如何将R语言data.frame中的节点路径转换为树结构？有无简便函数？

我有一个data.frame，其中一列代表节点路径，希望将其转换为树结构。请问是否存在简便的实现函数？以下是一个简单示例：

data <- data.frame(
  Name = c("A", "A1", "A2", "A1a", "A1b", "A2a", "A2b", "A2c"),
  Path = c("1", "1,1", "1,2", "1,1,1", "1,1,2", "1,2,1", "1,2,2", "1,2,3")
)

期望转换为如下形式：

nodes <- list(
  list(
    text = "A",
    li_attr = list(id = "1"),
    state = list(opened = TRUE),
    
    children = list(
      list(
        text = "A1",
        li_attr = list(id = "1,1"),
        state = list(opened = TRUE),
        
        children = list(
          list(
            text = "A1a",
            li_attr = list(id = "1,1,1")),
          list(
            text = "A1b",
            li_attr = list(id = "1,1,2"))
        )),
      list(
        text = "A2",
        li_attr = list(id = "1,2"),
        state = list(opened = TRUE),
        
        children = list(
          list(
            text = "A2a",
            li_attr = list(id = "1,2,1")),
          list(
            text = "A2b",
            li_attr = list(id = "1,2,2")),
          list(
            text = "A2c",
            li_attr = list(id = "1,2,3"))
        )
      )
    )
  )
)

解决方案：使用data.tree包 #

R的data.tree包专门处理层级树结构，能快速实现路径到树的转换，再输出成你需要的嵌套列表格式，步骤如下：

1. 安装并加载包 #

install.packages("data.tree")
library(data.tree)

2. 构建基础树结构 #

先把Path列的逗号分隔符替换为data.tree默认的斜杠，再直接转换为树节点：

# 转换路径分隔符
data$tree_path <- gsub(",", "/", data$Path)

# 从data.frame生成树
tree <- as.Node(data, pathName = "tree_path", pathDelimiter = "/")

3. 转换为目标列表格式 #

写一个递归函数，把树节点转换成你需要的嵌套结构：

tree_to_target_list <- function(node) {
  # 基础节点属性
  res <- list(
    text = node$Name,
    li_attr = list(id = gsub("/", ",", node$path))
  )
  
  # 非叶子节点添加展开状态和子节点
  if (node$childrenCount > 0) {
    res$state <- list(opened = TRUE)
    res$children <- lapply(node$children, tree_to_target_list)
  }
  
  return(res)
}

# 生成最终结果
nodes <- list(tree_to_target_list(tree))

运行后，nodes的结构就和你期望的完全一致。你可以用str(nodes)查看结构细节。

内容的提问来源于stack exchange，提问作者Kamaloka