r - 来自 child/parent 关系的分层数据框

Question

我有一个 child - parent data.frame，我想将其转换为包含所有级别和级别编号的完整层次列表。下面的示例数据分为三个级别，但可能更多。我可以使用什么函数来转换数据？

来源：

data.frame(name = c("land", "water", "air", "car", "bicycle", "boat", "balloon",
  "airplane", "helicopter", "Ford", "BMW", "Airbus"), parent = c(NA, NA, NA, 
  "land", "land", "water", "air", "air", "air", "car", "car", "airplane"))

         name   parent
1        land     <NA>
2       water     <NA>
3         air     <NA>
4         car     land
5     bicycle     land
6        boat    water
7     balloon      air
8    airplane      air
9  helicopter      air
10       Ford      car
11        BMW      car
12     Airbus airplane

目的地：

data.frame(level1 = c("land", "water", "air", "land", "land", "water", "air", 
  "air", "air", "land", "land", "air"), level2 = c(NA, NA, NA, "car", "bicylcle", 
  "boat", "balloon", "airplane", "helicopter", "car", "car", "airplane"),
  level3 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, "Ford", "BMW", "Airbus"), 
  level_number = c(1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3))

   level1     level2 level3 level_number
1    land       <NA>   <NA>            1
2   water       <NA>   <NA>            1
3     air       <NA>   <NA>            1
4    land        car   <NA>            2
5    land   bicylcle   <NA>            2
6   water       boat   <NA>            2
7     air    balloon   <NA>            2
8     air   airplane   <NA>            2
9     air helicopter   <NA>            2
10   land        car   Ford            3
11   land        car    BMW            3
12    air   airplane Airbus            3

Answer 1

使用 data.table 您可以执行以下操作：

require(data.table)
l <- list() # initialize empty list
setDT(dat) 
setkey(dat, parent) # setting up the data as keyed data.table
current_lvl <- dat[is.na(parent), .(level_number = 1), keyby=.(level1 = name)]

不current_lvl看起来如下（按level1键）

   level1 level_number
1:    air            1
2:   land            1
3:  water            1

现在的诀窍是加入 dat 和 current_lvl 并适当修改结果：

  current_lvl <- current_lvl[dat][ # Join the data.tables
!is.na(level_number)][ #exclude non-child-rows
  ,level_number := level_number + 1] # increment level_number
setnames(current_lvl, "name", paste0("level",ind+1)) # rename column
setkeyv(current_lvl, paste0("level",ind+1)) # set key

哪个给你（由 level2 键入）

   level1 level_number     level2
1:    air            2   airplane
2:    air            2    balloon
3:   land            2    bicycle
4:  water            2       boat
5:   land            2        car
6:    air            2 helicopter

按如下方式在 while 循环中运行：

while(nrow(current_lvl) > 0){
  ind <- length(l) + 1
  l[[ind]] <- current_lvl
  current_lvl <- current_lvl[dat][!is.na(level_number)][,level_number := level_number + 1]
  if(nrow(current_lvl) == 0L){
    break
  }
  setnames(current_lvl, "name", paste0("level",ind+1))
  setkeyv(current_lvl, paste0("level",ind+1))
}

你可以看看l看看结果。通过 rbindlist 结合这个给你你想要的东西

res <- rbindlist(l, fill=TRUE)
setcolorder(res, sort(names(res)))
res

结果是什么

> res
    level_number level1     level2 level3
 1:            1    air         NA     NA
 2:            1   land         NA     NA
 3:            1  water         NA     NA
 4:            2    air   airplane     NA
 5:            2    air    balloon     NA
 6:            2   land    bicycle     NA
 7:            2  water       boat     NA
 8:            2   land        car     NA
 9:            2    air helicopter     NA
10:            3    air   airplane Airbus
11:            3   land        car    BMW
12:            3   land        car   Ford

Answer 2

使用 data.tree 包，您可以执行以下操作：

library(data.tree)
df <- data.frame(name = c("land", "water", "air", "car", "bicycle", "boat", "balloon", "airplane", "helicopter", "Ford", "BMW", "Airbus"), 
                 parent = c("root", "root", "root", "land", "land", "water", "air", "air", "air", "car", "car", "airplane"))

请注意，我将 NA 替换为 "root"，这使得转换为 data.tree 更加容易。即：

tree <- FromDataFrameNetwork(df)

获取所需的格式变得很简单，因为我们可以使用 data.tree:

中的层次结构基础结构

ToDataFrameTree(tree, 
                level1 = function(x) x$path[2],
                level2 = function(x) x$path[3],
                level3 = function(x) x$path[4],
                level_number = function(x) x$level - 1)[-1,-1]

Answer 3

不要使用 "root" 作为顶级记录的父值。使用 data.tree 包的解决方案很棒，但是，在较新的版本中，"root" 是节点的保留名称。虽然它被自动替换为 "root2"，但对 FromDataFrameNetwork(df) 的调用并没有 return 想要的树。

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