如何將串列串列作為列保存到R中的data.frame物件中

可重現的代碼：

    Meal <- function(name, ingredients, time_to_cook = NA, time_of_day = "Dinner", cuisine = NA) {
m <- list(name = name, ingredients = ingredients, time = time_to_cook, meal = time_of_day, cuisine = cuisine)
  m 
}

burritos <- Meal("Burrito", ingredients = c("Tortilla", "Chicken Breast", "Sour Cream", "Pepper", "Onion", "Chili Paste"), time_to_cook = "Fast", cuisine = "Mexican")
pizza <- Meal("Pizza", ingredients = c("Flour", "Tinned Tomato", "Mozarella", "Garlic"))

rbind(data.frame(pizza), data.frame(burritos))

      name    ingredients time   meal cuisine
1    Pizza          Flour <NA> Dinner    <NA>
2    Pizza  Tinned Tomato <NA> Dinner    <NA>
3    Pizza      Mozarella <NA> Dinner    <NA>
4    Pizza         Garlic <NA> Dinner    <NA>
5  Burrito       Tortilla Fast Dinner Mexican
6  Burrito Chicken Breast Fast Dinner Mexican
7  Burrito     Sour Cream Fast Dinner Mexican
8  Burrito         Pepper Fast Dinner Mexican
9  Burrito          Onion Fast Dinner Mexican
10 Burrito    Chili Paste Fast Dinner Mexican

我更喜歡的行為是，如果成分沒有轉換成長格式，即保持成分作為一個串列，而不是每個成分的一行。有什么辦法可以得到這種行為嗎？

我不想一直使用關系資料庫，想知道是否有一種簡單的方法來處理這個問題。

uj5u.com熱心網友回復：

小題大做

您可以將它們存盤為tibble非常擅長存盤list列的 a，只需修改Meal().

library(tibble)

Meal <- function(name, ingredients, time_to_cook = NA, time_of_day = "Dinner", cuisine = NA) {
  m <- tibble(name = name, ingredients = list(ingredients), time = time_to_cook, meal = time_of_day, cuisine = cuisine)
  m 
}

burritos <- Meal("Burrito", ingredients = c("Tortilla", "Chicken Breast", "Sour Cream", "Pepper", "Onion", "Chili Paste"), time_to_cook = "Fast", cuisine = "Mexican")
pizza <- Meal("Pizza", ingredients = c("Flour", "Tinned Tomato", "Mozarella", "Garlic"))

rbind(pizza, burritos)
#> # A tibble: 2 × 5
#>   name    ingredients time  meal   cuisine
#>   <chr>   <list>      <chr> <chr>  <chr>  
#> 1 Pizza   <chr [4]>   <NA>  Dinner <NA>   
#> 2 Burrito <chr [6]>   Fast  Dinner Mexican

從原始解決方案嵌套

或者，您可以完全保留原始解決方案，tibble然后通過使用nest將其ingredients轉回list.

library(tidyr)

rbind(data.frame(pizza), data.frame(burritos)) %>%
  nest(ingredients = ingredients)
#> # A tibble: 2 × 5
#>   name    time  meal   cuisine ingredients     
#>   <chr>   <chr> <chr>  <chr>   <list>          
#> 1 Pizza   <NA>  Dinner <NA>    <tibble [4 × 1]>
#> 2 Burrito Fast  Dinner Mexican <tibble [6 × 1]>

基數R

最后，如果你想堅持使用基礎 R，你可以I(list())在@Akrun 的評論中使用like。

Meal <- function(name, ingredients, time_to_cook = NA, time_of_day = "Dinner", cuisine = NA) {
  m <- data.frame(name = name, ingredients = I(list(ingredients)), time = time_to_cook, meal = time_of_day, cuisine = cuisine)
  m 
}

burritos <- Meal("Burrito", ingredients = c("Tortilla", "Chicken Breast", "Sour Cream", "Pepper", "Onion", "Chili Paste"), time_to_cook = "Fast", cuisine = "Mexican")
pizza <- Meal("Pizza", ingredients = c("Flour", "Tinned Tomato", "Mozarella", "Garlic"))
  
rbind(pizza, burritos)
#>      name  ingredients time   meal cuisine
#> 1   Pizza Flour, T.... <NA> Dinner    <NA>
#> 2 Burrito Tortilla.... Fast Dinner Mexican

uj5u.com熱心網友回復：

這是一種快速而骯臟的方法

Meal <- function(name, ingredients, time_to_cook = NA, time_of_day = "Dinner", cuisine = NA) {
  m <- data.frame(name = name, time = time_to_cook, meal = time_of_day, cuisine = cuisine)
  m$ingredients <- list(ingredients)
  m
}

burritos <- Meal("Burrito", ingredients = c("Tortilla", "Chicken Breast", "Sour Cream", "Pepper", "Onion", "Chili Paste"), time_to_cook = "Fast", cuisine = "Mexican")
pizza <- Meal("Pizza", ingredients = c("Flour", "Tinned Tomato", "Mozarella", "Garlic"))

基于這個答案

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