命名列中的项目，该列表由`nest`（tidyr）使用自定义函数创建
name items in column which list created by `nest` (tidyr)using a custom function
我有一个 tibble 我想 nest() 然后 unnest_wider()，同时还以 tibble 格式维护嵌套数据的副本。我知道这听起来不是很优雅，但这是目前针对我的用例的最佳解决方案。但是，当我使用 unnest_wider() 函数时，name_repair 会创建丑陋的 ...1、...2 等名称。如何使用某些 purrr 函数 (https://community.rstudio.com/t/how-to-handle-lack-of-names-with-unnest-wider/40496) 命名列表中的项目（它们的长度不同）？这样当我 unnest_wider() 时，列的名称就更好了。
我正在寻找的一个小例子：
library(tidyverse)
mpg %>%
select(manufacturer, model, cty ) %>%
group_by(manufacturer, model) %>%
nest() %>%
mutate(vars_in_tibble = data) %>%
ungroup() %>%
unnest_wider(data) %>%
unnest_wider(cty)
给予
# A tibble: 38 x 14
manufacturer model ...1 ...2 ...3 ...4 ...5 ...6 ...7 ...8 ...9 ...10 ...11 vars_in_tibble
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble[,1] [7 x 1]>
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble[,1] [8 x 1]>
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble[,1] [3 x 1]>
4 chevrolet c1500 suburban 2wd 14 11 14 13 12 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble[,1] [4 x 1]>
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble[,1] [11 x 1]>
9 dodge dakota pickup 4wd 15 14 13 14 14 14 9 11 11 NA NA <tibble[,1] [9 x 1]>
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble[,1] [7 x 1]>
但我想要
# A tibble: 38 x 14
manufacturer model car_1 car_2 car_3 car_4 car_5 car_6 car_7 car_8 car_9 car_10 car_11 vars_in_tibble
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble[,1] [7 x 1]>
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble[,1] [8 x 1]>
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble[,1] [3 x 1]>
4 chevrolet c1500 suburban 2wd 14 11 14 13 12 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble[,1] [4 x 1]>
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble[,1] [11 x 1]>
9 dodge dakota pickup 4wd 15 14 13 14 14 14 9 11 11 NA NA <tibble[,1] [9 x 1]>
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble[,1] [7 x 1]>
我认为您可以将最后一行添加到您的代码中：
library(stringr)
mpg %>%
select(manufacturer, model, cty ) %>%
group_by(manufacturer, model) %>%
nest() %>%
mutate(vars_in_tibble = data) %>%
ungroup() %>%
unnest_wider(data) %>%
unnest_wider(cty) %>%
rename_with(~ str_replace(., "\.\.\.", "car_"), contains("..."))
# A tibble: 38 x 14
manufacturer model car_1 car_2 car_3 car_4 car_5 car_6 car_7 car_8 car_9 car_10 car_11
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA
4 chevrolet c1500 suburb~ 14 11 14 13 12 NA NA NA NA NA NA
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA
6 chevrolet k1500 tahoe ~ 14 11 11 14 NA NA NA NA NA NA NA
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16
9 dodge dakota picku~ 15 14 13 14 14 14 9 11 11 NA NA
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA
# ... with 28 more rows, and 1 more variable: vars_in_tibble <list>
相同的输出不同的方法-
library(dplyr)
library(tidyr)
mpg %>%
select(manufacturer, model, cty) %>%
group_by(manufacturer, model) %>%
mutate(col = paste0('cty', row_number())) %>%
ungroup %>%
pivot_wider(names_from = col, values_from = cty) %>%
rowwise() %>%
mutate(vars_in_tibble = list(tibble(cty = na.omit(c_across(starts_with('cty')))))) %>%
ungroup
