R help - Feb 2004 - efficient matrix approx

Hello,

I am looking for a highly efficient matrix version of linear
interpolation (like approx). As an example I have data like follows:

x<-data.frame(time=1:20, x=(1:20)/10, y=runif(20))
t <- seq(1.5, 15.5 ,by=0.5)

# and I found the following solution:

nam <- names(x)
app <- lapply(x[2:3],approx,x=x$time, xout=t)
r <- c(1, 2*(1:(ncol(x)-1)))
x.new <- as.data.frame(unlist(app[1:2], recursive=F)[r])
names(x.new) <- nam


As this routine should be called several times during an optimization
procedure I wonder if there is a more direct one which I have overlooked?

Thank you in advance

Thomas P.

Thomas Petzoldt <petzoldt at rcs.urz.tu-dresden.de> writes:
> Hello,
> 
> I am looking for a highly efficient matrix version of linear
> interpolation (like approx). As an example I have data like follows:
> 
> x<-data.frame(time=1:20, x=(1:20)/10, y=runif(20))
> t <- seq(1.5, 15.5 ,by=0.5)
> 
> # and I found the following solution:
> 
> nam <- names(x)
> app <- lapply(x[2:3],approx,x=x$time, xout=t)
> r <- c(1, 2*(1:(ncol(x)-1)))
> x.new <- as.data.frame(unlist(app[1:2], recursive=F)[r])
> names(x.new) <- nam
> 
> 
> As this routine should be called several times during an optimization
> procedure I wonder if there is a more direct one which I have overlooked?
I wouldn't know about efficiency, but the last bit of code is
equivalent to

app <- lapply(x[-1], approx, x=x$time, xout=t)
data.frame(time=t,lapply(app,"[[","y"))

which should at least be somewhat easier to read...

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   O__  ---- Peter Dalgaard             Blegdamsvej 3  
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