It's fairly clear from the documentation that approxfun() will not
extrapolate.
help.search("extrapolate")
library(Hmisc)
?approxExtrap
Some sort of minimization approach:
> approxExtrap(x=c(0,5,10,15,20), y=c(16,45,77,101,125),xout=c(-4,0,4))
$x
[1] -4 0 4
$y
[1] -7.2 16.0 39.2
> approxExtrap(x=c(0,5,10,15,20),
y=c(16,45,77,101,125),xout=seq(-2.8,-2.6, by=0.01))
$x
[1] -2.80 -2.79 -2.78 -2.77 -2.76 -2.75 -2.74 -2.73 -2.72 -2.71
-2.70 -2.69 -2.68
[14] -2.67 -2.66 -2.65 -2.64 -2.63 -2.62 -2.61 -2.60
$y
[1] -0.240 -0.182 -0.124 -0.066 -0.008 0.050 0.108 0.166 0.224
0.282 0.340
[12] 0.398 0.456 0.514 0.572 0.630 0.688 0.746 0.804 0.862
0.920
How accurate do you need the answer?
I tried Hmisc's inverseFunction(), but it returned 0 for an argument
of zero:
> invF <- inverseFunction(x=c(0,5,10,15,20), y=c(16,45,77,101,125))
> invF(0)
So I then hacked Harrell's inverseFunction by substituting
approxExtrap in every in instance
where approx appeared, creating invFunc2:
then
> invF <- invFunc2(x=c(0,5,10,15,20), y=c(16,45,77,101,125))
>
> invF(0)
[1] -2.758621
--
David Winsemius
On Jan 13, 2009, at 10:57 AM, e-letter wrote:
What is the problem that you are trying to solve?
From the data I provided: x=c(0,5,10,15,20) y=c(16,45,77,101,125); I
want to obtain the value of x when y=0.
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