CODEWITHSUNDEEP
rlist
ego_
ego_

Reputation: 1491

Extract nested list element where element names are similar

I have a nested liste, resulted from a function. Where the top element names are reapeated in the element names further down.

$`1`
$`1`$`1`
[1] 0 0 0 0 0 0 0 1 0

$`1`$`2`
[1] 0 0 0 0 0 0 0 0 0

$`2`
$`2`$`1`
[1] 0 0 0 1 1 0 0 0 0

$`2`$`2`
[1] 0 1 0 0 0 1 0 0 0

Is there a way to use an apply function (or whatever) to extract those vectors where the element and subelement names match. E.g. $1$1 and $2$2. I have a huge list (4000 elements with 4000 subelements) so efficiency is thus a must.

Alternatively - I have figured out a way out of this mess by using ´melt()´, but it's too consuming for the size of my set. But if anyone know how to replicate the effect - giving a dataframe with 3 columns one for elementname, one for subelement name and one for the vector - that will also work.

Regards and thanks :)

Upvotes: 3

Views: 3817

Answers (2)

James
James

Reputation: 66844

You can unlist them without recursion to remove the top level list structure, and then use regex-assisted subsetting on the names of this result.

l <- list(`1`=list(`1`=rpois(6,1),`2`=rep(0,6)),`2`=list(`1`=rep(0,6),`2`=rpois(6,1)))
l2 <- unlist(l,recursive=F)
l2[grepl("([0-9]+)[.]\\1",names(l2))]
$`1.1`
[1] 2 0 2 4 1 0

$`2.2`
[1] 0 0 0 2 1 0

Upvotes: 2

Sven Hohenstein
Sven Hohenstein

Reputation: 81693

This is a way to get a list of the vectors you want:

lapply(names(dat), function(x) dat[[x]][[x]])

In a data frame:

do.call("rbind",
        lapply(names(dat),
               function(x) data.frame(element = x,
                                      subelement = x,
                                      values = dat[[x]][[x]])
        )
)

Upvotes: 4

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