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r - 在 R 中建立家庭嵌套树父/子关系

转载 作者:行者123 更新时间:2023-12-02 08:12:21 24 4
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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 的示例

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”(没有父亲的顶级人物)的级别排序:

  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 进行递归查询来构建此表:

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)

大到长格式转换:

    library(tidyr)

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

long_family

递归查询查找“Guillou Arthur”(没有父亲的最高人物)的后代:

    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)创建包含所有家谱的 data.frame 对象。每棵树都以一个像“Cleach Marc”这样的族长(没有父亲)开始。 (用R方法或sqldf方法)

最佳答案

我们构建一个递归函数来获取父亲线,从那里一切都很容易。

首先,我们使用 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

关于r - 在 R 中建立家庭嵌套树父/子关系,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49607454/

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