Hello,
Dr. Simon Wood told me how to force a cubic spline passing through a
point. The code is as following. Anyone who knows how I can change the code
to force the first derivative to be certain value. For example, the first
derivative of the constrained cubic spline equals 2 at point (0, 0.6).
I really appreciate your help!
Thanks!
Best
Victor
Here is the initial reply and code provided by Dr. Simon Wood:
"Actually, you might as well use "gam" directly for this (has the
advantage that the smoothing parameter will be chosen correctly subject
to the constraint). Here is some code. Key idea is to set the basis and
penalty for the spline up first, apply the constraint, and then use gam
to fit it..."
best,
Simon
## Example constraining spline to pass through a
## particular point (0,.6)...
## Fake some data...
library(mgcv)
set.seed(0)
n <- 100
x <- runif(n)*4-1;x <- sort(x);
f <- exp(4*x)/(1+exp(4*x));y <- f+rnorm(100)*0.1;plot(x,y)
dat <- data.frame(x=x,y=y)
## Create a spline basis and penalty, making sure there is a knot
## at the constraint point, (0 here, but could be anywhere)
knots <- data.frame(x=seq(-1,3,length=9)) ## create knots
## set up smoother...
sm <- smoothCon(s(x,k=9,bs="cr"),dat,knots=knots)[[1]]
## 3rd parameter is value of spline at knot location 0,
## set it to 0 by dropping...
X <- sm$X[,-3] ## spline basis
S <- sm$S[[1]][-3,-3] ## spline penalty
off <- y*0 + .6 ## offset term to force curve through (0, .6)
## fit spline constrained through (0, .6)...
b <- gam(y ~ X - 1 + offset(off),paraPen=list(X=list(S)))
lines(x,predict(b))
## compare to unconstrained fit...
b.u <- gam(y ~ s(x,k=9),data=dat,knots=knots)
lines(x,predict(b.u))
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