Revert list structure

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EDIT -- working from @Josh O'Briens suggestion and my own improvemes

The problem was that do.call rbind was not calling rbind.data.frame which does some matching of names. rbind.data.frame should work, because data.frames are lists and each sublist is a list, so we could just call it directly.

apply(do.call(rbind.data.frame, z), 1, as.list)

However, while this may be succicint, it is slow because do.call(rbind.data.frame, ...) is inherently slow.

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Something like (in two steps)

 # convert each component of z to a data.frame
 # so rbind.data.frame so named elements are matched
 x <- data.frame((do.call(rbind, lapply(z, data.frame))))
 # convert each column into an appropriately named list
 o <- lapply(as.list(x), function(i,nam) as.list(`names<-`(i, nam)), nam = rownames(x))
 o
$a
$a$z1
[1] 1

$a$z2
[1] 1


$b
$b$z1
[1] 2

$b$z2
[1] 4


$c
$c$z1
[1] 3

$c$z2
[1] 0

And an alternative

# unique names
nn <- Reduce(unique,lapply(z, names))
# convert from matrix to list `[` used to ensure correct ordering
as.list(data.frame(do.call(rbind,lapply(z, `[`, nn))))

Edit:

Here's a more flexible version that will work on lists whose elements don't necessarily contain the same set of sub-elements.

fun <-  function(ll) {
    nms <- unique(unlist(lapply(ll, function(X) names(X))))
    ll <- lapply(ll, function(X) setNames(X[nms], nms))
    ll <- apply(do.call(rbind, ll), 2, as.list)
    lapply(ll, function(X) X[!sapply(X, is.null)])
}

## An example of an 'unbalanced' list
z <- list(z1 = list(a = 1, b = 2), 
          z2 = list(b = 4, a = 1, c = 0))
## Try it out
fun(z)

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Original answer

z <- list(z1 = list(a = 1, b = 2, c = 3), z2 = list(b = 4, a = 1, c = 0))

zz <- lapply(z, `[`, names(z[[1]]))   ## Get sub-elements in same order
apply(do.call(rbind, zz), 2, as.list) ## Stack and reslice