Question

我正在尝试创建一个员工表，他们向谁报告以及他们是什么级别。

我一直在使用类似的表，@ TonakShah足够帮助我计算最低级别的位置，并且上面的级别正在使用下面的解决方案。

我的员工表如下：

input = structure(list(Level.1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
1L), .Label = "Board", class = "factor"), Level.2 = structure(c(2L, 
2L, 2L, 1L, 1L, 3L, 3L), .Label = c("Aasha", "Grace", "Marisol"
), class = "factor"), Level.3 = structure(c(4L, 4L, 3L, 1L, 1L, 
2L, 2L), .Label = c("Alex", "Chandler", "Millie", "Tushad"), class = "factor"), 
    Level.4 = structure(c(2L, 2L, 6L, 1L, 5L, 3L, 4L), .Label = c("#", 
    "Frank", "Joey", "Rachel", "Sarah", "Tony"), class = "factor"), 
    Level.5 = structure(c(3L, 2L, 1L, 1L, 1L, 4L, 1L), .Label = c("#", 
    "Lela", "Millie", "Ross"), class = "factor"), Level.6 = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L), .Label = "#", class = "factor")), class = "data.frame", row.names = c(NA, 
-7L))

并使用Ronak在此处描述的技术（stackoverflow.com/questions/56903188/create-a-table-from-a-hierarchy/）

as.data.frame(t(apply(input, 1, function(x) 
       {new_x = x[x != "###"]; c(rev(tail(new_x, 2)), length(new_x)) })))

我可以获得所需的大部分表格。但是我很难找到“老板”（例如那些有雇员但没有“董事会”的人）。

我的理想输出看起来像这样（我添加了colnames以便于理解）：

structure(list(Subordinate = structure(c(9L, 4L, 14L, 5L, 7L, 
13L, 9L, 2L, 1L, 12L, 11L, 6L, 3L, 8L, 10L), .Label = c("Aasha", 
"Alex", "Chandler", "Frank", "Grace", "Joey", "Lela", "Marisol", 
"Millie", "Rachel", "Ross", "Sarah", "Tony", "Tushad"), class = "factor"), 
    Boss = structure(c(5L, 10L, 6L, 3L, 5L, 9L, 6L, 1L, 3L, 2L, 
    7L, 4L, 8L, 3L, 4L), .Label = c("Aasha", "Alex", "Board", 
    "Chandler", "Frank", "Grace", "Joey", "Marisol", "Millie", 
    "Tushad"), class = "factor"), Level = c(5L, 4L, 3L, 2L, 5L, 
    4L, 3L, 3L, 2L, 4L, 5L, 4L, 3L, 2L, 4L)), class = "data.frame", row.names = c(NA, 
-15L))

我想我可能会循环执行，但这似乎不是最好的答案。谁能提供其他提示？

Answer 1

无法提出一个更漂亮的解决方案，但这可行。在先前使用的while调用中使用apply循环，我们可以完成

output <- do.call(rbind.data.frame, apply(input, 1, function(x) {
   new_x = as.character(x[x != "#"])
   list_df <- list()
   i = 1
   while(length(new_x) >= 2) {
      #Get last 2 eneteries
      list_df[[i]] <- c(rev(tail(new_x, 2)), length(new_x))
      #Go one level deeper
      new_x = head(new_x, -1)
      i  = i +1
   }
   do.call(rbind, list_df)
}))

#To remove duplicate enteries
output[!duplicated(output), ]

#         V1       V2 V3
#1    Millie    Frank  5
#2     Frank   Tushad  4
#3    Tushad    Grace  3
#4     Grace    Board  2
#5      Lela    Frank  5
#9      Tony   Millie  4
#10   Millie    Grace  3
#12     Alex    Aasha  3
#13    Aasha    Board  2
#14    Sarah     Alex  4
#17     Ross     Joey  5
#18     Joey Chandler  4
#19 Chandler  Marisol  3
#20  Marisol    Board  2
#21   Rachel Chandler  4

R函数计算出先前的值

1 个答案: