R 按组前缀旋转较长的轴

我需要按列字符串前缀分组。下面的玩具示例有两个组A和B，但我需要一个通用的tidyverse解决方案，用于任何数量的前缀组

#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 

---

以下是两行显示的两步流程，用于演示。首先，长透视以创建k、stat name和value的列，然后宽透视以创建所需的结果

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")



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 

希望这能奏效：

df %>% pivot_longer(cols = contains(".")) %>% 
       mutate(k = substr(name,1,1), name = substr(name,3,nchar(name))) %>% 
       pivot_wider(names_from = name, values_from = value) %>% 
       arrange(k)

例如：

df
# A tibble: 5 x 7
#  date       A.mean A.median A.min B.mean B.median B.min
#  <date>      <dbl>    <dbl> <dbl>  <dbl>    <dbl> <dbl>
#1 2020-01-01 17.9       40.2 12.6    32.7   17.9    14.3
#2 2020-01-02 49.5       29.8 50.0    36.5    0.788  49.7
#3 2020-01-03  0.375     48.2 20.7    14.9   33.0    12.1
#4 2020-01-04  5.42      10.1 16.8    35.5   49.4    10.7
#5 2020-01-05 17.9       28.2  5.64   25.8   31.3    10.8


df %>% pivot_longer(cols = contains(".")) %>% 
       mutate(k = substr(name,1,1), name = substr(name,3,nchar(name))) %>% 
       pivot_wider(names_from = name, values_from = value) %>% 
       arrange(k)

# A tibble: 10 x 5
#  date       k       mean median   min
   <date>     <chr>  <dbl>  <dbl> <dbl>
# 1 2020-01-01 A     17.9   40.2   12.6 
# 2 2020-01-02 A     49.5   29.8   50.0 
# 3 2020-01-03 A      0.375 48.2   20.7 
# 4 2020-01-04 A      5.42  10.1   16.8 
# 5 2020-01-05 A     17.9   28.2    5.64
# 6 2020-01-01 B     32.7   17.9   14.3 
# 7 2020-01-02 B     36.5    0.788 49.7 
# 8 2020-01-03 B     14.9   33.0   12.1 
# 9 2020-01-04 B     35.5   49.4   10.7 
#10 2020-01-05 B     25.8   31.3   10.8 

---

这很有效。在我的实际例子中，.前面有可变的字符串长度，所以这更灵活，mutatek=str\u extractname，.*？=\\，name=str\u extractname，？