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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