try this:
l <- vector("list", 3)
l[[1]] <- list(4, "hello")
l[[2]] <- list(7, "world")
l[[3]] <- list(9, " !!!! ")
lis <- lapply(l, "names<-", value = c("V1", "V2"))
do.call("rbind", lapply(lis, data.frame, stringsAsFactors = FALSE))
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]:
> Hello,
>
>
> Given a list with all elements having identical layout, e.g.:
>
>
> l = NULL
> l[[1]] = list(4, "hello")
> l[[2]] = list(7, "world")
> l[[3]] = list(9, " !!!! ")
>
>
> is there an easy way to collapse this list into a data.frame with each
> row being the elements of the list ?
> I.e. in this case I want to convert the list into a data.frame with 3
> rows and 2 columns, where column 1 holds the integer values, and column
> 2 the character values.
>
> I can get it done by looping over all elements and rbind them together
> to the final result, but that is quite slow (for large sets) and ugly,
> so I was wondering if there's an easy syntax.
>
> thanks a lot in advance,
> Thomas
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>
>
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