Hi,
I tried both ideas, but it isn't that what I'm looking for.
I want to avoid for loop, because the matrix is of big size(1200*1200
entries)
With a loop I would do:
for ( i in seq(along = SmoothList))
{
Xarry[i,] <- predict(SmoothList[[i]],Xarry[i,])$y
}
Actually I want to do more than just to predict a value, but it isn't
important for the initial question...
Gunther
-----Ursprüngliche Nachricht-----
Von: Petr Pikal [mailto:[EMAIL PROTECTED]
Gesendet: Montag, 18. September 2006 11:44
An: Gunther Höning
Cc: r-help@stat.math.ethz.ch
Betreff: Re: AW: [R] Question on apply() with more information...
Hi
If I am correct apply do not choose from SmoothList as you expected.
Instead probably
lapply(SmoothList, predict,Xarray)
or
mapply(predict,SmoothList, Xarray)
can give you probably what you want.
HTH
Petr
On 18 Sep 2006 at 9:26, Gunther Höning wrote:
From: Gunther Höning <[EMAIL PROTECTED]>
To: "'Petr Pikal'" <[EMAIL PROTECTED]>,
<r-help@stat.math.ethz.ch>
Subject: AW: [R] Question on apply() with more information...
Date sent: Mon, 18 Sep 2006 09:26:01 +0200
> Ok.
> I tried this too, but it still doesn't work.
> Here some more information to try out, but just an excerpt of Xarray
>
> x <- c(0.11,0.25,0.45,0.65,0.80,0.95,1)
> Y <-
> matrix(c(15,83,57,111,150,168,175,37,207,142,277,375,420,437),nrow=2)
>
> sm <- function(y,x){smooth.spline(x,y)} SmoothList <- apply(Y,1,sm,x)
> NewValues <- function(x,LIST){predict(LIST,x)} Xarray <-
> matrix(c(0.15,0.56,0.66,0.45,0.19,0.17,0.99,0.56,0.77,0.41,0.11,0.63,0
> .42,0. 43),nrow=2)
>
>
> apply(Xarray, 2, NewValues,SmoothList) apply(Xarray, 2,
> NewValues,LIST=SmoothList)
>
>
>
> -----Ursprüngliche Nachricht-----
> Von: Petr Pikal [mailto:[EMAIL PROTECTED]
> Gesendet: Montag, 18. September 2006 08:43
> An: Gunther Höning; r-help@stat.math.ethz.ch
> Betreff: Re: [R] Question on apply()
>
> Hi
>
> not much information about what can be wrong. As nobody knows your
> Xarray and SmoothList it is hard to guess. You even omitted to show
> what "does not work" So here are few guesses.
>
> predict usually expects comparable data apply(Xarray, 2,
> NewValues,LIST=SmoothList)
>
>
> HTH
> Petr
>
>
>
>
> On 18 Sep 2006 at 8:05, Gunther Höning wrote:
>
> From: Gunther Höning <[EMAIL PROTECTED]>
> To: <r-help@stat.math.ethz.ch>
> Date sent: Mon, 18 Sep 2006 08:05:28 +0200
> Subject: [R] Question on apply()
>
> > Dear list,
> >
> > I try to do the following:
> > I have an list of length n, with elements done by smooth.spline
> > (SmoothList). Now I have a matrix with n rows and m columns with
> > x-values(Xarray) Now I want ot predict the y-values. Therefor I want
> > to take the first element of SmoothList and the first row of Xarray
> > and predict for each element in Xarray the y value. And then take
> > the second element of SmoothList and second row of Xarray, third row
> > of SmoothList and third row of Xarray and so on....
> >
> > I tried following:
> >
> > NewValues <- function(x,LIST){predict(LIST,x)} apply(Xarray, 2,
> > NewValues,SmoothList)
> >
> > But it don't work.
> >
> > Could anybody help please ?
> >
> > Gunther
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html and provide commented,
> > minimal, self-contained, reproducible code.
>
> Petr Pikal
> [EMAIL PROTECTED]
>
Petr Pikal
[EMAIL PROTECTED]
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