I think this will do it:
sum(sqrt(diff(x.pred)^2 + diff(y.pred)^2))
Michael
On Tue, Nov 8, 2011 at 11:34 AM, Nicolas Schuck
<nico.sch...@googlemail.com> wrote:
> Dear R-community,
>
> I have a fitted bivariate polynomial, i.e:
>
> fit = lm(cbind(x, y)~poly(t, 15))
>
> and I would like to determine the length of the line in the interval t =
> [a, b]. Obviously, I could use predict and go through all the points, i.e.
>
> for (t in a:(b-1)) {
> length = length + sqrt((x.pred[t] - x.pred[t+1])^2 + (y.pred[t] -
> y.pred[t+1])^2)
> }
>
> but that would take very long given the amount of data I have. Do you know
> of any better solutions?
>
> Many thanks!
> Nicolas
>
> [[alternative HTML version deleted]]
>
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