Thanks, Henrique
I *do* need to preserve the type of each list element (num or chr).
Thus, a first step might be
> as.data.frame(testlist, stringsAsFactors=FALSE)
shape cell.fill back.fill scale.max
1 0 red white 100
2 0 blue white 100
3 2 green
This gives the result you asked for:
DF <- as.data.frame(testlist)
lapply(split(DF, 1:nrow(DF)), unclass)
although it might be good enough to just do this depending on what you need:
DF <- as.data.frame(testlist)
split(DF, 1:nrow(DF)
On Fri, Jan 15, 2010 at 8:09 AM, Michael Friendly wrote:
>
Try this:
#'1) But apply converts all to 'character'
apply(as.data.frame(testlist), 1, as.list)
#2)
lapply(split(x <- as.data.frame(testlist), 1:nrow(x)), as.list)
On Fri, Jan 15, 2010 at 11:09 AM, Michael Friendly wrote:
> I have a list of vectors, all forced to be the same length:
>
> testlis
fingers too fast; didn't need the enclosing 'list':
> testlist <- list(
+ shape=c(0, 0, 2),
+ cell.fill=c("red","blue","green"),
+ back.fill=rep("white",3),
+ scale.max=rep(100,3)
+ )
>
> wanted <- lapply(seq(length(testlist[[1]])), function(x){
+ # iterate for each element of the lis
Does this do it:
> testlist <- list(
+ shape=c(0, 0, 2),
+ cell.fill=c("red","blue","green"),
+ back.fill=rep("white",3),
+ scale.max=rep(100,3)
+ )
>
> wanted <- lapply(seq(length(testlist[[1]])), function(x){
+ # iterate for each element of the list
+ result <- lapply(names(test
I have a list of vectors, all forced to be the same length:
testlist <- list(
shape=c(0, 0, 2),
cell.fill=c("red","blue","green"),
back.fill=rep("white",3),
scale.max=rep(100,3)
)
> str(testlist)
List of 4
$ shape: num [1:3] 0 0 2
$ cell.fill: chr [1:3] "red" "blue" "green"
$ bac
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