将tibble指定为tibble中的嵌套单元格

Question

如果有两只老虎，我想把一只老虎指定为另一只老虎的一个细胞。然而，当前的方法不断地将我的子Ti球转换成不同的格式，这会导致错误。

我知道我想做什么是可能的：

library(dplyr)

desired = tibble(
  row = 1:4,
  nested = list(
    tibble(text = c("a","b","c")),
    tibble(numbers = 1:4),
    tibble(label = c("high", "med", "low"), value = c(10, 5, 0)),
    NA
  )
)

这段代码产生了两列的tibble。第二列包含不同大小和内容的碎片。

但是，我不能用递增的步骤来构造它：

library(dplyr)

# outer tibble
df = tibble(row = 1:4)
# inner tibbles
t1 = tibble(text = c("a","b","c"))
t2 = tibble(numbers = 1:4)
t3 = tibble(label = c("high", "med", "low"), value = c(10, 5, 0))

# setup column of type tibble to assign into
df = df %>%
  mutate(nested = purrr::map(row, ~{ tibble() }))

# assign (fails)
df$nested[1] = t1 # no error but df$nested[1] is of type character not tibble
df$nested[2] = t2 # no error but df$nested[2] is of type int not tibble
df$nested[3] = t3 # gives warning and only one column of t3 is assigned

我收到的警告是：

df$nested3 3中的

= t3 :要替换的项数不是替换长度的倍数

这种方法的结果如下：

> df
# A tibble: 4 x 2
    row nested          
  <int> <list>          
1     1 <chr [3]>       
2     2 <int [4]>       
3     3 <chr [3]>       
4     4 <tibble [0 × 0]>

我如何在现有的(外部) tibble中将(内部的) tibbles分配给单个的单元格，从而最终得到嵌套的tibbles？

Answer 1

使用[[而不是[可以这样做：

library(dplyr, w = FALSE)

df <- tibble(row = 1:4, nested = vector("list", 4))

df$nested[[1]] <- t1
df$nested[[2]] <- t2
df$nested[[3]] <- t3
df$nested[[4]] <- NA

df
#> # A tibble: 4 × 2
#>     row nested          
#>   <int> <list>          
#> 1     1 <tibble [3 × 1]>
#> 2     2 <tibble [4 × 1]>
#> 3     3 <tibble [3 × 2]>
#> 4     4 <lgl [1]>