将嵌套列表转换为数据框：仅提取感兴趣的特定元素
Convert nested list to dataframe: extract only specific elements of interest
类似的题我看过很多，但总不能适应我的情况。我有嵌套列表形式的数据，想以某种方式将其转换为数据框。
my_data_object <-
list(my_variables = list(
age = list(
type = "numeric",
originType = "slider",
originSettings = structure(list(), .Names = character(0)),
originIndex = 5L,
title = "what is your age?",
valueDescriptions = NULL
),
med_field = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 6L,
title = "what medical branch are you at?",
valueDescriptions = list(card = "Cardiology", ophth = "Ophthalmology",
derm = "Dermatology")
),
covid_vaccine = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 8L,
title = "when do you plan to get vaccinated?",
valueDescriptions = list(
next_mo = "No later than next month",
within_six_mo = "No later than six months from now",
never = "I will not get vaccinated"
)
)
))
期望的输出
var_name type originType title
<chr> <chr> <chr> <chr>
1 age numeric slider what is your age?
2 med_field string choice what medical branch are you at?
3 covid_vaccine string choice when do you plan to get vaccinated?
我的失败尝试
library(tibble)
library(tidyr)
my_data_object %>%
enframe() %>%
unnest_longer(value) %>%
unnest(value)
## # A tibble: 18 x 3
## name value value_id
## <chr> <named list> <chr>
## 1 my_variables <chr [1]> age
## 2 my_variables <chr [1]> age
## 3 my_variables <named list [0]> age
## 4 my_variables <int [1]> age
## 5 my_variables <chr [1]> age
## 6 my_variables <NULL> age
## 7 my_variables <chr [1]> med_field
## 8 my_variables <chr [1]> med_field
## 9 my_variables <named list [0]> med_field
## 10 my_variables <int [1]> med_field
## 11 my_variables <chr [1]> med_field
## 12 my_variables <named list [3]> med_field
## 13 my_variables <chr [1]> covid_vaccine
## 14 my_variables <chr [1]> covid_vaccine
## 15 my_variables <named list [0]> covid_vaccine
## 16 my_variables <int [1]> covid_vaccine
## 17 my_variables <chr [1]> covid_vaccine
## 18 my_variables <named list [3]> covid_vaccine
我正在尝试使用 tidyverse 函数来获取它，但到目前为止，我似乎没有朝着正确的方向前进。多谢指导
编辑
与我最初提供的示例数据不同，实际上我的数据的层次结构有点不同。我认为一旦我掌握了方法，这将很容易概括，但事实并非如此。因此，如果我们考虑数据如下，但实际上我只关心 my_variables 子列表。
my_data_object_2 <-
list(
other_variables = list(
whatever_var_1 = list(
type = "numeric",
originType = "slider",
originSettings = structure(list(), .Names = character(0)),
originIndex = 5L,
title = "blah question",
valueDescriptions = NULL
)
),
my_variables = list(
age = list(
type = "numeric",
originType = "slider",
originSettings = structure(list(), .Names = character(0)),
originIndex = 5L,
title = "what is your age?",
valueDescriptions = NULL
),
med_field = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 6L,
title = "what medical branch are you at?",
valueDescriptions = list(card = "Cardiology", ophth = "Ophthalmology",
derm = "Dermatology")
),
covid_vaccine = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 8L,
title = "when do you plan to get vaccinated?",
valueDescriptions = list(
next_mo = "No later than next month",
within_six_mo = "No later than six months from now",
never = "I will not get vaccinated"
)
)
)
)
那么我如何“放大”/“提取”my_variables 和 只有这样 才能得到我在上面的“期望输出”中指定的 table ？
您可以flatten对象，使用enframe和unnest_wider创建新列。
library(tidyverse)
my_data_object %>%
flatten() %>%
tibble::enframe() %>%
unnest_wider(value)
# name type originType originIndex title valueDescriptions
# <chr> <chr> <chr> <int> <chr> <list>
#1 age numeric slider 5 what is your age? <NULL>
#2 med_field string choice 6 what medical branch are you at? <named list [3]>
#3 covid_vaccine string choice 8 when do you plan to get vaccinated? <named list [3]>
然后您可以删除不需要的列。
仅使用 my_data_object_2$my_variables :
my_data_object_2$my_variables %>%
tibble::enframe() %>%
unnest_wider(value)
遍历 my_data_object tiblifying 指定的列并使用 map_dfr 将它们放在一起（或者 fun(my_data_object$my_variables) 就足够了，这取决于一般情况）。示例数据中没有缺失字段，但如果 3 个规范字段中的任何一个缺失，则将 .default = NA 作为 lcol_chr 参数添加到该字段规范。
library(purrr)
library(tibblify)
spec <- lcols(
lcol_chr("type"),
lcol_chr("originType"),
lcol_chr("title")
)
fun <- function(x) cbind(var_name = names(x), tibblify(x, spec))
map_dfr(my_data_object, fun)
给予：
var_name type originType title
1 age numeric slider what is your age?
