[R] Kolmogorov-Smirnov test and the plot of max distance between two ecdf curves
Rui Barradas
ruipbarradas at sapo.pt
Sat May 26 15:49:39 CEST 2012
Hello,
Try the following.
(i've changed the color of the first ecdf.)
loga <- log10(a+1) # do this
logb <- log10(b+1) # only once
f.a <- ecdf(loga)
f.b <- ecdf(logb)
# (2) max distance D
x <- seq(min(loga, logb), max(loga, logb), length.out=length(loga))
x0 <- x[which( abs(f.a(x) - f.b(x)) == max(abs(f.a(x) - f.b(x))) )]
y0 <- f.a(x0)
y1 <- f.b(x0)
plot(f.a, verticals=TRUE, do.points=FALSE, col="blue")
plot(f.b, verticals=TRUE, do.points=FALSE, col="green", add=TRUE)
## alternatine, use standard R plot of ecdf
#plot(f.a, col="blue")
#lines(f.b, col="green")
points(c(x0, x0), c(y0, y1), pch=16, col="red")
segments(x0, y0, x0, y1, col="red", lty="dotted")
## alternative, down to x axis
#segments(x0, 0, x0, y1, col="red", lty="dotted")
Hope this helps,
Rui Barradas
maxbre wrote
>
> Hi all,
>
> given this example
>
> #start
>
> a<-c(0,70,50,100,70,650,1300,6900,1780,4930,1120,700,190,940,
>
> 760,100,300,36270,5610,249680,1760,4040,164890,17230,75140,1870,22380,5890,2430)
> length(a)
>
> b<-c(0,0,10,30,50,440,1000,140,70,90,60,60,20,90,180,30,90,
> 3220,490,20790,290,740,5350,940,3910,0,640,850,260)
> length(b)
>
> out<-ks.test(log10(a+1),log10(b+1))
>
> # max distance D
> out$statistic
>
> f.a<-ecdf(log10(a+1))
> f.b<-ecdf(log10(b+1))
>
> plot(f.a, verticals=TRUE, do.points=FALSE, col="red")
> plot(f.b, verticals=TRUE, do.points=FALSE, col="green", add=TRUE)
>
> #inverse of ecdf a
> x.a<-get("x", environment(f.a))
> y.a<-get("y", environment(f.a))
>
> # inverse of ecdf b
> x.b<-get("x", environment(f.b))
> y.b<-get("y", environment(f.b))
>
>
> #end
>
> I want to plot the max distance between the two ecdf curves as in the
> above given chart
>
> Is that possible and how?
>
>
> Thanks for your help
>
> PS: this is an amended version of a previous thread (but no reply
> followed) that I’ve deleted from Nabble repository because I realised it
> was not enough clear (now I hope it’s a little better, sorry for that)
>
--
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