使用深度名称向量作为索引替换嵌套列表 [英] Replacing nested list using a vector of names of depths as an index

问题描述

获取一个简单的嵌套列表L:

Take a simple nested list L:

L <- list(lev1 = list(lev2 = c("bit1","bit2")), other=list(yep=1))
L
#$lev1
#$lev1$lev2
#[1] "bit1" "bit2"
#
# 
#$other
#$other$yep
#[1] 1

以及一个向量，该向量为我要从L中选择的每个部分提供一系列深度:

And a vector giving a series of depths for each part I want to select from L:

sel <- c("lev1","lev2")

索引时我想要的结果是:

The result I want when indexing is:

L[["lev1"]][["lev2"]]
#[1] "bit1" "bit2"

我可以像这样使用Reduce进行概括:

Which I can generalise using Reduce like so:

Reduce(`[[`, sel, init=L)
#[1] "bit1" "bit2"

现在，我想扩展此逻辑以进行替换，例如:

Now, I want to extend this logic to do a replacement, like so:

L[["lev1"]][["lev2"]] <- "new val"

，但是我对如何生成递归[[选择的方式感到很困惑，这使得我也可以对其进行赋值.

, but I am genuinely stumped as to how to generate the recursive [[ selection in a way that will allow me to then assign to it as well.

推荐答案

为什么你不能这样做

L[[sel]] <- "new val"

好吧，如果您想做很多事，那么 您仍然可以将Reduce与modifyList一起使用，或者可以使用[[<-.这是modifyList的示例:

well if you want to do the long way then You could still use Reduce with modifyList or you could use [[<-. Here is an example with modifyList:

modifyList(L,Reduce(function(x,y)setNames(list(x),y),rev(sel),init = "new val"))
$lev1
$lev1$lev2
[1] "new val"


$other
$other$yep
[1] 1

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