r语言 - 为什么在列表列上使用c()不能与dplyr summarize一起工作?

Question

我有一个列表列，我想为每个组使用c()，以将这些列表组合在summarize中。这应该导致每组一行，但事实并非如此(注意代码是使用dplyr>= 1.1.0编写的):

library(dplyr)
df <- tibble::tibble(group = c("A", "A", "B"),
list_col = list(list("One"), list("Two"), list("Three")))
df |> 
summarize(list_col = c(list_col),
.by = group)

这回报:

group list_col  
<chr> <list>    
1 A     <list [1]>
2 A     <list [1]>
3 B     <list [1]>
Warning message:
Returning more (or less) than 1 row per `summarise()` group was deprecated in dplyr 1.1.0.
i Please use `reframe()` instead.
i When switching from `summarise()` to `reframe()`, remember that `reframe()` always
returns an ungrouped data frame and adjust accordingly.
Call `lifecycle::last_lifecycle_warnings()` to see where this warning was generated. 

预期输出

output <- tibble::tibble(group = c("A", "B"),
list_col = list(list("One", "Two"), list("Three")))
group list_col  
<chr> <list>    
1 A     <list [2]>
2 B     <list [1]>
output$list_col[[1]]
[[1]]
[1] "One"
[[2]]
[1] "Two"

替代解决方案

您可以执行如下代码。然而，A)它改变了列的逐行类型和B)我想具体知道为什么c()不起作用:

df |>
summarize(list_col = list(unlist(list_col)),
.by = group)
group list_col 
<chr> <list>   
1 A     <chr [2]>
2 B     <chr [1]>

---

在第一组(A)中，我预计会发生以下事情，将两个列表合并为一个列表:

c(list("One"), list("Two"))
[[1]]
[1] "One"
[[2]]
[1] "Two"

那么，为什么这不起作用呢?这是一个错误，还是有什么与语法我失踪?

Answer 1

library(dplyr)
out <- df %>% 
reframe(list_col = list(as.list(unlist(list_col))), .by = group)

与产出

> out
# A tibble: 2 × 2
group list_col  
<chr> <list>    
1 A     <list [2]>
2 B     <list [1]>
> out$list_col[[1]]
[[1]]
[1] "One"
[[2]]
[1] "Two"

op的预期

> output$list_col[[1]]
[[1]]
[1] "One"
[[2]]
[1] "Two"

---

考虑到c和unlist的不同，recursive的默认参数为FALSE/TRUE

c(…， recursive = FALSE, use.names = TRUE)

unlist(x, recursive = TRUE, use.names = TRUE)

ie。基本的区别是

> c(list("a"))
[[1]]
[1] "a"
> unlist(list("a"))
[1] "a"

对于多于两个元素的list,...可变参数长度为1，因为它是传递给c的单个列表。

> c(list("a", "b"))
[[1]]
[1] "a"
[[2]]
[1] "b"

c不做任何事情，除非我们将它与do.call一起使用，其中list的每个元素都作为单独的参数传递

> do.call(c, list("a", "b"))
[1] "a" "b"

用OP的例子

> df$list_col[1:2]
[[1]]
[[1]][[1]]
[1] "One"

[[2]]
[[2]][[1]]
[1] "Two"
> c(df$list_col[1:2])
[[1]]
[[1]][[1]]
[1] "One"

[[2]]
[[2]][[1]]
[1] "Two"
> do.call(c, df$list_col[1:2])
[[1]]
[1] "One"
[[2]]
[1] "Two"

。如果我们这样做

out2 <- df %>% 
reframe(list_col = list(do.call(c, list_col)), .by = group)

与产出

> out2$list_col[[1]]
[[1]]
[1] "One"
[[2]]
[1] "Two"