Question

我有以下tibble数据，

 h <- structure(list(label = list(list(structure(list(id = 431676528L, 
    url = "https://api.github.com/repos/emergenzeHack/terremotocentro/labels/per%20sviluppatori", 
    name = "per sviluppatori", color = "d4c5f9", default = FALSE), .Names = c("id", 
"url", "name", "color", "default")), structure(list(id = 442034204L, 
    url = "https://api.github.com/repos/emergenzeHack/terremotocentro/labels/sito%20principale", 
    name = "sito principale", color = "5319e7", default = FALSE), .Names = c("id", 
"url", "name", "color", "default"))), list(structure(list(id = 442051239L, 
    url = "https://api.github.com/repos/emergenzeHack/terremotocentro/labels/mappa", 
    name = "mappa", color = "0052cc", default = FALSE), .Names = c("id", 
"url", "name", "color", "default")), structure(list(id = 431676528L, 
    url = "https://api.github.com/repos/emergenzeHack/terremotocentro/labels/per%20sviluppatori", 
    name = "per sviluppatori", color = "d4c5f9", default = FALSE), .Names = c("id", 
"url", "name", "color", "default")), structure(list(id = 442034204L, 
    url = "https://api.github.com/repos/emergenzeHack/terremotocentro/labels/sito%20principale", 
    name = "sito principale", color = "5319e7", default = FALSE), .Names = c("id", 
"url", "name", "color", "default"))), list(NA_character_)), mainId = c("216226960", 
"215647494", "242390063")), .Names = c("label", "mainId"), row.names = c(NA, 
-3L), class = c("tbl_df", "tbl", "data.frame"))

我希望将标签中的值与mainId配对，以便我可以将labe l中的每个子元素与其主ID 相关联。我正在尝试使用标题为tibble：label，url，name，color，mainId。

a similar question的解决方案正常工作，除非NA

的子元素中嵌套label

map_df(h, flatten_dfr)

bind_rows_（x，.id）出错：参数1必须具有名称

Answer 1

您可以先删除遗失mainId的{{1}}，然后将其添加回label（如果full_join，则只需bind_rows s是独一无二的。）

mainId

Answer 2

这是一种方法，它将包含NA_character_的元素替换为NA列表，其名称与第一行的第一个元素相同。在此之后，bind_rows和unnest将正常工作。

library(tidyverse)

nested_names <- names(pluck(h, 'label', 1, 1))

h2 <- h %>% 
    mutate(label = map(label, map_if, 
                       ~is.null(names(.x)), 
                       ~setNames(rep(list(NA), length(nested_names)), 
                                 nested_names)), 
           label = map(label, bind_rows)) %>% 
    unnest()

h2
#> # A tibble: 6 x 6
#>   mainId           id url                           name     color default
#>   <chr>         <int> <chr>                         <chr>    <chr> <lgl>  
#> 1 216226960 431676528 https://api.github.com/repos… per svi… d4c5… FALSE  
#> 2 216226960 442034204 https://api.github.com/repos… sito pr… 5319… FALSE  
#> 3 215647494 442051239 https://api.github.com/repos… mappa    0052… FALSE  
#> 4 215647494 431676528 https://api.github.com/repos… per svi… d4c5… FALSE  
#> 5 215647494 442034204 https://api.github.com/repos… sito pr… 5319… FALSE  
#> 6 242390063        NA <NA>                          <NA>     <NA>  NA

用NAs压平dplyr中的tibble

2 个答案: