将列转换为列表

Question

我在 2 列（Region 和 GroupedCateg）中有数据。请参阅下面的数据框。我想将其转换为嵌套列表。我尝试使用 dplyr 的 group_by 和 do() 函数，然后转换为列表，但它不起作用。

df <- read.table(header = T,
                 text = '
Region  GroupedCateg
Beja    Alentejo
Evora   Alentejo
Portalegre  Alentejo
Faro    Algarve
Aveiro  Central
"Castelo Branco"    Central
Coimbra Central
Leiria  Central
Santarem    Central
Acores  Islands
Madeira Islands
Lisboa  Lisbon
Setubal Lisbon
Braga   North
Braganca    North
"Viana do Castelo"  North
"Vila Real" North
Porto   Porto
')

列表中的所需输出。地区将在名称中。嵌套列表中的各个 GroupedCateg

list(
  list(
    name =  "Alentejo",
    categories = list("Beja", "Evora", "Portalegre")
  ),
  list(
    name =  "Algarve",
    categories = list("Faro")
  ),
  list(
    name =  "Central",
    categories = list("Aveiro", "Castelo Branco", "Coimbra", "Leiria", "Santarem" )
  ),
  
  list(
    name =  "North",
    categories = list("Braga", "Braganca", "Viana do Castelo", "Vila Real")
  ),
  list(
    name =  "Lisbon",
    categories = list("Lisboa", "Setubal")
  ),
  list(
    name =  "Islands",
    categories = list("Acores", "Madeira")
  ),
  
  list(
    name =  "Porto",
    categories = list("Porto")
  )
  
)

Answer 1

您可以在purrr 中使用pmap()。

library(dplyr)
library(purrr)

x <- df %>%
  group_by(GroupedCateg) %>% 
  summarise(Region = list(Region)) %>%
  pmap(~ list(name = .x, categories = as.list(.y)))

---

对应的baseR版本：

y <- apply(aggregate(Region ~ GroupedCateg, df, c),
     1, function(y) list(name = y[[1]], categories = as.list(y[[2]])))

---

all.equal(x, y)
# [1] TRUE

---

输出

[[1]]
[[1]]$name
[1] "Alentejo"

[[1]]$categories
[[1]]$categories[[1]]
[1] "Beja"

[[1]]$categories[[2]]
[1] "Evora"

[[1]]$categories[[3]]
[1] "Portalegre"



[[2]]
[[2]]$name
[1] "Algarve"

[[2]]$categories
[[2]]$categories[[1]]
[1] "Faro"



[[3]]
[[3]]$name
[1] "Central"

[[3]]$categories
[[3]]$categories[[1]]
[1] "Aveiro"

[[3]]$categories[[2]]
[1] "Castelo Branco"

[[3]]$categories[[3]]
[1] "Coimbra"

[[3]]$categories[[4]]
[1] "Leiria"

[[3]]$categories[[5]]
[1] "Santarem"



[[4]]
[[4]]$name
[1] "Islands"

[[4]]$categories
[[4]]$categories[[1]]
[1] "Acores"

[[4]]$categories[[2]]
[1] "Madeira"



[[5]]
[[5]]$name
[1] "Lisbon"

[[5]]$categories
[[5]]$categories[[1]]
[1] "Lisboa"

[[5]]$categories[[2]]
[1] "Setubal"



[[6]]
[[6]]$name
[1] "North"

[[6]]$categories
[[6]]$categories[[1]]
[1] "Braga"

[[6]]$categories[[2]]
[1] "Braganca"

[[6]]$categories[[3]]
[1] "Viana do Castelo"

[[6]]$categories[[4]]
[1] "Vila Real"



[[7]]
[[7]]$name
[1] "Porto"

[[7]]$categories
[[7]]$categories[[1]]
[1] "Porto"

Answer 2

在列上使用基础 split，然后 lapply 根据需要重新格式化：

x <- split(df$Region, df$GroupedCateg)

res <- lapply(names(x), function(i){
  list(name = i,
       categories = as.list(x[[ i ]]))
})

Answer 3

在 Base-R 中

apply(aggregate(Region~GroupedCateg,df,c),1, function(x) list(name=x[1], category=as.list(x[2]$Region))) 
    
[[1]]
[[1]]$name
[[1]]$name$GroupedCateg
[1] "Alentejo"


[[1]]$category
[[1]]$category[[1]]
[1] "Beja"

[[1]]$category[[2]]
[1] "Evora"

[[1]]$category[[3]]
[1] "Portalegre"



[[2]]
[[2]]$name
[[2]]$name$GroupedCateg
[1] "Algarve"


[[2]]$category
[[2]]$category[[1]]
[1] "Faro"



[[3]]
[[3]]$name
[[3]]$name$GroupedCateg
[1] "Central"


[[3]]$category
[[3]]$category[[1]]
[1] "Aveiro"

[[3]]$category[[2]]
[1] "Castelo Branco"

[[3]]$category[[3]]
[1] "Coimbra"

[[3]]$category[[4]]
[1] "Leiria"

[[3]]$category[[5]]
[1] "Santarem"



[[4]]
[[4]]$name
[[4]]$name$GroupedCateg
[1] "Islands"


[[4]]$category
[[4]]$category[[1]]
[1] "Acores"

[[4]]$category[[2]]
[1] "Madeira"



[[5]]
[[5]]$name
[[5]]$name$GroupedCateg
[1] "Lisbon"


[[5]]$category
[[5]]$category[[1]]
[1] "Lisboa"

[[5]]$category[[2]]
[1] "Setubal"



[[6]]
[[6]]$name
[[6]]$name$GroupedCateg
[1] "North"


[[6]]$category
[[6]]$category[[1]]
[1] "Braga"

[[6]]$category[[2]]
[1] "Braganca"

[[6]]$category[[3]]
[1] "Viana do Castelo"

[[6]]$category[[4]]
[1] "Vila Real"



[[7]]
[[7]]$name
[[7]]$name$GroupedCateg
[1] "Porto"


[[7]]$category
[[7]]$category[[1]]
[1] "Porto"