[R] Kendall Theil line as fit?

[Peter Ehlers]

On 2011-03-11 14:43, jonbfish wrote:
> Thanks for the response, sorry I didn't post it initially.
>
> kt.mat<-
> function(x,y,z){
> for(i in 1:length(x)){for(j in
> 1:length(y)){z[i,j]<-(y[j]-y[i])/(x[j]-x[i])}}
> return(z)}
>
>
> kt.slope<-
> function(x,y,z,s){
> count<-0
> for(i in 1:length(x)){for(j in 1:length(y)){
> if(j>= i+1) {
> count<-count+1
> s[count]<-z[i,j]}
> }}
> print(count)
> return(s)}
>
> #Site23
>
> x<- c(1999,2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010)
> y<-
> c(17.942,3.43,14.062,14.814,13.778,13.706,9.748,13.088,12.1309728,9.644646671,9.134,8.84)
>
> z<-matrix(0:0,length(x),length(y))
> z<-kt.mat(x,y,z)
> z
>
> s<-c(1:(length(x)*(length(x)-1)/2))
> s<-kt.slope(x,y,z,s)
> s
> slope=median(s)
> intercept=median(y)-slope*median(x)
> cbind(slope,intercept)
> plot(x,y)
>
> abline(intercept,slope)

Okay, you're using abline() for the KT line.
But I still don't know what you're after.
 From your original post:

   Is there a way to make it appear like a regression fit instead
   of a line that extends from the edges of the plot? I would like
   to have the OLS appear as a dotted line and the KT a solid line
   but as it is the KT line is longer.

So how are plotting your 'regression fit'?
abline( lm( y ~ x ) ) would also extend across the plot.
I suppose that you could use segments() with the
range of x-values.

BTW, here's a shorter version of your code:

  yy <- outer(y, y, "-")
  xx <- outer(x, x, "-")
  z  <- yy / xx
  s  <- z[lower.tri(z)]
  slope <- median(s)
  intercept <- median(y) - slope * median(x)

Peter Ehlers

>
>
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