将数据框转换为树网络兼容列表

Question

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

是否有巧妙的方法将其转换为 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")))))))

为了绘制具有任意级别集的 Reingold-Tilford 树：

我已经尝试了几个 sub-optimal 例程，包括 for 循环的混乱组合，但我无法以所需的格式获得它。

理想情况下，该函数将缩放以将第一列视为 root（起点），而其他列将是 children.

的不同级别

编辑

有人就同一主题询问了 similar question，@MrFlick 提供了一个有趣的递归函数。原始数据框有一组固定的级别。我引入了 NAs 以添加 @MrFlick 初始解决方案中未解决的另一个复杂级别（任意级别集）。

数据

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"))

Answer 1

对于这种情况更好的策略可能是递归 split() 下面是这样的一个实现。首先，这是样本数据

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"))

请注意，我已将 "NA" 字符串替换为真实的 NA 值。现在，函数

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))
}

那我们可以运行

rsplit(dd)

它似乎与测试数据一起工作。唯一不同的是子节点的排列顺序。

将数据框转换为树网络兼容列表

Convert a data frame to a treeNetwork compatible list

r

networkd3