It appears the answer to your goal after a discursive exploration of
"interpolation", which was really extrapolation, is that you need to
look at the predict methods for linear (and other sorts as well) models.
?predict
?predict.lm
> y <- c(16,45,77,101,125)
> x <- c(0,5,10,15,20)
>
> lmmo
On Jan 15, 2009, at 11:31 AM, e-letter wrote:
Perhaps a coding error on my part (or on your part). Perhaps
different
methods (none of which you describe)?
I suspect that my method only used the first two points (I just
checked by plotting and -2.7 is closer to the paper and pen result I
g
> Perhaps a coding error on my part (or on your part). Perhaps different
> methods (none of which you describe)?
>
> I suspect that my method only used the first two points (I just
> checked by plotting and -2.7 is closer to the paper and pen result I
> get than is -3.28. Perhaps you made an extra
On Jan 15, 2009, at 10:04 AM, e-letter wrote:
On 13/01/2009, David Winsemius wrote:
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),
On 13/01/2009, David Winsemius wrote:
> 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
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
> approxE
>
> 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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
approxfun returns a function; that is not an error message:
> x=c(0,5,10,15,20)
> y=c(16,45,77,101,125)
>
> approx(x,y,method="linear")
$x
[1] 0.000 0.4081633 0.8163265 1.2244898 1.6326531 2.0408163
2.4489796 2.8571429 3.2653061
[10] 3.6734694 4.0816327 4.4897959 4.8979592 5.306
On 08/01/2009, Greg Snow wrote:
> If you want to just linearly interpolate, then use the functions approx or
> approxfun from the stats package (one of those that is loaded by default).
I have read the guide for approx and approxfun functions. Below is my data.
x=c(0,5,10,15,20)
y=c(16,45,77,101,
8, 2009 9:22 AM
> To: r-help@r-project.org
> Subject: [R] interpolation to abscissa
>
> Readers,
>
> I have looked at various documents hosted on the web site; I couldn't
> find anything on interpolation. So I started r and accessed the help
> (help.start()). (by the
Readers,
I have looked at various documents hosted on the web site; I couldn't
find anything on interpolation. So I started r and accessed the help
(help.start()). (by the way is it possible to configure r to open help
in opera instead of firefox?) Initially I read the help for the akima
package b
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