2 med_field string choice what medical branch are you at?
3 covid_vaccine string choice when do you plan to get vaccinated?
根据一般情况，@mgirlich 进行的这种简化（类似于此答案介绍中的备选方案）可能会奏效。 spec来自上方。
library(tibblify)
cbind(
var_name = names(my_data_object[[1]]),
tibblify(my_data_object[[1]], spec)
)
像往常一样对 select 特定列使用 lapply，只是 rbind 它们。
res <- do.call(rbind.data.frame,
lapply((my_data_object)[[1]], `[`, c("type", "originType", "title")))
res
# type originType title
# age numeric slider what is your age?
# med_field string choice what medical branch are you at?
# covid_vaccine string choice when do you plan to get vaccinated?
如果您希望行名位于第一列，请执行：
`rownames<-`(cbind(var=rownames(res), res), NULL)
# var type originType title
# 1 age numeric slider what is your age?
# 2 med_field string choice what medical branch are you at?
# 3 covid_vaccine string choice when do you plan to get vaccinated?
类似的题我看过很多，但总不能适应我的情况。我有嵌套列表形式的数据，想以某种方式将其转换为数据框。
my_data_object <-
list(my_variables = list(
age = list(
type = "numeric",
originType = "slider",
originSettings = structure(list(), .Names = character(0)),
originIndex = 5L,
title = "what is your age?",
valueDescriptions = NULL
),
med_field = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 6L,
title = "what medical branch are you at?",
valueDescriptions = list(card = "Cardiology", ophth = "Ophthalmology",
derm = "Dermatology")
),
covid_vaccine = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 8L,
title = "when do you plan to get vaccinated?",
valueDescriptions = list(
next_mo = "No later than next month",
within_six_mo = "No later than six months from now",
never = "I will not get vaccinated"
)
)
))
期望的输出
var_name type originType title
<chr> <chr> <chr> <chr>
1 age numeric slider what is your age?
2 med_field string choice what medical branch are you at?
3 covid_vaccine string choice when do you plan to get vaccinated?
