B样条混乱 [英] B Spline confusion
问题描述
我意识到板上有一些有关B样条曲线的帖子，但实际上使我更加困惑，所以我认为有人可能会帮助我.
我具有x值从0到1的模拟数据.我想将其结点为0、0.1、0.2，...，0.9、1的三次样条(degree = 3)拟合到我的数据中我也想使用B样条基础和OLS进行参数估计(我不是在寻找罚样条).
我想我需要spline包中的bs函数，但是我不太确定，我也不知道该怎么喂它.
我还要绘制所得的多项式样条曲线.
谢谢！
## simulate some data - from mgcv::magic
set.seed(1)
n <- 400
x <- 0:(n-1)/(n-1)
f <- 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10
y <- f + rnorm(n, 0, sd = 2)
## load the splines package - comes with R
require(splines)
根据需要使用OLS估算值，可以在lm的公式中使用bs()函数. bs提供由结，多项式的阶数等给定的基函数.
mod <- lm(y ~ bs(x, knots = seq(0.1, 0.9, by = 0.1)))
您可以像对待线性模型一样对待它.
> anova(mod)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
bs(x, knots = seq(0.1, 0.9, by = 0.1)) 12 2997.5 249.792 65.477 < 2.2e-16 ***
Residuals 387 1476.4 3.815
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
关于结位置的一些指针. bs有一个参数Boundary.knots，默认值为Boundary.knots = range(x)-因此，当我在上面指定knots参数时，我没有包括边界结.
阅读?bs了解更多信息.
生成拟合样条线的图
在评论中，我讨论了如何绘制拟合的样条曲线.一种选择是按照协变量对数据进行排序.这适用于单个协变量，但不必适用于2个或多个协变量.另一个问题是，您只能在观察到的x值上评估拟合的样条-如果对协变量进行了密集采样，这很好，但是如果没有，则样条可能看起来很奇怪，具有长的线性截面. >
更通用的解决方案是使用predict从模型为协变量或协变量的新值生成预测.在下面的代码中，我演示了如何对上述模型执行此操作，并预测了x范围内的100个均匀间隔的值.
pdat <- data.frame(x = seq(min(x), max(x), length = 100))
## predict for new `x`
pdat <- transform(pdat, yhat = predict(mod, newdata = pdat))
## now plot
ylim <- range(pdat$y, y) ## not needed, but may be if plotting CIs too
plot(y ~ x)
lines(yhat ~ x, data = pdat, lwd = 2, col = "red")
产生
I realise that there are posts on the topic of B-Splines on this board but those have actually made me more confused so I thought someone might be able to help me.
I have simulated data for x-values ranging from 0 to 1. I'd like to fit to my data a cubic spline (degree = 3) with knots at 0, 0.1, 0.2, ... , 0.9, 1. I'd also like to use the B-Spline basis and OLS for parameter estimation (I'm not looking for penalised splines).
I think I need the bs function from the spline package but I'm not quite sure and I also don't know what exactly to feed it.
I'd also like to plot the resulting polynomial spline.
Thanks!
## simulate some data - from mgcv::magic
set.seed(1)
n <- 400
x <- 0:(n-1)/(n-1)
f <- 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10
y <- f + rnorm(n, 0, sd = 2)
## load the splines package - comes with R
require(splines)
You use the bs() function in a formula to lm as you want OLS estimates. bs provides the basis functions as given by the knots, degree of polynomial etc.
mod <- lm(y ~ bs(x, knots = seq(0.1, 0.9, by = 0.1)))
You can treat that just like a linear model.
> anova(mod)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
bs(x, knots = seq(0.1, 0.9, by = 0.1)) 12 2997.5 249.792 65.477 < 2.2e-16 ***
Residuals 387 1476.4 3.815
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Some pointers on knot placement. bs has an argument Boundary.knots, with default Boundary.knots = range(x) - hence when I specified the knots argument above, I did not include the boundary knots.
Read ?bs for more information.
Producing a plot of the fitted spline
In the comments I discuss how to draw the fitted spline. One option is to order the data in terms of the covariate. This works fine for a single covariate, but need not work for 2 or more covariates. A further issue is that you can only evaluate the fitted spline at the observed values of x - this is fine if you have densely sampled the covariate, but if not, the spline may look odd, with long linear sections.
A more general solution is to use predict to generate predictions from the model for new values of the covariate or covariates. In the code below I show how to do this for the model above, predicting for 100 evenly-spaced values over the range of x.
pdat <- data.frame(x = seq(min(x), max(x), length = 100))
## predict for new `x`
pdat <- transform(pdat, yhat = predict(mod, newdata = pdat))
## now plot
ylim <- range(pdat$y, y) ## not needed, but may be if plotting CIs too
plot(y ~ x)
lines(yhat ~ x, data = pdat, lwd = 2, col = "red")
That produces
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