按组前缀旋转更长的时间

我需要按列字符串前缀分组更长的时间。下面的玩具示例有两个组“A”和“B”，但我需要一个针对任意数量的前缀组的通用 tidyverse 解决方案。

#toy df
set.seed(1)
df <- data.table(
  date = rep(seq(as.Date("2020-01-01"),as.Date("2020-01-05"),by="day"),each=6),
  k = rep(c("A.mean","A.median","A.min","B.mean","B.median","B.min"),5),
  v = runif(30,0,50)
  ) %>%
  pivot_wider(names_from = k, values_from = v)

df %>% head

  date       A.mean A.median  A.min B.mean B.median B.min
  <date>      <dbl>    <dbl>  <dbl>  <dbl>    <dbl> <dbl>
1 2020-01-01   13.3     18.6 28.6    45.4      10.1 44.9 
2 2020-01-02   47.2     33.0 31.5     3.09     10.3  8.83
3 2020-01-03   34.4     19.2 38.5    24.9      35.9 49.6 
4 2020-01-04   19.0     38.9 46.7    10.6      32.6  6.28
5 2020-01-05   13.4     19.3  0.670  19.1      43.5 17.0 

#pivot longer by group prefix
df %>%
  select(date,matches("A\\.")) %>%
  rename_with(~str_replace(.x,"A\\.","")) %>%
  mutate( k = "A") %>%
  bind_rows(
    df %>%
      select(date,matches("B\\.")) %>%
      rename_with(~str_replace(.x,"B\\.","")) %>%
      mutate( k = "B")
  )

   date        mean median    min k    
   <date>     <dbl>  <dbl>  <dbl> <chr>
 1 2020-01-01 13.3    18.6 28.6   A    
 2 2020-01-02 47.2    33.0 31.5   A    
 3 2020-01-03 34.4    19.2 38.5   A    
 4 2020-01-04 19.0    38.9 46.7   A    
 5 2020-01-05 13.4    19.3  0.670 A    
 6 2020-01-01 45.4    10.1 44.9   B    
 7 2020-01-02  3.09   10.3  8.83  B    
 8 2020-01-03 24.9    35.9 49.6   B    
 9 2020-01-04 10.6    32.6  6.28  B    
10 2020-01-05 19.1    43.5 17.0   B 

这是一个两步过程（出于演示目的，以两行显示）。首先旋转更长的时间来创建 k、统计名称和值的列，然后旋转更宽的位置来创建所需的结果。

下面编辑的代码，通过在“names_to”说明符中使用“.value”通配符选项，只需一步即可获得答案。

library(tidyr)
set.seed(1)
df <- data.frame(
   date = rep(seq(as.Date("2020-01-01"),as.Date("2020-01-05"),by="day"),each=6),
   k = rep(c("A.mean","A.median","A.min","B.mean","B.median","B.min"),5),
   v = runif(30,0,50)
) %>%
   pivot_wider(names_from = k, values_from = v)


#temp <- pivot_longer(df, -date, names_sep = "\\.", names_to = c("k", "stat"))
#answer <- pivot_wider(temp, id_cols = c("date", "k"), names_from= "stat", values_from="value")

#updated answer simplified down to just the pivot longer function
answer <- pivot_longer(df, -date, names_sep = "\\.", names_to = c("k", ".value"))

print(head(answer))
# A tibble: 6 x 5
date       k      mean median   min
<date>     <chr> <dbl>  <dbl> <dbl>
1 2020-01-01 A     13.3    18.6 28.6 
2 2020-01-01 B     45.4    10.1 44.9 
3 2020-01-02 A     47.2    33.0 31.5 
4 2020-01-02 B      3.09   10.3  8.83
5 2020-01-03 A     34.4    19.2 38.5 
6 2020-01-03 B     24.9    35.9 49.6