Education", ylab="Seniority", zlab="Income")
obs <- trans3d(income_2$Education, income_2$Seniority,income_2$Income,p)
pred <- trans3d(income_2$Education, income_2$Seniority,fitted(model1),p)
points(obs, col="red",pch=16)
segments(obs$x, obs$y, pred$x, pred$y)

双重问题

1. 我是否用gam创建了合适的光滑薄板？ 我使用的是smooth.terms bs="tp" ，文档中说：“它们是薄板样条线的降阶版本，并使用薄板样条线罚分。”
2. 如何创建粗糙的薄板样条拟合，使训练数据零误差？ （上面的第一个数字）

Answer 1

income_2 <- structure(list(Education = c(21.5862068965517, 18.2758620689655, 
12.0689655172414, 17.0344827586207, 19.9310344827586, 18.2758620689655, 
19.9310344827586, 21.1724137931034, 20.3448275862069, 10, 13.7241379310345, 
18.6896551724138, 11.6551724137931, 16.6206896551724, 10, 20.3448275862069, 
14.1379310344828, 16.6206896551724, 16.6206896551724, 20.3448275862069, 
18.2758620689655, 14.551724137931, 17.448275862069, 10.4137931034483, 
21.5862068965517, 11.2413793103448, 19.9310344827586, 11.6551724137931, 
12.0689655172414, 17.0344827586207), Seniority = c(113.103448275862, 
119.310344827586, 100.689655172414, 187.586206896552, 20, 26.2068965517241, 
150.344827586207, 82.0689655172414, 88.2758620689655, 113.103448275862, 
51.0344827586207, 144.137931034483, 20, 94.4827586206897, 187.586206896552, 
94.4827586206897, 20, 44.8275862068966, 175.172413793103, 187.586206896552, 
100.689655172414, 137.931034482759, 94.4827586206897, 32.4137931034483, 
20, 44.8275862068966, 168.965517241379, 57.2413793103448, 32.4137931034483, 
106.896551724138), Income = c(99.9171726114381, 92.579134855529, 
34.6787271520874, 78.7028062353695, 68.0099216471551, 71.5044853814318, 
87.9704669939115, 79.8110298331255, 90.00632710858, 45.6555294997364, 
31.9138079371295, 96.2829968022869, 27.9825049000603, 66.601792415137, 
41.5319924201478, 89.00070081522, 28.8163007592387, 57.6816942573605, 
70.1050960424457, 98.8340115435447, 74.7046991976891, 53.5321056283034, 
72.0789236655191, 18.5706650327685, 78.8057842852386, 21.388561306174, 
90.8140351180409, 22.6361626208955, 17.613593041445, 74.6109601985289
)), .Names = c("Education", "Seniority", "Income"), row.names = c(NA, 
-30L), class = "data.frame")

library(mgcv)

首先，您可以将s(Education, Seniority, bs = 'tp')用于二元薄板样条曲线，而不必使用张量积构造。 薄板样条线在任何尺寸上都是明确定义的。

其次， mgcv不会进行回归而不是插值，因此，如果不进行调整，就无法使拟合的样条曲线遍历所有点。 薄板样条的调整包括：

1. 通过将k设置为唯一采样点的确切数量（如果您有超过2000个唯一数据位置，则为xt来禁用bs = 'tp'后面的低秩逼近；
2. 设置sp = 0禁用样条曲线的惩罚。

您的数据集income_2中唯一采样位置的数量为

xt <- unique(income_2[c("Education", "Seniority")]) 
nrow(xt)
#[1] 30

因为30小于2000，所以我们可以将k = 30设置k = 30不需要将xt传递给s(, bs = 'tp') 。

interpolation_model <- gam(Income ~ s(Education, Seniority, k = 30, sp = 0),
                           data = income_2)
interpolation_model$residuals
# [1]  2.131628e-13  2.728484e-12  4.561684e-12  1.264766e-12  3.495870e-12
# [6]  4.177991e-12 -1.023182e-12  1.193712e-12  2.231104e-12  6.878054e-12
#[11]  6.309619e-12  6.679102e-13  7.574386e-12  3.637979e-12  4.227729e-12
#[16]  1.790568e-12  4.376943e-12  5.130119e-12  8.242296e-13 -6.536993e-13
#[21]  2.771117e-12  1.811884e-12  3.495870e-12  9.141132e-12  2.117417e-12
#[26]  7.243983e-12 -3.979039e-13  6.352252e-12  6.203038e-12  3.652190e-12

现在您看到所有残差均为零。

您还可以查找直接执行薄板样条插值的其他程序包。

---

薄板花键是各向同性的/径向的，如果变量的比例不同，请小心！

感谢您的解释和解决我的问题。 您知道为什么花键表面看起来起伏更大，而脊却更少吗？

因为您的两个变量的比例差异很大。 您要首先标准化两个变量，然后拟合薄板样条。

## this is how your original data look like on the 2D domain
with(income_2, plot(Education, Seniority, asp = 1))

## let's scale it
xt_scaled <- scale(xt)
dat <- data.frame(xt_scaled, Income = income_2$Income)

with(dat, plot(Education, Seniority, asp = 1))

## fit a model on scaled data
interpolation_model <- gam(Income ~ s(Education, Seniority, k = 30, sp = 0),
                           data = dat)

## grid on the transformed space
x <- range(dat$Education)
x <- seq(x[1], x[2], length.out=30)
y <- range(dat$Seniority)
y <- seq(y[1], y[2], length.out=30)

## prediction on the transformed space
newdat <- expand.grid(Education = x, Seniority = y)
z <- matrix(predict(interpolation_model, newdat), nrow = length(x))

现在要生成图，我们想将网格反向转换为其原始比例。 注意，这不需要转换预测值。

## back transform the grid
scaled_center <- attr(xt_scaled, "scaled:center")
#Education Seniority 
# 16.38621  93.86207 
scaled_scale <- attr(xt_scaled, "scaled:scale")
#Education Seniority 
# 3.810622 55.715623 
xx <- x * scaled_scale[1] + scaled_center[1]
yy <- y * scaled_scale[2] + scaled_center[2]

## use `xx`, `yy` and `z`
p <- persp(xx, yy, z, theta = 30, phi = 30,
           col = "yellow",expand = 0.5, shade = 0.2,
           xlab = "Education", ylab = "Seniority", zlab = "Income")
obs <- trans3d(income_2$Education, income_2$Seniority, income_2$Income, p)
pred <- trans3d(income_2$Education, income_2$Seniority, fitted(interpolation_model), p)
points(obs, col="red",pch=16)
segments(obs$x, obs$y, pred$x, pred$y)