R help - Sep 2010 - Question: Form a new list with the index replicated equal to the number of elements in that index

Dear R-Helpers,
I have a list l1 like:

l1[[1]]
a b c

l1[[2]]
d

l1[[3]]
e f

I want an output res like:

res[[1]]
1 1 1

res[[2]]
2

res[[3]]
3 3

Essentially, I want to replicate each index equal to the number of elements
present in that index.

Below is what I do to accomplish this:

l1 <- list(c("a", "b", "c"), "d",
c("e", "f"))
res <- mapply(rep, seq_along(l1),times=(lapply(l1, length)))

Is there a more elegant way of doing this (possibly using one (l/m)apply
function)?

Thanks in advance for the help.

Best Regards,
S.

	[[alternative HTML version deleted]]

On 09/13/2010 08:41 PM, Sunny Srivastava wrote:
> Dear R-Helpers,
> I have a list l1 like:
> 
> l1[[1]]
> a b c
> 
> l1[[2]]
> d
> 
> l1[[3]]
> e f
> 
> I want an output res like:
> 
> res[[1]]
> 1 1 1
> 
> res[[2]]
> 2
> 
> res[[3]]
> 3 3
> 
> Essentially, I want to replicate each index equal to the number of elements
> present in that index.
> 
> Below is what I do to accomplish this:
> 
> l1 <- list(c("a", "b", "c"),
"d", c("e", "f"))
> res <- mapply(rep, seq_along(l1),times=(lapply(l1, length)))
> 
> Is there a more elegant way of doing this (possibly using one (l/m)apply
> function)?
Don't know about elegance, but how about

lapply(seq_along(l1), function(i)rep(i, length(l1[[j]])) )

-- 
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

Try this:

relist(rep(1:length(l1), sapply(l1, length)), l1)


On Mon, Sep 13, 2010 at 3:41 PM, Sunny Srivastava
<research.baba@gmail.com>wrote:
> Dear R-Helpers,
> I have a list l1 like:
>
> l1[[1]]
> a b c
>
> l1[[2]]
> d
>
> l1[[3]]
> e f
>
> I want an output res like:
>
> res[[1]]
> 1 1 1
>
> res[[2]]
> 2
>
> res[[3]]
> 3 3
>
> Essentially, I want to replicate each index equal to the number of elements
> present in that index.
>
> Below is what I do to accomplish this:
>
> l1 <- list(c("a", "b", "c"),
"d", c("e", "f"))
> res <- mapply(rep, seq_along(l1),times=(lapply(l1, length)))
>
> Is there a more elegant way of doing this (possibly using one (l/m)apply
> function)?
>
> Thanks in advance for the help.
>
> Best Regards,
> S.
>
>        [[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.
>


-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

	[[alternative HTML version deleted]]

Sunny -
    I don't think mapply is needed:
>  lapply(1:length(mylist),function(x)rep(x,length(mylist[[x]])))
[[1]]
[1] 1 1 1

[[2]]
[1] 2

[[3]]
[1] 3 3

 					- Phil Spector
 					 Statistical Computing Facility
 					 Department of Statistics
 					 UC Berkeley
 					 spector at stat.berkeley.edu


On Mon, 13 Sep 2010, Sunny Srivastava wrote:
> Dear R-Helpers,
> I have a list l1 like:
>
> l1[[1]]
> a b c
>
> l1[[2]]
> d
>
> l1[[3]]
> e f
>
> I want an output res like:
>
> res[[1]]
> 1 1 1
>
> res[[2]]
> 2
>
> res[[3]]
> 3 3
>
> Essentially, I want to replicate each index equal to the number of elements
> present in that index.
>
> Below is what I do to accomplish this:
>
> l1 <- list(c("a", "b", "c"),
"d", c("e", "f"))
> res <- mapply(rep, seq_along(l1),times=(lapply(l1, length)))
>
> Is there a more elegant way of doing this (possibly using one (l/m)apply
> function)?
>
> Thanks in advance for the help.
>
> Best Regards,
> S.
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.
>