考虑以下示例数据

library(tidyverse)
df <- tibble(group = c("a", "b", "b"), val = list(1:3, 4:6, 7:12))
## A tibble: 3 × 2
#  group val      
#  <chr> <list>   
#1 a     <int [3]>
#2 b     <int [3]>
#3 b     <int [6]>

我想合并group组合列val中的条目，以给出预期的输出

df_out <- tibble(group = c("a", "b"), val = list(1:3, 4:12))

我正在寻找一个tidyverse解决方案，但一直不成功。比如说

df %>% group_by(group) %>% summarise(val = map(val, c), .groups = "drop")

不会连接两个"b"行中的val项，而是生成警告

# A tibble: 3 × 2
  group val      
  <chr> <list>   
1 a     <int [3]>
2 b     <int [3]>
3 b     <int [6]>
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.

我理解这个警告，但我不明白 * 为什么 *“每个summarise()组超过1行”被返回。有人能解释一下并提供解决方案吗？
我希望有一个简单的两步group_by + summarise解决方案（即避免nest ing等）。
澄清一下：val中的数字不是按顺序排列的，也不是一个序列。一组不同的val数可能是c(1, 10, 2)和c(4, 7, 7)。预期的组合输出将是c(1, 10, 2, 4, 7, 7)。所以：

df <- tibble(group = c("a", "b", "b"), val = list(1:3, c(1, 10, 2), c(4, 7, 7)))
df_out <- tibble(group = c("a", "b"), val = list(1:3, c(1, 10, 2, 4, 7, 7)))