R:重新列出平面列表

Question

This question有一个很好的扁平列表解决方案，同时保留其数据类型(unlist 没有):

flatten = function(x, unlist.vectors=F) {
    while(any(vapply(x, is.list, logical(1)))) {
        if (! unlist.vectors)
            x = lapply(x, function(x) if(is.list(x)) x else list(x))
        x = unlist(x, recursive=F)
    }
    x
}

如果我给它以下列表，它会按预期运行:

> a = list(c(1,2,3), list(52, 561), "a")
> flatten(a)
[[1]]
[1] 1 2 3

[[2]]
[1] 52

[[3]]
[1] 561

[[4]]
[1] "a"

现在我想像 a 一样重组平面列表. relist惨败:

> relist(flatten(a), skeleton=a)
[[1]]
[[1]][[1]]
[1] 1 2 3

[[1]][[2]]
[1] 52

[[1]][[3]]
[1] 561

[[2]]
[[2]][[1]]
[[2]][[1]][[1]]
[1] "a"

[[2]][[2]]
[[2]][[2]][[1]]
NULL

[[3]]
[[3]][[1]]
NULL

现在，我当然可以做 relist(unlist(b), a)但这又会丢失数据类型。重组平面列表的好方法是什么？

如果它处理与 unlist.vectors 类似的属性，则加分。正确。

Answer 1

一种方法是:

relist2 = function(x, like, relist.vectors=F) {
    if (! relist.vectors) 
        like = rapply(a, function(f) NA, how='replace')
    lapply(relist(x, skeleton=like), function(e) unlist(e, recursive=F))
}

这保留了类并区分列表和向量:

> relist2(flatten(a), like=a)
[[1]]
[1] 1 2 3

[[2]]
[[2]][[1]]
[1] 52

[[2]][[2]]
[1] 561

[[3]]
[1] "a"

> relist2(flatten(a, unlist.vectors=T), like=a, relist.vectors=T)
[[1]]
[1] 1 2 3

[[2]]
[[2]][[1]]
[1] 52

[[2]][[2]]
[1] 561

[[3]]
[1] "a"