R 如何将数据帧转换为嵌套列表

提供以下结构的

data.frame（）

var1.性别

var1.score.raw

var1.score.raw.lower

var1.score.raw.upper

[…]

var2.性别

var2.score.raw

var2.score.raw.lower

var2.score.raw.upper

[…]

如何将其转换为多维列表，按

拆分

样本数据：

---

df稍后我将使用一些可以查看列名中句点的内容（更复杂）来编辑此内容，但是，在不自动化的情况下，您可以创建嵌套列表，如下所示：

df <- data.frame('var1.gender' = c(1,1,3,3), 'var1.score.raw' = c(12.3, 12.4, 14.5, 13.2), 'var1.score.raw.lower' = c(11,11,13,12), 'var1.score.raw.upper' = c(13,13,15,14), 'var2.gender' = c(1,1,3,3), 'var2.score.raw' = c(12.3, 12.4, 14.5, 13.2), 'var2.score.raw.lower' = c(11,11,13,12), 'var2.score.raw.upper' = c(13,13,15,14))
df

# changed your naming here to remove the not-needed ".raw."
colnames(df) <- c("var1.gender", "var1.score.raw", "var1.score.lower", "var1.score.upper", "var2.gender", "var2.score.raw", "var2.score.lower", "var2.score.upper")

nested <- with(df, expr = {list(var1 = list(gender = var1.gender, 
                                            score = list(raw = var1.score.raw, 
                                                         lower = var1.score.lower, 
                                                         upper = var1.score.upper)),
                                var2 = list(gender = var2.gender, 
                                            score = list(raw = var2.score.raw, 
                                                         lower = var2.score.lower, 
                                                         upper = var2.score.upper)))})
nested
$var1
$var1$gender
[1] 1 1 3 3

$var1$score
$var1$score$raw
[1] 12.3 12.4 14.5 13.2

$var1$score$lower
[1] 11 11 13 12

$var1$score$upper
[1] 13 13 15 14



$var2
$var2$gender
[1] 1 1 3 3

$var2$score
$var2$score$raw
[1] 12.3 12.4 14.5 13.2

$var2$score$lower
[1] 11 11 13 12

$var2$score$upper
[1] 13 13 15 14

df通过构造“df”的方式，构建通缉名单的一种简单方法是为“df”的每一列评估一个调用，如
list[[“X”][[“Y”]][[“Z”][…]=df$X.Y.Z.
。这可以通过操纵“语言”对象动态完成

定义一个接受列表、名称/索引的字符向量和在该级别分配的值的函数，我们有：

assign_list_element = function(x, inds, val)
{
    cl = bquote(x[[.(inds[1])]])
    for(s in inds[-1]) cl = bquote(.(cl)[[.(s)]])

    cl = call("<-", cl, bquote(.(val))) 
    print(cl); flush.console() 

    eval(cl)  

    return(x)
}

使用这种方法，列表在每次迭代时都会重新构造，尽管它的元素不会被复制。
请提供我不清楚所需的输出是什么。你所说的“多维列表”到底是什么意思？你的输出抛出df$var1.score？首先，你提供的df有一个.raw.lower，但你不想再被raw分割，只需要分数。你确定要按时间段具体划分吗？这很好，除了bquote阅读之外，+1，你有没有推荐一些阅读材料让我熟悉R编程的黑暗面？@EvanFriedland:在同样的背景下，你可以看一下和一般的“惰性评估”概念。
nester <- function(df, splitby = "."){
  separated <- strsplit(colnames(df), paste0("[", splitby, "]"))
  # in order to rbind this into a matrix, we have to make all vectors the same length
  n <- max(rapply(separated, length))
  separated <- do.call(rbind, rapply(separated, function(x) {length(x) <- n; x }, how = "replace"))
  separated <- ifelse(is.na(separated), "empty", separated)
  listnames <- apply(separated, 2, unique)
  L <- list()
  # Assumes n is 3. 
  for(L1 in listnames[[1]]){
    L[[L1]] <- list() # create List level 1
    for(L2 in listnames[[2]]){
      L[[L1]][[L2]] <- list() # create List level 2
      for(L3 in listnames[[3]]){
        L[[L1]][[L2]][[L3]] <- list() # create list level 3
        # If no data exists for that list combination ...
        if(length(df[,which(separated[,1] == L1 & separated[,2] == L2 & separated[,3] == L3)]) == 0){
          L[[L1]][[L2]][[L3]] <- NULL # then remove that nested list.
        } else {
          # otherwise go ahead and put that column in as a list
          L[[L1]][[L2]][[L3]] <- df[,which(separated[,1] == L1 & separated[,2] == L2 & separated[,3] == L3)]
          # if data is sitting in a list$empty ...
          if( L3 == "empty" ){
            z <- unname(unlist(L[[L1]][[L2]][[L3]]))
            L[[L1]][[L2]][[L3]] <- as.vector(z) # save the empty L3 to the L2
            #L[[L1]][[L2]][[L3]] <- NULL # and delete the L3
          }  
        }
      }
    }
  }
  return(L)
}
df.List <- nester(df, splitby = ".")
df.List

assign_list_element = function(x, inds, val)
{
    cl = bquote(x[[.(inds[1])]])
    for(s in inds[-1]) cl = bquote(.(cl)[[.(s)]])

    cl = call("<-", cl, bquote(.(val))) 
    print(cl); flush.console() 

    eval(cl)  

    return(x)
}

nms = strsplit(names(df), ".", TRUE)
l = list()
for(i in seq_along(nms)) l = assign_list_element(l, nms[[i]], df[[i]])
#x[["var1"]][["gender"]] <- c(1, 1, 3, 3)
#x[["var1"]][["score"]][["raw"]] <- c(12.3, 12.4, 14.5, 13.2)
#x[["var1"]][["score"]][["lower"]] <- c(11, 11, 13, 12)
#x[["var1"]][["score"]][["upper"]] <- c(13, 13, 15, 14)
#x[["var2"]][["gender"]] <- c(1, 1, 3, 3)
#x[["var2"]][["score"]][["raw"]] <- c(12.3, 12.4, 14.5, 13.2)
#x[["var2"]][["score"]][["lower"]] <- c(11, 11, 13, 12)
#x[["var2"]][["score"]][["upper"]] <- c(13, 13, 15, 14)

str(l)
#List of 2
# $ var1:List of 2
#  ..$ gender: num [1:4] 1 1 3 3
#  ..$ score :List of 3
#  .. ..$ raw  : num [1:4] 12.3 12.4 14.5 13.2
#  .. ..$ lower: num [1:4] 11 11 13 12
#  .. ..$ upper: num [1:4] 13 13 15 14
# $ var2:List of 2
#  ..$ gender: num [1:4] 1 1 3 3
#  ..$ score :List of 3
#  .. ..$ raw  : num [1:4] 12.3 12.4 14.5 13.2
#  .. ..$ lower: num [1:4] 11 11 13 12
#  .. ..$ upper: num [1:4] 13 13 15 14