我的失败尝试
library(tibble)
library(tidyr)
my_data_object %>%
enframe() %>%
unnest_longer(value) %>%
unnest(value)
## # A tibble: 18 x 3
## name value value_id
## <chr> <named list> <chr>
## 1 my_variables <chr [1]> age
## 2 my_variables <chr [1]> age
## 3 my_variables <named list [0]> age
## 4 my_variables <int [1]> age
## 5 my_variables <chr [1]> age
## 6 my_variables <NULL> age
## 7 my_variables <chr [1]> med_field
## 8 my_variables <chr [1]> med_field
## 9 my_variables <named list [0]> med_field
## 10 my_variables <int [1]> med_field
## 11 my_variables <chr [1]> med_field
## 12 my_variables <named list [3]> med_field
## 13 my_variables <chr [1]> covid_vaccine
## 14 my_variables <chr [1]> covid_vaccine
## 15 my_variables <named list [0]> covid_vaccine
## 16 my_variables <int [1]> covid_vaccine
## 17 my_variables <chr [1]> covid_vaccine
## 18 my_variables <named list [3]> covid_vaccine
我正在尝试使用 tidyverse 函数来获取它，但到目前为止，我似乎没有朝着正确的方向前进。多谢指导
编辑
与我最初提供的示例数据不同，实际上我的数据的层次结构有点不同。我认为一旦我掌握了方法，这将很容易概括，但事实并非如此。因此，如果我们考虑数据如下，但实际上我只关心 my_variables 子列表。
my_data_object_2 <-
list(
other_variables = list(
whatever_var_1 = list(
type = "numeric",
originType = "slider",
originSettings = structure(list(), .Names = character(0)),
originIndex = 5L,
title = "blah question",
valueDescriptions = NULL
)
),
my_variables = list(
age = list(
type = "numeric",
originType = "slider",
originSettings = structure(list(), .Names = character(0)),
originIndex = 5L,
title = "what is your age?",
valueDescriptions = NULL
),
med_field = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 6L,
title = "what medical branch are you at?",
valueDescriptions = list(card = "Cardiology", ophth = "Ophthalmology",
derm = "Dermatology")
),
covid_vaccine = list(
type = "string",
originType = "choice",
originSettings = structure(list(), .Names = character(0)),
originIndex = 8L,
title = "when do you plan to get vaccinated?",
valueDescriptions = list(
next_mo = "No later than next month",
within_six_mo = "No later than six months from now",
never = "I will not get vaccinated"
)
)
)
)
那么我如何“放大”/“提取”my_variables 和 只有这样 才能得到我在上面的“期望输出”中指定的 table ？
您可以flatten对象，使用enframe和unnest_wider创建新列。
library(tidyverse)
my_data_object %>%
flatten() %>%
tibble::enframe() %>%
unnest_wider(value)
# name type originType originIndex title valueDescriptions
# <chr> <chr> <chr> <int> <chr> <list>
#1 age numeric slider 5 what is your age? <NULL>
#2 med_field string choice 6 what medical branch are you at? <named list [3]>
#3 covid_vaccine string choice 8 when do you plan to get vaccinated? <named list [3]>
然后您可以删除不需要的列。
仅使用 my_data_object_2$my_variables :
my_data_object_2$my_variables %>%
tibble::enframe() %>%
unnest_wider(value)
遍历 my_data_object tiblifying 指定的列并使用 map_dfr 将它们放在一起（或者 fun(my_data_object$my_variables) 就足够了，这取决于一般情况）。示例数据中没有缺失字段，但如果 3 个规范字段中的任何一个缺失，则将 .default = NA 作为 lcol_chr 参数添加到该字段规范。
library(purrr)
library(tibblify)
spec <- lcols(
lcol_chr("type"),
lcol_chr("originType"),
lcol_chr("title")
)
fun <- function(x) cbind(var_name = names(x), tibblify(x, spec))
map_dfr(my_data_object, fun)
给予：
var_name type originType title
1 age numeric slider what is your age?
2 med_field string choice what medical branch are you at?
3 covid_vaccine string choice when do you plan to get vaccinated?
根据一般情况，@mgirlich 进行的这种简化（类似于此答案介绍中的备选方案）可能会奏效。 spec来自上方。
library(tibblify)
cbind(
var_name = names(my_data_object[[1]]),
tibblify(my_data_object[[1]], spec)
)
像往常一样对 select 特定列使用 lapply，只是 rbind 它们。
res <- do.call(rbind.data.frame,
lapply((my_data_object)[[1]], `[`, c("type", "originType", "title")))
res
# type originType title
# age numeric slider what is your age?
# med_field string choice what medical branch are you at?
# covid_vaccine string choice when do you plan to get vaccinated?
如果您希望行名位于第一列，请执行：
`rownames<-`(cbind(var=rownames(res), res), NULL)
# var type originType title
# 1 age numeric slider what is your age?
# 2 med_field string choice what medical branch are you at?
# 3 covid_vaccine string choice when do you plan to get vaccinated?