Try this:

 lapply(mydf[-6], ccf, y = mydf[6])

On Tue, Jun 1, 2010 at 5:50 PM, Bunny, lautloscrew.com <
bunny@lautloscrew.com> wrote:

> Dear all,
>
> I am trying to avoid a for loop here and wonder if the following is
> possible:
>
> I have a data.frame with 6 columns and i want to get a cross-correlogram
> (by using ccf) . Obivously ccf only accepts two columns at once and then
> returms a list.
> In fact, with a for loop i“d do the following
>
>
> for (i in 1:6) {
>
>  x[[i]]=ccf(mydf[,i],mydf[,6])
>
>
> }
>
> Is there any chance to the same with lapply? e.g. lapply(mydf,"ccf", .... )
> with ... respresenting the changing arguments for ccf functions (note only
> the first argument does actually change)
>
> thx for any suggestions in advance
>
> best
>
> matt
> ______________________________________________
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paranį-Brasil
25° 25' 40" S 49° 16' 22" O

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