在bs-call中指定钳位结向量 [英] Specify clamped knot vector in bs-call

问题描述

我打算将钳制的b样条曲线拟合到R中的一组控制点，但是难以理解bs中结参数的使用。给定一组控制点：

I intend to fit a clamped b-spline to a set of control points in R, but have trouble understanding the use of the knots parameter in bs. Given a set of control points:

path <- data.frame(
    x = c(3, 3.5, 4.6875, 9.625, 5.5625, 19.62109375, 33.6796875, 40.546875, 36.59375, 34.5, 33.5, 33),
    y = c(0, 1, 4, 5, 6, 8, 7, 6, 5, 2, 1, 0)
)

我独立地拟合x和y沿路径的距离：

I fit x and y independently against the distance along the path:

path$distance <- c(0, cumsum(sqrt(diff(path[,1])^2 + diff(path[,2])^2)))
path$distance
##  [1]  0.000000  1.118034  4.344511  9.382259 13.566026 27.766169 41.860284 48.799899 52.877545 56.535931 57.950145
## [12] 59.068179

，但我想提供一个开放的统一结向量来锚定适合第一个和最后一个点-使用 df 不支持此功能。

but I want to provide an open uniform knot vector in order to anchor the fit to the first and last point - using df does not support this.

据我了解，对于我给定一组点，样条曲线的度数为3，则必须有（12-1）+ 3 + 2 = 16 节（每m = n + p + 1，对于＃knots = m + 1，＃control = n + 1，度= p），因此对于受限制的样条曲线，这应该是一个不错的结向量：

As I understand it, for my given set of points, and a degree of 3 for the spline, there must be (12-1)+3+2 = 16 knots (per m=n+p+1, for #knots=m+1, #control=n+1, degree=p), so for a clamped spline, this should be a nice knot vector:

knots <- seq(path$distance[1], path$distance[12], length.out = 10)
knots <- c(rep(knots[1], 3), knots, rep(knots[10], 3))
knots
##  [1]  0.000000  0.000000  0.000000  0.000000  6.563131 13.126262 19.689393 26.252524 32.815655
## [10] 39.378786 45.941917 52.505048 59.068179 59.068179 59.068179 59.068179

尽管如此，使用它会产生一些疯狂的数字，以及关于等级不足的警告，因此很明显我必须以某种方式弄错了：

using this gives some crazy numbers though, as well as a warning about rank-deficiency, so clearly I must have it wrong somehow:

pred_df  <-  data.frame(x=0,y=0,distance=seq(min(path$distance), max(path$distance), length.out=100))
xPath <- predict(lm(x~bs(distance, knots=knots, degree = 3), path), pred_df)
## Warning message:
## In predict.lm(lm(x ~ bs(distance, knots = knots, degree = degree),  :
##   prediction from a rank-deficient fit may be misleading
summary(xPath)
##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
## -2133.000     3.468    16.700  -161.900    64.590   857.800 

指定的正确方法是什么

推荐答案

根据您的参数和结数据判断，的确违反了Schoenberg-Whitney条件。我对R中的语法不熟悉，但似乎您正在path $ distance [1]和path $ distance [12]之间统一生成结值。这个不对。您应该通过以下方式生成结值：

Judging from your 'parameter' and 'knots' data, indeed they violate the Schoenberg-Whitney condition. I am not familiar with the syntax in R, but it seems that you are generating the knot values uniformly between path$distance[1] and path$distance[12]. This is not right. You should generate your knot values in the following way:

0）将参数表示为p [i]，其中i = 0至（n-1），p [ 0] = 0.0，n是点数。对于您的情况，n = 12

1）创建结值

0) Denote the parameters as p[i], where i = 0 to (n-1), p[0]=0.0 and n is number of points. For your case, n=12
1) Create the knot values as

knot [0] =（p [1] + p [2 ] + p [3]）/ 3

结[1] =（p [2] + p [3] + p [4]）/ 3

结[2] =（p [3] + p [4] + p [5]）/ 3

......

这些是内部结点值。您应注意，在此步骤中将不使用p [0]和p [n-1]。您应该为您的案例获得8个内部结点值。

2）现在，将p [0]添加到结点值的前面4次（对于度= 3）并添加p [n-1]到结点值的末尾4次，您就完成了。对于您的情况，您应该总共获得16个结点值。

knot[0] = (p[1]+p[2]+p[3])/3
knot[1] = (p[2]+p[3]+p[4])/3
knot[2] = (p[3]+p[4]+p[5])/3
......
These are the interior knot values. You should notice that p[0] and p[n-1] will not be used in this step. You should obtain 8 interior knot values for your case.
2) Now, add p[0] to the front of the knot values 4 times (for degree=3) and add p[n-1] to the end of the knot values 4 times and you are done. For your case, you should get 16 knot values in total.

请注意，这不是生成有效结点矢量的 only 方法。但是以这种方式生成的结向量将始终满足Shoenberg-Whitney条件。

Please note that this is not the only way to generate a valid knot vector. But knot vector generated in this way will always satisfy the Shoenberg-Whitney condition.

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