check the following:

sapply(comb, "[[", "a")
# or
data.frame(
    "names" = names(comb),
    "value" = sapply(comb, "[[", "a"),
    row.names = seq_along(comb)
)


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


Quoting [EMAIL PROTECTED]:

Using R 2.6.2, say I have the following list of lists, "comb":

data1 <- list(a = 1, b = 2, c = 3)
data2 <- list(a = 4, b = 5, c = 6)
data3 <- list(a = 3, b = 6, c = 9)
comb <- list(data1 = data1, data2 = data2, data3 = data3)

So that all names for the lowest level list are common.  How can I most
efficiently access all of the sublist items "a" indexed by the outer
list names?  For example, I can loop through comb[[i]], unlisting as I
go, and then look up the field "a", as below, but there has got to be a
cleaner way.

finaldata <- double(0)
for(i in 1:length(names(comb))) {
        test <- unlist(comb[[i]])
        finaldata <- c(finaldata, test[which(names(test) == "a")])
}
data.frame(names(comb), finaldata)

Gives what I want:
  names.comb. finaldata
1       data1         1
2       data2         4
3       data3         3

Any help you can give would be greatly appreciated.  Thanks.
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