如何将数据框转换为嵌套列表

Question

提供以下结构的data.frame() var1.gender var1.score.raw var1.score.raw.lower var1.score.raw.upper [...] var2.gender var2.score.raw var2.score.raw.lower var2.score.raw.upper [...] 如何将其转换为按.拆分的多维列表？

示例数据： 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)) 生成的列表应该类似于这样：

$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

- Comfort Eagle

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请提供可重现的示例。参考链接。 - Sotos

我不清楚您希望得到的输出是什么。您所说的“多维列表”具体指什么？ - MrFlick

你的输出抛出了 df$var1.score？ - Frank

首先，你提供的 df 具有 .raw.lower，但你不想再按 raw 分割，只需要 score。你确定要特别按句点分割吗？ - Evan Friedland

Answer 1

顺便提一下，“df”是一种构建所需列表的简单方法，对于“df”的每一列，可以通过评估类似list[["X"]][["Y"]][["Z"]][...] = df$X.Y.Z...的调用来构建所需列表。这可以通过操作“语言”对象动态完成。

定义一个接受列表、名称/索引的字符向量和要分配在该级别的值的函数，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)
}

有些bquote调用可以更简单或使用substitute替代，但是如上所述，使用它构造的指数在排版方面更好。

然后，针对"df"的每一列，重新构造一个-初始为空的-列表：

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

使用这种方法，列表在每次迭代时被重新构造，但它的元素并不被复制。

- alexis_laz

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这太棒了，+1 除了bquote阅读之外，您还推荐哪些阅读来熟悉R编程的这一黑暗面呢？ - Evan Friedland

@EvanFriedland：在相同的上下文中，您可以查看相应的手册和一般的“惰性求值”概念。 - alexis_laz

Answer 2

我稍后会编辑一些内容，用于查看列名称中的句点（更加复杂），但是在不自动化的情况下，您可以创建嵌套列表，例如：

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

尝试制作一个动态版本，但在考虑递归时迷失了方向。无论如何，如果您扩展数据集中的varX数量，这可能会起作用。它不像手动操作那样干净，并且仍然有一个$empty list。

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

- Evan Friedland