n = 50
set.seed(100)
x = matrix(runif(n, -2, 2), nrow=n)
y = 2 + 0.75*sin(x) - 0.75*cos(x) + rnorm(n, 0, 0.2)
In R, I want to estimate above spline function by Least Square method.
(在R中，我想通过最小二乘法估算上述样条函数。)
The following hint wants me to generate design matrix:
(以下提示希望我生成设计矩阵：)
nknots = 12
mat.tpb = function(x, nknots){
n = length(x)
knots = seq(-2, 2, len=nknots)
zb = cbind(1, x ,x^2, x^3)
zt = matrix(NA, nrow=n, ncol=nknots)
for(i in 1:nknots){zt[,i] = ifelse(x>knots[i], (x-knots[i]) ^3, 0)}
cbind(zb, zt)}
z = mat.tpb(x, nknots)
I simplify the above for easier to read:
(我简化以上内容以便于阅读：)
mat.tpb = function(x, 12){
zb = cbind(1, x ,x^2, x^3)
zt = matrix(NA, nrow=length(x), ncol=12)
for(i in 1:12){zt[,i] = ifelse(x>seq(-2, 2, len=12)[i], (x-seq(-2, 2, len=12)[i])^3, 0)}
cbind(zb, zt)}
z = mat.tpb(x, 12)
I think the "t_j" in function "g_k+j+1(x)" means "seq(-2, 2, len=12)" in hint.
(我认为函数“ g_k + j + 1（x）”中的“ t_j”在提示中表示“ seq（-2，2，len = 12）”。)
Then since
(然后因为)
> seq(-2, 2, len=12)
[1] -2.0000000 -1.6363636 -1.2727273 -0.9090909 -0.5454545 -0.1818182 0.1818182 0.5454545 0.9090909 1.2727273 1.6363636 2.0000000'
So that I can extend the estimator function as
(这样我就可以将estimator函数扩展为) 
Which in word is
(换句话说)
E(y|x)=λ_1×1+λ_2×x+λ_3×x^2+λ_4×x^3+
(E（y | x）=λ_1×1 +λ_2×x +λ_3×x ^ 2 +λ_4×x ^ 3 +)
λ_5×max?{x-(-2+(4/11)×0),0}^3+λ_6×max?{(x-(-2+(4/11)×1),0}^3+λ_7×max?{(x-(-2+(4/11)×2),0}^3+λ_8×max?{(x-(-2+(4/11)×3),0}^3+λ_9×max?{(x-(-2+(4/11)×4),0}^3+λ_10×max?{(x-(-2+(4/11)×5),0}^3+λ_11×max?{(x-(-2+(4/11)×6),0}^3+λ_12×max?{(x-(-2+(4/11)×7),0}^3+λ_13×max?{(x-(-2+(4/11)×8),0}^3+λ_14×max?{(x-(-2+(4/11)×9),0}^3+λ_15×max?{(x-(-2+(4/11)×10),0}^3+λ_16×max?{(x-(-2(4/11)×11),0}^3
(λ_5×max?{x-（-2+（4/11）×0），0} ^ 3 +λ_6×max?{（x-（-2+（4/11）×1），0} ^ 3 +λ_7×max?{（x-（-2+（4/11）×2），0} ^ 3 +λ_8×max?{（x-（-2+（4/11）×3），0} ^ 3 +λ_9×max?{（x-（-2+（4/11）×4），0} ^ 3 +λ_10×max?{（x-（-2+（4/11）×5）， 0} ^ 3 +λ_11×max?{（x-（-2+（4/11）×6），0} ^ 3 +λ_12×max?{（x-（-2+（4/11）×7 ），0} ^ 3 +λ_13×max?{（x-（-2+（4/11）×8），0} ^ 3 +λ_14×max?{（x-（-2+（4/11） ×9），0} ^ 3 +λ_15×max?{（x-（-2+（4/11）×10），0} ^ 3 +λ_16×max?{（x-（-2（4/11 ）×11），0} ^ 3)
After simplification, what I want want to estimate is
(简化后，我想估计的是) 
Which in word is
(换句话说)
λ_1×1+λ_2×x+λ_3×x^2+λ_4×x^3+ λ_5×max{?(x+2),0}^3+λ_6×max?{(x+18/11),0}^3+λ_7×max?{(x+14/11),0}^3+λ_8×max?{(x+10/11),0}^3+λ_9×max?{(x+6/11),0}^3+λ_10×max?{(x+2/11),0}^3+λ_11×max?{(x-2/11),0}^3+λ_12×max?{(x-6/11),0}^3+λ_13×max?{(x-10/11),0}^3+λ_14×max?{(x-14/11),0}^3+λ_15×max?{(x-18/11),0}^3+λ_16×max?{(x-2),0}^3
(λ_1×1 +λ_2×x +λ_3×x ^ 2 +λ_4×x ^ 3 +λ_5×max {?（x + 2），0} ^ 3 +λ_6×max?{（x + 18/11），0 } ^ 3 +λ_7×max?{（x + 14/11），0} ^ 3 +λ_8×max?{（x + 10/11），0} ^ 3 +λ_9×max?{（x + 6 / 11），0} ^ 3 +λ_10×max?{（x + 2/11），0} ^ 3 +λ_11×max?{（x-2 / 11），0} ^ 3 +λ_12×max?{{ x-6 / 11），0} ^ 3 +λ_13×max?{（x-10 / 11），0} ^ 3 +λ_14×max?{（x-14 / 11），0} ^ 3 +λ_15× max?{（x-18 / 11），0} ^ 3 +λ_16×max?{（x-2），0} ^ 3)
But I do not know why the hint wants me to generate design matrix,
(但是我不知道为什么提示要我生成设计矩阵，)
so I am just going to estimate the function I simplified.
(所以我只是估计一下我简化的功能。)
But it will like this:
(但是它会像这样：)
> summary(lm(y ~ x+I(x^2)+I(x^3)+I(max(x+2, 0))^3+ I(max(x+(18/11), 0))^3+ I(max(x+(14/11), 0))^3+ I(max(x+(10/11), 0))^3+ I(max(x+(6/11), 0))^3+ I(max(x+(2/11), 0))^3+ I(max(x-(2/11), 0))^3+ I(max(x-(6/11), 0))^3+ I(max(x-(10/11), 0))^3+ I(max(x-(14/11), 0))^3+ I(max(x-(18/11), 0))^3+ I(max(x-2, 0))^3))
Error in model.frame.default(formula = y ~ x + I(x^2) + I(x^3) + I(max(x + :
variable lengths differ (found for 'I(max(x + 2, 0))')
Please tell me in R, what function or package I can use to estimate the function I extend,
(请在R中告诉我，我可以使用哪些函数或包来估算我扩展的函数，)
or you know what the hint means and have other way to estimate please also tell me.
(或者您知道提示的含义，并有其他估算方法，请也告诉我。)
Thank you very much.
(非常感谢你。)
ask by abba translate from so 
