Re: List question

["Stewart, William" <william.stewart@rbc.com>]

The following will solve the immediate problem:

        lapply(x, FUN=function(x){
                x[[3]] <- matrix(rnorm(100,10, 2), 10,10)
                return(x)
                })


A more generalized version (if you actually wanted control over what was being 
placed into x[[k]][[3]]) would be:

        (I've simplified your x but kept the same structure)
        x <- list(list(1, list("a"),list()),
                  list(2, list("b"),list()),
                  list(3, list("c"),list()))
          
        (for your problem, replace v1, etc. with your randome matrices or 
anything else)
        value <- list("v1", "v2", "v3")

        y <- lapply(1:length(x), FUN=function(k, x, value){
                        x[[k]][[3]] <- value[[k]]
                        return(x[[k]])
                        }, x=x, value=value)


-------------------------
William A. Stewart
Manager, Operational Risk
RBC Financial Group
william.stewart@rbc.com
phone: 416-974-3019



-----Original Message-----
From: s-news-owner@lists.biostat.wustl.edu
[mailto:s-news-owner@lists.biostat.wustl.edu]On Behalf Of Dave Evens
Sent: Wednesday, August 17, 2005 9:39 AM
To: s-news@lists.biostat.wustl.edu
Subject: [S] List question




Dear all,

My query involves a list. For example, suppose I have
a list x

x <- list(list(rpois(100,2), list(LETTERS[1:20]),
list()),
          list(rpois(150,2), list(LETTERS[1:12]),
list()),
          list(rpois(200,2), list(LETTERS[1:19]),
list()))


and I want to replace the empty list in each sub-list
of x, this is easy using the for-loop

for(k in 1:length(x)) x[[k]][[3]] <- matrix(rnorm(100,
10, 2), 10,10)

but how do I the equivalent of this using the lapply
command?

This is the syntax of the lapply command

lapply(1:length(x), function(k) matrix(rnorm(100, 10,
2), 10,10))

but how do I get it in x[[k]][[3]]?

I've tried  using the assign command but it doesn't
appear to work.

Thanks for any help in advance.

Dave







                
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