Fitting Cubic Splines

Splines.R

#FITTING SPLINES IN R
# D Menezes 20170531


#0. PRELIMINARIES
#install.packages("pacman") #only needed if you don't have it installed
pacman::p_load("splines")

#1. INPUT VALUES
input.x<-c(seq(from=0,to=120,by=12)) #assuming we have annual spline points
input.y<-c(0,0.48,1.08,1.03,1.01,1,1,1,1,1,1) # % Developed say
reqd.x<-c(seq(from=0,to=120,by=3)) #suppose we want quarterly spline output
n=length(input.x)

#2 KEY IDEA

#build basic spline
option.1<-spline(input.x, 100* input.y) #simplest - fits one cubic between each input x value points
option.1b<-spline(input.x, 100* input.y,xout=reqd.x) #fits one cubic between each required x value

#Compare

plot(input.x, 100*input.y, 
     main = paste("LDF Example: Spline Fit vs Actuals"), 
     xlab="Months Developed", 
     ylab="% Dev", 
     pch=16, 
     cex=1, 
     xaxt="n")
axis(1, at=seq(0,120,by=12), las=1)

points(option.1b)
lines(option.1b, col="red")


legend('bottomright', 
       legend=c("Annual-input", 
                "Quarterly - output"),  
       pch=c(16,1)
)


#3 ALTERNATIVES: Other spline options
option.2<-spline(input.x, 100* input.y,n=200) #creates a cubic spline with 200 points
option.3<-spline(input.x, 100* input.y,method="fmm") #different 
option.4<-spline(input.x, 100* input.y,method="natural")
option.5<-spline(input.x, 100* input.y,method="periodic")
option.6<-smooth.spline(input.x, 100* input.y) #too smooth

#compare:

plot(option.1$x, option.1$y, 
     main = paste("Alternative Examples: Spline Fit vs Actuals"), 
     xlab="Months Developed", 
     ylab="% Dev", 
     pch=16, 
     cex=1, 
     col=1)
lines(option.1,col=1, lwd=1)
lines(option.2,col=2, lwd=1)
lines(option.3,col=3, lwd=1)
lines(option.4,col=4, lwd=1)
lines(option.5,col=5, lwd=1)
lines(option.6,col=6, lwd=1)
legend('bottom', 
       legend=c("Default", 
                "200 point", 
                "FMM", 
                "Natural", 
                "Periodic", 
                "Smoothed"), 
       lty=1,
       lwd=2.5, 
       col=1:6 
       )


#4 RETURNING THE VALUE OF ARBITRARY X VALUE - USE interSpline
option.7<-interpSpline(input.x, 100* input.y)
predict(option.7,x=57)
  

#5 Blog specific chart output
plot(input.x, 100*input.y, 
     main = paste("Raw data"), 
     xlab="Months Developed", 
     ylab="% Dev", 
     pch=16, 
     cex=1, 
     xaxt="n")
axis(1, at=seq(0,120,by=12), las=1)