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.tur...@xtra.co.nz
> To: oti...@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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