Is the following 'g' what you want? A better example might be with
A2a <- lapply(A1, function(x)x+seq_along(x)/(100*length(x)))
g <- function (x, y) {
xLengths <- vapply(x, FUN = length, FUN.VALUE = 0L)
yLengths <- vapply(y, FUN = length, FUN.VALUE = 0L)
stopifnot(identical(xLengths, yLengths))
split(unlist(y, use.names = FALSE), unlist(x, use.names = FALSE))
}
Used as
> g(A1,A2)
$`1`
[1] 2.718282
$`2`
[1] 7.389056 7.389056
$`3`
[1] 20.08554
$`4`
[1] 54.59815 54.59815 54.59815
$`5`
[1] 148.4132 148.4132
$`13`
[1] 442413.4
$`23`
[1] 9744803446
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jul 9, 2014 at 2:04 AM, Lorenzo Alfieri <alfio...@hotmail.com>
wrote:
> Thanks Bill and the other guys for the variety of useful replies!
> In fact I'm working with pretty big lists (with ~35000 sublists) and
> Bill's solution is the fastest one in terms of computing time.
> Now comes the second part of the question... :-)
> I've my usual list of values and time indices to sort:
> A1<-list(c(1:4),c(2,4,5),23,c(4,5,13))
> and then another list A2 with variables which have to be paired with the
> values of A1:
> A2<-sapply(A1, "exp") #(in my case there's no exp relation between A1
> and A2, they're completely uncorrelated. That's just an example )
> > A2
> [[1]]
> [1] 2.718282 7.389056 20.085537 54.598150
>
> [[2]]
> [1] 7.389056 54.598150 148.413159
>
> [[3]]
> [1] 9744803446
>
> [[4]]
> [1] 54.59815 148.41316 442413.39201
>
> Now I'd like to reorder the elements of A2 according to the same rule
> applied for A1:
>
> f <- function (x) {
> lengths <- vapply(x, FUN = length, FUN.VALUE = 0L)
> split(rep(seq_along(x), lengths), unlist(x, use.names = FALSE))
> }
> B1<-f(A1)
>
> and thus obtain a list B2 which looks like this:
> > B2
> $`1`
> [1] 2.718282
>
> $`2`
> [1] 7.389056 7.389056
>
> $`3`
> [1] 20.08554
>
> $`4`
> [1] 54.59815 54.59815 54.59815
>
> $`5`
> [1] 148.4132 148.4132
>
> $`13`
> [1] 442413.4
>
> $`23`
> [1] 9744803446
>
> (In this example each element is the exp() of the sublist name, but in a
> general case they would be uncorrelated, and the resulting elements of each
> sublist would be different)
> Any idea?
> Alfio
>
>
> > From: wdun...@tibco.com
> > Date: Tue, 8 Jul 2014 12:11:09 -0700
> > Subject: Re: [R] reorder a list
> > To: alfio...@hotmail.com
> > CC: r-help@r-project.org
>
> >
> > f <- function (x) {
> > lengths <- vapply(x, FUN = length, FUN.VALUE = 0L)
> > split(rep(seq_along(x), lengths), unlist(x, use.names = FALSE))
> > }
> > f(A1) # gives about what you want (has, e.g., name 23, not position
> > 23, in output)
> > Bill Dunlap
> > TIBCO Software
> > wdunlap tibco.com
> >
> >
> > On Tue, Jul 8, 2014 at 9:39 AM, Lorenzo Alfieri <alfio...@hotmail.com>
> wrote:
> > > Hi,
> > > I'm trying to find a way to reorder the elements of a list.
> > > Let's say I have a list like this:
> > > A1<-list(c(1:4),c(2,4,5),23,c(4,5,13))
> > >
> > >> A1
> > > [[1]]
> > > [1] 1 2 3 4
> > >
> > > [[2]]
> > > [1] 2 4 5
> > >
> > > [[3]]
> > > [1] 23
> > >
> > > [[4]]
> > > [1] 4 5 13
> > >
> > > All the elements included in it are values, while each sublist is a
> time index
> > > Now, I'd like to reorder the list (without looping) so to obtain one
> sublist for each value, which include all the time indices where each value
> appears.
> > > In other words, the result should look like this:
> > >>A2
> > > [[1]]
> > > [1] 1
> > >
> > > [[2]]
> > > [1] 1 2 #because value "2" appears in the time index [[1]] and [[2]]
> of A1
> > >
> > > [[3]]
> > > [1] 1
> > >
> > > [[4]]
> > > [1] 1 2 4
> > >
> > > [[5]]
> > > [1] 2 4
> > >
> > > [[13]]
> > > [1] 4
> > >
> > > [[23]]
> > > [1] 3
> > >
> > > Any suggestion?
> > > Thanks
> > > Alfio
> > >
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > ______________________________________________
> > > R-help@r-project.org mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
>
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