[R] Constrained cubic smoothing spline
Hello everone, Anyone who knows how to force a cubic smoothing spline to pass through a particular point? I found on website someone said that we can use "cobs package" to force the spline pass through certain points or impose shape constraints (increasing, decreasing). However, this package is using B-spline and can only do linear and quadratic B-spline. In my research, I need to force a cubic smoothing spline to pass a point. Thanks! Victor [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] About basic logical operators
Hello everyone, I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78 [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54 [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30 [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16 -0.14 -0.12 -0.10 -0.08 -0.06 [49] -0.04 -0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 [61] 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42 [73] 0.44 0.46 0.48 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.66 [85] 0.68 0.70 0.72 0.74 0.76 0.78 0.80 0.82 0.84 0.86 0.88 0.90 [97] 0.92 0.94 0.96 0.98 1.00 > x[x<=0.02] [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78 [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54 [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30 [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16 -0.14 -0.12 -0.10 -0.08 -0.06 [49] -0.04 -0.02 0.00 > x[x<0.2] [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78 [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54 [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30 [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16 -0.14 -0.12 -0.10 -0.08 -0.06 [49] -0.04 -0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 [61] 0.20 > Why does "x[x<=0.02]" return no 0.02 but "x[x<0.2]" return a subsample with 0.02? Anyone who can tell me why? Thanks! Victor [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Use pcls in "mgcv" package to achieve constrained cubic spline
Hello everyone, Dr. wood told me that I can adapting his example to force cubic spline to pass through certain point. I still have no idea how to achieve this. Suppose we want to force the cubic spline to pass (1,1), how can I achieve this by adapting the following code? # Penalized example: monotonic penalized regression spline . # Generate data from a monotonic truth. x<-runif(100)*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) # Show regular spline fit (and save fitted object) f.ug<-gam(y~s(x,k=10,bs="cr"));lines(x,fitted(f.ug)) # Create Design matrix, constraints etc. for monotonic spline sm<-smoothCon(s(x,k=10,bs="cr"),dat,knots=NULL)[[1]] F<-mono.con(sm$xp); # get constraints G<-list(X=sm$X,C=matrix(0,0,0),sp=f.ug$sp,p=sm$xp,y=y,w=y*0+1) G$Ain<-F$A;G$bin<-F$b;G$S<-sm$S;G$off<-0 p<-pcls(G); # fit spline (using s.p. from unconstrained fit) fv<-Predict.matrix(sm,data.frame(x=x))%*%p lines(x,fv,col=2) lines(x,f,col="blue") Thanks a lot!! Victor [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Impose constraint on first order derivative at a point for cubic smoothing spline
Hello, Dr. Simon Wood told me how to force a cubic spline passing through a point. The code is as following. Anyone who knows how I can change the code to force the first derivative to be certain value. For example, the first derivative of the constrained cubic spline equals 2 at point (0, 0.6). I really appreciate your help! Thanks! Best Victor Here is the initial reply and code provided by Dr. Simon Wood: "Actually, you might as well use "gam" directly for this (has the advantage that the smoothing parameter will be chosen correctly subject to the constraint). Here is some code. Key idea is to set the basis and penalty for the spline up first, apply the constraint, and then use gam to fit it..." best, Simon ## Example constraining spline to pass through a ## particular point (0,.6)... ## Fake some data... library(mgcv) set.seed(0) n <- 100 x <- runif(n)*4-1;x <- sort(x); f <- exp(4*x)/(1+exp(4*x));y <- f+rnorm(100)*0.1;plot(x,y) dat <- data.frame(x=x,y=y) ## Create a spline basis and penalty, making sure there is a knot ## at the constraint point, (0 here, but could be anywhere) knots <- data.frame(x=seq(-1,3,length=9)) ## create knots ## set up smoother... sm <- smoothCon(s(x,k=9,bs="cr"),dat,knots=knots)[[1]] ## 3rd parameter is value of spline at knot location 0, ## set it to 0 by dropping... X <- sm$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)) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
