[R] performing function on lists where each element is a data frame

[Phil Spector]

Are you saying that

    sapply(xx,function(x)weighted.mean(x$H,x$N))

assuming the list of data frames was called xx, 
doesn't give the result you want?  Could you please
elaborate as to why sapply doesn't give the correct
result?

 					- Phil Spector
 					 Statistical Computing Facility
 					 Department of Statistics
 					 UC Berkeley
 					 spector at stat.berkeley.edu


On Wed, 14 Oct 2009, ewaters wrote:

>
> I have browsed the archive, and played around with a whole bunch of ways of
> approaching the problem, and I cannot get the answer that I want.
>
> I have a list with 2000 elements. Each element is a data frame with 2
> columns. As an example, see element 1999 below.
>
> [[1999]]
>   H   N
> 6    0.54  26
> 10   0.31  13
> 11   0.12  34
> 13   0.00  38
> 4    0.59  37
> 16   0.44   9
> 19   0.08  13
> 14   0.64  39
> 8    0.75  12
> 24   0.42  26
> 9    0.61 111
> 20   0.22   9
>
> Now, I want to caluculate the weighted mean for H in each list element, with
> N providing the weights. I can do this fine for an individual list element:
>
>> weighted.mean(list[[1999]]$H,list[[1999]]$N)
> [1] 0.4456403
>
> or even:
>> weighted.mean(list[[i]]$H,list[[i]]$N)
> [1] 0.5992683
>
> but when I try and apply it in a lopp to give me back the weighted mean for
> each list element, I get errors:
>
>> list_blah<-matrix(1:2000)
>> for(i in 1:2000){
> + list_blah[i]=weighted.mean(list[[i]]$H,list[[i]]$N)}
> Error in list[[i]] : subscript out of bounds
>
> I am clearly not understanding a basic syntax issue, and I am not overly
> skilled at using loops. I have also tried using lapply instead of a for
> loop, with the same problem.
>
> Asking for list[i] instead of list[[i]] gives NA results, as would be
> expected.
>
> Can anyone shed light?
>
>
> weighted.mean
>
> --
> View this message in context: http://www.nabble.com/performing-function-on-lists-where-each-element-is-a-data-frame-tp25902745p25902745.html
> Sent from the R help mailing list archive at Nabble.com.
>
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