R 将数据帧转换为与树网络兼容的列表

考虑以下数据框：

   Country     Provinces          City Zone
1   Canada   Newfondland      St Johns    A
2   Canada           PEI Charlottetown    B
3   Canada   Nova Scotia       Halifax    C
4   Canada New Brunswick   Fredericton    D
5   Canada        Quebec            NA   NA
6   Canada        Quebec   Quebec City   NA
7   Canada       Ontario       Toronto    A
8   Canada       Ontario        Ottawa    B
9   Canada      Manitoba      Winnipeg    C
10  Canada  Saskatchewan        Regina    D

是否有一种聪明的方法将其转换为

treeNetwork

兼容列表（来自

networkD3

包），格式如下：

CanadaPC <- list(name = "Canada",
                 children = list(
                   list(name = "Newfoundland",
                        children = list(list(name = "St. John's",
                                             children = list(list(name = "A"))))),
                   list(name = "PEI",
                        children = list(list(name = "Charlottetown",
                                             children = list(list(name = "B"))))),
                   list(name = "Nova Scotia",
                        children = list(list(name = "Halifax",
                                             children = list(list(name = "C"))))),
                   list(name = "New Brunswick",
                        children = list(list(name = "Fredericton",
                                             children = list(list(name = "D"))))),
                   list(name = "Quebec",
                        children = list(list(name = "Quebec City"))),
                   list(name = "Ontario",
                        children = list(list(name = "Toronto",
                                             children = list(list(name = "A"))),
                                        list(name = "Ottawa",
                                             children = list(list(name = "B"))))),
                   list(name = "Manitoba",
                        children = list(list(name = "Winnipeg",
                                             children = list(list(name = "C"))))),
                   list(name = "Saskatchewan",
                        children = list(list(name = "Regina",
                                             children = list(list(name = "D")))))))

---

对于这种情况，更好的策略可能是递归的

split（）

，下面是这样一个实现。首先，这里是示例数据

dd<-structure(list(Country = c("Canada", "Canada", "Canada", "Canada", 
"Canada", "Canada", "Canada", "Canada", "Canada", "Canada"), 
    Provinces = c("Newfondland", "PEI", "Nova Scotia", "New Brunswick", 
    "Quebec", "Quebec", "Ontario", "Ontario", "Manitoba", "Saskatchewan"
    ), City = c("St Johns", "Charlottetown", "Halifax", "Fredericton", 
    NA, "Quebec City", "Toronto", "Ottawa", "Winnipeg", "Regina"
    ), Zone = c("A", "B", "C", "D", NA, NA, "A", "B", "C", 
    "D")), class = "data.frame", row.names = c(NA, -10L), .Names = c("Country", 
"Provinces", "City", "Zone"))

然后我们就可以跑了

rsplit(dd)

---

它似乎与测试数据一起工作。唯一的区别是子节点的排列顺序。

我认为它没有提供正确的嵌套，因为

treeNetwork（）

函数没有呈现预期的输出。我编辑了这个问题以反映正确的

dput（）

structure输入有什么具体错误？您不包含任何代码来测试结果。实际上，此解决方案不假设一个根节点，因此它返回一个列表。请尝试

rsplit（dd）[[1]]

rsplit <- function(x) {
    x <- x[!is.na(x[,1]),,drop=FALSE]
    if(nrow(x)==0) return(NULL)
    if(ncol(x)==1) return(lapply(x[,1], function(v) list(name=v)))
    s <- split(x[,-1, drop=FALSE], x[,1])
    unname(mapply(function(v,n) {if(!is.null(v)) list(name=n, children=v) else list(name=n)}, lapply(s, rsplit), names(s), SIMPLIFY=FALSE))
}

rsplit(dd)