[R] ccf time units

[Gabor Grothendieck]

The unit of time for a "ts" class object is deltat(ldeaths).
See the
   ?deltat
help page.

On 3/29/07, tom soyer <tom.soyer at gmail.com> wrote:
> Hi,
>
> I am using ccf but I could not figure out how to calculate the actual lag in
> number of periods from the returned results. The documentation for ccf
> says:"The lag is returned and plotted in units of time". What does "units of
> time" mean? For example:
>
> > x=ldeaths
> > x1=lag(ldeaths,1)
> > results=ccf(x,x1)
> > results
>
> Autocorrelations of series 'X', by lag
>
> -1.2500 -1.1667 -1.0833 -1.0000 -0.9167 -0.8333 -0.7500 -0.6667 -0.5833 -
> 0.5000
>  -0.297   0.011   0.380   0.651   0.738   0.664   0.392  -0.011  -0.383  -
> 0.618
> -0.4167 -0.3333 -0.2500 -0.1667 -0.0833  0.0000  0.0833  0.1667  0.2500
> 0.3333
>  -0.693  -0.619  -0.362   0.020   0.405   0.770   0.981   0.749   0.376  -
> 0.004
>  0.4167  0.5000  0.5833  0.6667  0.7500  0.8333  0.9167  1.0000  1.0833
> 1.1667
>  -0.366  -0.610  -0.686  -0.602  -0.368  -0.012   0.388   0.650   0.705
> 0.614
>  1.2500
>  0.341
> I see that at time unit 0.0833, the correlation is close to 1. But how do I
> know time unit 0.0833 is actually lag 1? Is there a function to do the
> conversion?
>
> Thanks,
>
> Tom
>
>        [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.
>