
Rolf, many thanks for these suggestions, I will definitely give these a go.
I ended up resorting to locator() to ascertain the x y coordinates I required, but obviously this isn't optimal.
Thanks again,Oli

> Date: Thu, 7 Feb 2013 12:38:22 +1300
> From: rolf.turner@xtra.co.nz
> To: otills@hotmail.com
> CC: r-help@r-project.org
> Subject: Re: [R] Predictions from the Segmented Package
> 
> 
> 
> Since no-one else seems to have responded to your question, I'll
> chip in with my two bob's worth:
> 
> (1)  You can probably get what you might want to get from predict()
> by looking at X$fitted.values --- where X is the object returned by
> segmented().
> 
> (2) But predict() is probably the wrong way to go.  You can get
> the break points (or point if you have just two segments) by
> looking at X$psi[,2].
> 
> You can get the slopes of the segments from slope(X).  You
> can get the intercepts of the segments from intercept(X).
> 
> This should give you all the information that you need.  Said
> he, optimistically.
> 
> HTH
> 
>      cheers,
> 
>          Rolf Turner
> 
> P. S. Does anyone know why or whither Vito Muggeo has disappeared?
> 
>          R. T.
> 
> On 02/06/2013 11:30 AM, oli tills wrote:
> > Hi,
> > I would like to calculate the area under segmented regression lines (single breakpoints). I had thought that I could do this using the predict() function to ascertain some key x y values in order to determine the dimensions for two trapezoids (under each segment of my regression). However, I've not had any success with this and have just read online that the predict function does not work for segmented. Has anyone successfully done what I describe? Or is able to provide some advice on how best to approach this task?
 		 	   		  
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