R help - Jun 2010 - lapply with functions with changing parameters

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

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)
You don't give a reproducible example, but since you want to apply a 
function to a list, the answer is yes.  Just defining the function and 
the list is the trick, untested:

lapply(mydf[, 1:5], function(x) ccf(x, mydf[, 6]))

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
> ______________________________________________
> R-help@r-project.org 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.
>


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

	[[alternative HTML version deleted]]

Henrique, 

thx, your suggestion worked perfectly fine for me. 


On 01.06.2010, at 23:01, Henrique Dallazuanna wrote:
>  lapply(mydf[-6], ccf, y = mydf[6])