# manufacturer model cty1 cty2 cty3 cty4 cty5 cty6 cty7 cty8 cty9 cty10 cty11 vars_in_tibble
# <chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
# 1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble [7 × 1]>
# 2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble [8 × 1]>
# 3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble [3 × 1]>
# 4 chevrolet c1500 suburban … 14 11 14 13 12 NA NA NA NA NA NA <tibble [5 × 1]>
# 5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble [5 × 1]>
# 6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble [4 × 1]>
# 7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble [5 × 1]>
# 8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble [11 × 1…
# 9 dodge dakota pickup 4… 15 14 13 14 14 14 9 11 11 NA NA <tibble [9 × 1]>
#10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble [7 × 1]>
# … with 28 more rows
您可以像这样
使用purrr::map将嵌套列表转换为命名列表
mpg %>%
select(manufacturer, model, cty) %>%
group_by(manufacturer, model) %>%
nest() %>%
mutate(vars_in_tibble = data,
data = map(data, ~.x %>% mutate(id = paste0('cty_', row_number())) %>%
pivot_wider(names_from = id, values_from = cty)
)) %>%
ungroup() %>%
unnest(data)
# A tibble: 38 x 14
manufacturer model cty_1 cty_2 cty_3 cty_4 cty_5 cty_6 cty_7 cty_8 cty_9 cty_10 cty_11 vars_in_tibble
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble[,1] [7 x 1]>
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble[,1] [8 x 1]>
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble[,1] [3 x 1]>
4 chevrolet c1500 suburban 2wd 14 11 14 13 12 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble[,1] [4 x 1]>
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble[,1] [11 x 1~
9 dodge dakota pickup 4wd 15 14 13 14 14 14 9 11 11 NA NA <tibble[,1] [9 x 1]>
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble[,1] [7 x 1]>
# ... with 28 more rows
我有一个 tibble 我想 nest() 然后 unnest_wider()，同时还以 tibble 格式维护嵌套数据的副本。我知道这听起来不是很优雅，但这是目前针对我的用例的最佳解决方案。但是，当我使用 unnest_wider() 函数时，name_repair 会创建丑陋的 ...1、...2 等名称。如何使用某些 purrr 函数 (https://community.rstudio.com/t/how-to-handle-lack-of-names-with-unnest-wider/40496) 命名列表中的项目（它们的长度不同）？这样当我 unnest_wider() 时，列的名称就更好了。
我正在寻找的一个小例子：
library(tidyverse)
mpg %>%
select(manufacturer, model, cty ) %>%
group_by(manufacturer, model) %>%
nest() %>%
mutate(vars_in_tibble = data) %>%
ungroup() %>%
unnest_wider(data) %>%
unnest_wider(cty)
给予
# A tibble: 38 x 14
manufacturer model ...1 ...2 ...3 ...4 ...5 ...6 ...7 ...8 ...9 ...10 ...11 vars_in_tibble
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble[,1] [7 x 1]>
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble[,1] [8 x 1]>
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble[,1] [3 x 1]>
4 chevrolet c1500 suburban 2wd 14 11 14 13 12 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble[,1] [4 x 1]>
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble[,1] [11 x 1]>
9 dodge dakota pickup 4wd 15 14 13 14 14 14 9 11 11 NA NA <tibble[,1] [9 x 1]>
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble[,1] [7 x 1]>
但我想要
# A tibble: 38 x 14
manufacturer model car_1 car_2 car_3 car_4 car_5 car_6 car_7 car_8 car_9 car_10 car_11 vars_in_tibble
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble[,1] [7 x 1]>
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble[,1] [8 x 1]>
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble[,1] [3 x 1]>
4 chevrolet c1500 suburban 2wd 14 11 14 13 12 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble[,1] [4 x 1]>
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble[,1] [11 x 1]>
9 dodge dakota pickup 4wd 15 14 13 14 14 14 9 11 11 NA NA <tibble[,1] [9 x 1]>
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble[,1] [7 x 1]>
我认为您可以将最后一行添加到您的代码中：
library(stringr)
mpg %>%
select(manufacturer, model, cty ) %>%
group_by(manufacturer, model) %>%
nest() %>%
mutate(vars_in_tibble = data) %>%
ungroup() %>%
unnest_wider(data) %>%
unnest_wider(cty) %>%
rename_with(~ str_replace(., "\.\.\.", "car_"), contains("..."))
