用与R的B s()函数相同的方法在Matlab中计算B样条基

我在Matlab中寻找(一个理想的内置)函数,该函数以与R中相同的方式计算B样条基矩阵,例如,对于具有20个等距节点的3次样条基,我将在R中执行

require(splines)
B = bs(x = seq(0,1,length.out=100),
        knots = seq(0, 1, length.out=20), # 20 knots
        degree = 3,
        intercept = FALSE)
matplot(B,type="l")

为了在Matlab中得到同样的结果,我想我可以使用

B = spcol(linspace(0,1,20),3,linspace(0,1,100));
plot(B);

但可以看到,当时的边界节点已不复存在。 有没有想过在Matlab中使用什么等效语法来获得与R中相同的结果?

PS R用于的代码 bs() 有点简单:

basis <- function(x, degree, i, knots) {
  if(degree == 0){
    B <- ifelse((x >= knots[i]) & (x < knots[i+1]), 1, 0)
  } else {
    if((knots[degree+i] - knots[i]) == 0) {
      alpha1 <- 0
    } else {
      alpha1 <- (x - knots[i])/(knots[degree+i] - knots[i])
    }
    if((knots[i+degree+1] - knots[i+1]) == 0) {
      alpha2 <- 0
    } else {
      alpha2 <- (knots[i+degree+1] - x)/(knots[i+degree+1] - knots[i+1])
    }
    B <- alpha1*basis(x, (degree-1), i, knots) + alpha2*basis(x, (degree-1), (i+1), knots)
  }
  return(B)
}

bs <- function(x, degree=3, interior.knots=NULL, intercept=FALSE, Boundary.knots = c(0,1)) {
  if(missing(x)) stop("You must provide x")
  if(degree < 1) stop("The spline degree must be at least 1")
  Boundary.knots <- sort(Boundary.knots)
  interior.knots.sorted <- NULL
  if(!is.null(interior.knots)) interior.knots.sorted <- sort(interior.knots)
  knots <- c(rep(Boundary.knots[1], (degree+1)), interior.knots.sorted, rep(Boundary.knots[2], (degree+1)))
  K <- length(interior.knots) + degree + 1
  B.mat <- matrix(0,length(x),K)
  for(j in 1:K) B.mat[,j] <- basis(x, degree, j, knots)
  if(any(x == Boundary.knots[2])) B.mat[x == Boundary.knots[2], K] <- 1
  if(intercept == FALSE) {
    return(B.mat[,-1])
  } else {
    return(B.mat)
  }
}