[R] reorder a list
Greg Snow
538280 at gmail.com
Tue Jul 8 19:24:15 CEST 2014
And here is another approach:
> out <- vector('list',length(unique(unlist(A1))))
> names(out) <- sort(unique(unlist(A1)))
> for( i in seq_along(A1) ) {
+ for( j in as.character(A1[[i]]) ) {
+ out[[j]] <- c(out[[j]], i)
+ }
+ }
>
> out
$`1`
[1] 1
$`2`
[1] 1 2
$`3`
[1] 1
$`4`
[1] 1 2 4
$`5`
[1] 2 4
$`13`
[1] 4
$`23`
[1] 3
Which approach is faster/more efficient could depend on what your real
data looks like, how big the list is, how large/variable the vectors
within the list are. Both methods could probably also be improved,
but unless the list is big enough that either takes quite a bit of
time then it is probably not worth the time to optimize them.
On Tue, Jul 8, 2014 at 11:11 AM, Greg Snow <538280 at gmail.com> wrote:
> Here is one approach that gives almost the same answer as your example:
>
>> A1<-list(c(1:4),c(2,4,5),23,c(4,5,13))
>>
>> A2 <- sort(unique(unlist(A1)))
>> names(A2) <- A2
>> sapply(A2, function(x) which( sapply(A1, function(y) x %in% y) ),
> + simplify=FALSE, USE.NAMES=TRUE )
> $`1`
> [1] 1
>
> $`2`
> [1] 1 2
>
> $`3`
> [1] 1
>
> $`4`
> [1] 1 2 4
>
> $`5`
> [1] 2 4
>
> $`13`
> [1] 4
>
> $`23`
> [1] 3
>
> If you want the `23` to be in the 23rd element of the list (with empty
> values before it) then just change A2 to be a vector from 1 to the
> largest value
>
> On Tue, Jul 8, 2014 at 10:39 AM, Lorenzo Alfieri <alfios17 at 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]]
>>
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>
>
>
> --
> Gregory (Greg) L. Snow Ph.D.
> 538280 at gmail.com
--
Gregory (Greg) L. Snow Ph.D.
538280 at gmail.com
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