R中的内置家庭嵌套树父/子关系

Question

我正在研究家谱 :

我根据 sqldf https://www.r-bloggers.com/exploring-recursive-ctes-with-sqldf/

改编了 Bob Horton 的例子

我的数据：

      person            father
      Guillou Arthur    NA          
      Cleach Marc       NA          
      Guillou Eric      Guillou Arthur          
      Guillou Jacques   Guillou Arthur          
      Cleach Franck     Cleach Marc         
      Cleach Leo        Cleach Marc         
      Cleach Herbet     Cleach Leo          
      Cleach Adele      Cleach Herbet           
      Guillou Jean      Guillou Eric            
      Guillou Alan      Guillou Eric

我的结果，后代按 "Guillou Arthur" 级别排序（没有父亲的最高人）：

  name    parent_name              level
  Guillou Arthur    NA                  1       
  Guillou Eric      Guillou Arthur      2       
  Guillou Jacques   Guillou Arthur      2       
  Guillou Alan      Guillou Eric        3       
  Guillou Jean     Guillou Eric         3

您可以使用 sqldf 的递归查询构建此 table :

数据：

 person <- c("Guillou Arthur",
              "Cleach Marc",
              "Guillou Eric",
              "Guillou Jacques", 
              "Cleach Franck",
              "Cleach Leo",
              "Cleach Herbet",
              "Cleach Adele",
              "Guillou Jean",
              "Guillou Alan" )
 father <- c(NA, NA, "Guillou Arthur" , "Guillou Arthur", "Cleach Marc", "Cleach Marc", "Cleach Leo", "Cleach Herbet", "Guillou Eric", "Guillou Eric")


family <- data.frame(person, father)

从大到长的格式转换：

    library(tidyr)

    long_family <- gather(family, parent, parent_name, -person)

    long_family

查找 "Guillou Arthur" 的后代的递归查询（没有父亲的顶级人物）：

    library(sqldf)
      descendants_sql <- "
      WITH RECURSIVE descendants (name, parent_name, level) AS (
        SELECT person, parent_name, 1 FROM long_family 
          WHERE person = '%s'
          AND parent = '%s'

          UNION ALL
          SELECT F.person, F.parent_name, D.level + 1 
              FROM descendants D
              JOIN long_family F
              ON F.parent_name = D.name)

      SELECT * FROM descendants ORDER BY level, name
      "
      fam <- sqldf(sprintf(descendants_sql, 'Guillou Arthur', 'father'))
      fam

我的问题：
如何直接使用 R（而不是 sql）创建一个包含所有家谱的 data.frame 对象。每棵树都以 "Cleach Marc" 这样的族长（没有父亲）开头。（使用 R 方法或 sqldf 方法）

Answer 1

您或许可以使用图形工具来完成此操作。所以使用 igraph，你可以使用 ego 函数获得邻居。

速写（需要检查！）

library(igraph)

family[] = lapply(family, factor, levels=unique(unlist(family)))

g = graph_from_adjacency_matrix(table(family))

cg = connect.neighborhood(g, order=length(V(g)), mode="out")

cbind( V(cg)$name, 
       sapply(ego(g, mode="out", mindist=1), function(x) replace(names(x), length(names(x))==0, NA)),
       ego_size(cg, mode="out") )[grep("Guillou", V(cg)$name),]

[,1]                   [,2]             [,3]
[1,] "Guillou Arthur"  NA               "1" 
[2,] "Guillou Eric"    "Guillou Arthur" "2" 
[3,] "Guillou Jacques" "Guillou Arthur" "2" 
[4,] "Guillou Jean"    "Guillou Eric"   "3" 
[5,] "Guillou Alan"    "Guillou Eric"   "3"

事实上，也许您不需要创建邻域图并且可以通过：

cbind( V(g)$name, 
       sapply(ego(g, mode="out", mindist=1), function(x) replace(names(x), length(names(x))==0, NA)),
       ego_size(g, mode="out", order=length(V(g))) )[grep("Cleach", V(g)$name),]

Answer 2

我们构建一个递归函数来获取父行，从那里一切都很简单。

首先我们用 stringsAsFactors = FALSE 定义数据，以便更顺利地重新格式化。

family <- data.frame(person, father,stringsAsFactors = FALSE)

函数

father_line <- function(x){
dad <- subset(family,person==x)$father
if(is.na(dad)) return(x)
c(x,father_line(dad))
}

father_line ("Guillou Alan")
# [1] "Guillou Alan"   "Guillou Eric"   "Guillou Arthur"

用它来获得等级和其他东西

family$father_line <- lapply(family$person,father_line)
family$level       <- lengths(family$father_line)
family$patriarch   <- sapply(family$father_line,tail,1)

#             person         father                                          father_line level      patriarch
# 1   Guillou Arthur           <NA>                                       Guillou Arthur     1 Guillou Arthur
# 2      Cleach Marc           <NA>                                          Cleach Marc     1    Cleach Marc
# 3     Guillou Eric Guillou Arthur                         Guillou Eric, Guillou Arthur     2 Guillou Arthur
# 4  Guillou Jacques Guillou Arthur                      Guillou Jacques, Guillou Arthur     2 Guillou Arthur
# 5    Cleach Franck    Cleach Marc                           Cleach Franck, Cleach Marc     2    Cleach Marc
# 6       Cleach Leo    Cleach Marc                              Cleach Leo, Cleach Marc     2    Cleach Marc
# 7    Cleach Herbet     Cleach Leo               Cleach Herbet, Cleach Leo, Cleach Marc     3    Cleach Marc
# 8     Cleach Adele  Cleach Herbet Cleach Adele, Cleach Herbet, Cleach Leo, Cleach Marc     4    Cleach Marc
# 9     Guillou Jean   Guillou Eric           Guillou Jean, Guillou Eric, Guillou Arthur     3 Guillou Arthur
# 10    Guillou Alan   Guillou Eric           Guillou Alan, Guillou Eric, Guillou Arthur     3 Guillou Arthur

例如，要获得规定的预期输出：

subset(family,patriarch == "Guillou Arthur",select=c(person,father,level))
#             person         father level
# 1   Guillou Arthur           <NA>     1
# 3     Guillou Eric Guillou Arthur     2
# 4  Guillou Jacques Guillou Arthur     2
# 9     Guillou Jean   Guillou Eric     3
# 10    Guillou Alan   Guillou Eric     3

tidyverse 看起来像这样：

library(tidyverse)
family %>%
  mutate(family_line = map(person,father_line),
         level = lengths(family_line),
         patriarch = map(family_line,last)) %>%
  filter(patriarch == "Guillou Arthur") %>%
  select(person,father,level)

#            person         father level
# 1  Guillou Arthur           <NA>     1
# 2    Guillou Eric Guillou Arthur     2
# 3 Guillou Jacques Guillou Arthur     2
# 4    Guillou Jean   Guillou Eric     3
# 5    Guillou Alan   Guillou Eric     3

R中的内置家庭嵌套树父/子关系

Built Family nested tree parent / children relationship in R

tree

recursion

r

igraph

sqldf