R中是否有一种方法可以将扁平化的层次结构表转换为完全展开的输出以用于映射？

问题描述

我有一个扁平的层次结构，深约10层。有一个钥匙，但是很遗憾，这个钥匙没有被遵守，所以不能用于我的目的。

我需要把这张桌子放到一个很宽的层次结构中，我唯一的关系是行ID和父ID。

我的数据看起来像这样（简化）；

id  name            description     code    sub_level_name  parent_id
1   Parent Company  My Big Company  MBC     Company 
2   Franchise1      My Franchise1   MF1     Franchise       1
3   Store1          My Store1       MS1     Store           2
4   Store2          My Store2       MS2     Store           2
5   Store1Owner     My Store1Owner  MSO1    Store Owner     3
6   Store2Owner     My Store2Owner  MSO2    Store Owner     4

我希望结果看起来像

company_name    company_description   company_code  franchise_name  franchise_description   franchise_code  store_name  store_description   store_code  storeowner_name storeowner_description  storeowner_code
Parent Company  My Big Company        MBC           Franchise1      My Franchise1               MF1  Store1           My Store1        MS1      Store1Owner            My Store1Owner         MSO1
Parent Company  My Big Company        MBC           Franchise1      My Franchise1               MF1  Store2           My Store2        MS2      Store2Owner            My Store2Owner         MSO2

通常我会使用pivot_wider并使用sub_level_name列，但这是我所说的在较大范围内未遵循的内容（该表有7000行，并且可以深10行）

我觉得我需要做的是将父ID与ID匹配，并根据其sub_level_name将各列连接起来，以创建“ x.name，x.description，x.code”列的类型，然后执行逐行，直到达到层次结构的顶层。

非常有用的帮助-不必是R解决方案，SQL解决方案也很好。

编辑：Dput源数据

structure(list(id = c(1,2,3,4,5,6),name = c("Parent Company","Franchise1","Store1","Store2","Store1Owner","Store2Owner"
),description = c("My Big Company","My Franchise1","My Store1","My Store2","My Store1Owner","My Store2Owner"),code = c("MBC","MF1","MS1","MS2","MSO1","MSO2"),sub_level_name = c("Company","Franchise","Store","Store Owner","Store Owner"),parent_id = c(NA,1,4)),class = c("spec_tbl_df","tbl_df","tbl","data.frame"),row.names = c(NA,-6L),spec = structure(list(
    cols = list(id = structure(list(),class = c("collector_double","collector")),name = structure(list(),class = c("collector_character",description = structure(list(),code = structure(list(),sub_level_name = structure(list(),parent_id = structure(list(),"collector"))),default = structure(list(),class = c("collector_guess",skip = 1),class = "col_spec"))

解决方法

您可以尝试为每个id创建一个sub_level_name列，删除不需要的列并获取宽格式的数据。

library(dplyr)

df %>%
  group_by(sub_level_name) %>%
  mutate(id = row_number()) %>%
  select(-parent_id) %>%
  tidyr::pivot_wider(names_from = sub_level_name,values_from = c(name,description,code))