# A tibble: 38 x 14
manufacturer model car_1 car_2 car_3 car_4 car_5 car_6 car_7 car_8 car_9 car_10 car_11
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA
4 chevrolet c1500 suburb~ 14 11 14 13 12 NA NA NA NA NA NA
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA
6 chevrolet k1500 tahoe ~ 14 11 11 14 NA NA NA NA NA NA NA
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16
9 dodge dakota picku~ 15 14 13 14 14 14 9 11 11 NA NA
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA
# ... with 28 more rows, and 1 more variable: vars_in_tibble <list>
相同的输出不同的方法-
library(dplyr)
library(tidyr)
mpg %>%
select(manufacturer, model, cty) %>%
group_by(manufacturer, model) %>%
mutate(col = paste0('cty', row_number())) %>%
ungroup %>%
pivot_wider(names_from = col, values_from = cty) %>%
rowwise() %>%
mutate(vars_in_tibble = list(tibble(cty = na.omit(c_across(starts_with('cty')))))) %>%
ungroup
# manufacturer model cty1 cty2 cty3 cty4 cty5 cty6 cty7 cty8 cty9 cty10 cty11 vars_in_tibble
# <chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
# 1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble [7 × 1]>
# 2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble [8 × 1]>
# 3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble [3 × 1]>
# 4 chevrolet c1500 suburban … 14 11 14 13 12 NA NA NA NA NA NA <tibble [5 × 1]>
# 5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble [5 × 1]>
# 6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble [4 × 1]>
# 7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble [5 × 1]>
# 8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble [11 × 1…
# 9 dodge dakota pickup 4… 15 14 13 14 14 14 9 11 11 NA NA <tibble [9 × 1]>
#10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble [7 × 1]>
# … with 28 more rows
您可以像这样
使用purrr::map将嵌套列表转换为命名列表mpg %>%
select(manufacturer, model, cty) %>%
group_by(manufacturer, model) %>%
nest() %>%
mutate(vars_in_tibble = data,
data = map(data, ~.x %>% mutate(id = paste0('cty_', row_number())) %>%
pivot_wider(names_from = id, values_from = cty)
)) %>%
ungroup() %>%
unnest(data)
# A tibble: 38 x 14
manufacturer model cty_1 cty_2 cty_3 cty_4 cty_5 cty_6 cty_7 cty_8 cty_9 cty_10 cty_11 vars_in_tibble
<chr> <chr> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <list>
1 audi a4 18 21 20 21 16 18 18 NA NA NA NA <tibble[,1] [7 x 1]>
2 audi a4 quattro 18 16 20 19 15 17 17 15 NA NA NA <tibble[,1] [8 x 1]>
3 audi a6 quattro 15 17 16 NA NA NA NA NA NA NA NA <tibble[,1] [3 x 1]>
4 chevrolet c1500 suburban 2wd 14 11 14 13 12 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
5 chevrolet corvette 16 15 16 15 15 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
6 chevrolet k1500 tahoe 4wd 14 11 11 14 NA NA NA NA NA NA NA <tibble[,1] [4 x 1]>
7 chevrolet malibu 19 22 18 18 17 NA NA NA NA NA NA <tibble[,1] [5 x 1]>
8 dodge caravan 2wd 18 17 16 16 17 17 11 15 15 16 16 <tibble[,1] [11 x 1~
9 dodge dakota pickup 4wd 15 14 13 14 14 14 9 11 11 NA NA <tibble[,1] [9 x 1]>
10 dodge durango 4wd 13 13 9 13 11 13 11 NA NA NA NA <tibble[,1] [7 x 1]>
# ... with 28 more rows