# Calculation of elasticity for non-linear curve

I've written this code to calculate the elasticity for econometric analysis. Basically, this algorithm estimate the function:

$$\epsilon(x) = \frac{x}{f(x)}\ f'(x)$$

###############################################

elasticity<-function(y.p,x.p,band){
require(KernSmooth)
as.vector(y.p);as.vector(x.p)
loess.fun<-locpoly(x.p,y.p,bandwidth=band)
plot(loess.fun,type='l',lwd=2,lty=4,
ylim=c(0,100),xlab="Price",
ylab="Share", col=2,
main="Local Polynomials Estimation")
loess.der<-locpoly(x.p,y.p,drv=1,bandwidth=band,
gridsize = length(y.p))
y1<-as.vector(loess.der$y) eps<-matrix(NA,length(y.p),2) eps[,1]<-x.p for(i in 1:length(y.p)){ eps[i,2]<-x.p[i]/y.p[i]*y1[i] } j<-order(x.p) colnames(eps)<-c("Price","Elasticity") return(eps[j,]) } ###############################################  Can I improve this algorithm? ## migrated from stats.stackexchange.comJul 29 '16 at 11:47 This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. ## 1 Answer I'll provide feedback from top to bottom: 1. elasticity<-function(y.p,x.p,band){  It's recommended to use spaces to make your code easier on the eyes. Spaces also avoid ambiguous situations for people who are not experts in R's syntax rules; for example, when someone sees a<-b, is it assigning b to a, or is it asking if a is less than -b? Maybe another way to convince you is to make you notice that R's compiled code does contain spaces (run print(lapply) for example); it shows that R's authors agreed that code readability was more important than compactness. Now about the choice of arguments. Why not use the same argument names (and order) as locpoly since it is the main function being called here? So I'd recommend x, y, bandwidth. Or maybe x.price, y.price, bandwidth if it really helps knowing these are prices. 2. require(KernSmooth)  To convince you to use library instead of require, check http://www.r-bloggers.com/library-vs-require-in-r/. 3. as.vector(y.p);as.vector(x.p)  These two calls to as.vector do nothing if you do not assign the output. You probably meant to do: y.p <- as.vector(y.p) x.p <- as.vector(x.p) 11. y1<-as.vector(loess.der$y)


If you check using is.vector(), you will see that loess.der$y is already a vector, so as.vector is unnecessary. 12. eps<-matrix(NA,length(y.p),2)  Would a data.frame make more sense? Or if you really want a matrix, should use be using cbind? 14. for(i in 1:length(y.p)){ 15. eps[i,2]<-x.p[i]/y.p[i]*y1[i] 16. }  Here, you are really missing up on vectorization. This should be: eps[, 2] <- x.p / y.p * y1  Overall, if defining a data.frame, appreciate the simplicity and readability of: eps <- data.frame(Price = x.p, Elasticity = x.p / y.p * loess.der$y)


Or if you want a matrix, just replace data.frame() with cbind().

Taking all of my comments into account, your code could look like this:

elasticity <- function(x, y, bandwidth) {
library(KernSmooth)
x <- as.vector(x)
y <- as.vector(y)
loess.fun <- locpoly(x = x, y = y, bandwidth = bandwidth)
plot(loess.fun, type = 'l', lwd = 2, lty = 4,
ylim = c(0, 100), xlab = "Price", ylab = "Share", col = 2,
main = "Local Polynomials Estimation")
loess.der <- locpoly(x = x, y = y, drv = 1, bandwidth = bandwidth,
gridsize = length(y))

eps <- data.frame(Price = x,
Elasticity = x / y * loess.der$y) return(eps[order(eps$Price), ])
}

• Thank you very much, youre answer is complete and full of suggestions. But i've run the two algorithms and the results are the same, so this means that it works :D – Enzo D'Innocenzo Aug 1 '16 at 6:54
• That's right, my suggestions were only to improve on your coding style, as there was nothing wrong in terms of working code. – flodel Aug 1 '16 at 18:37