# [R] create a zero matrix & fill

jim holtman jholtman at gmail.com
Fri Jul 4 21:37:27 CEST 2008

```It would help if you follows the posting guide and provided commented,
minimal, self-contained, reproducible code.  You at least have to tell
what the structure of Results is (str(Results)).  It looks like a list
and if you are trying to plot the index vs. the mean, you might try:

plot(Results[[1]], Results[[2]])

On Fri, Jul 4, 2008 at 11:50 AM, mysimbaa <adel.tekari at sisltd.ch> wrote:
>
> Dear R user,
>
> I have written a function which returns max,min and variation of a power
> (see below)
> Power is a given matrix(1,n)
>
> I call the function
>>Variation<-VAR(p,(n-deltat))
>
> Now the problem is when I want plot(Results[1],Results[2]). Not possible!
> I become the following error (in english it means: Error in
> as.double.default(x) :Object cannot be transformed in double)
>
>> plot(Variation[1],Variation[2])
> Fehler in as.double.default(x) : (list) Objekt kann nicht nach 'double'
> umgewandelt werden
>
> Any suggestion?
>
> Hier is the function:
> #Computing variation of the power
> VAR<-function(power,length){
> tvar=pmean=pmin=pmax=varmax=varmin<-matrix(data=0,ncol=(length-tml0))
> for(i in tml0:length){
> tvar[i]=i
> pmean[i]=mean(power[i:(i+deltat)])
> pmin[i]=min(power[i:(i+deltat)])
> pmax[i]=max(power[i:(i+deltat)])
> varmax[i]=100*(pmax[i]-pmean[i])/pmean[i]
> varmin[i]=100*(pmean[i]-pmin[i])/pmean[i]
> Results=list(tvar,pmean,pmin,pmax,varmax,varmin)
>
> }}
>
>
> Thanks,
> --
> View this message in context: http://www.nabble.com/create-a-zero-matrix---fill-tp18281974p18281974.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> and provide commented, minimal, self-contained, reproducible code.
>

--
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

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