r - 我如何使用 pivot wider 按组进行汇总

Question

df <- data.frame( plant1= c('1','0','1','0','0','1','0','0','1'),
                  plant2= c('0','1','0','1','1','0','1','1','0'),
                  Public.1= c('1','0','1','0','0','0','0','0','1'),
                  Private.1= c('0','0','0','0','0','1','0','0','0'),
                  Public.2= c('0','0','0','1','1','0','0','1','0'),
                  Private.2= c('0','1','0','0','0','0','1','0','0'))
df

我如何根据 Plant1 和 2 使用 pivot wider 来总结 Public Private？逻辑背后:每个 Plant 元素只能注册 Public 或 Privateplant 1和Public.1 Private.1相关

预期输出:

 plant1 plant2 Public Private 
1      1      0      1       0
2      0      1      0       1      
3      1      0      1       0
4      0      1      1       0
5      0      1      1       0
6      1      0      0       1
7      0      1      0       1
8      0      1      1       0 
9      1      0      1       0

Answer 1

我们可以使用names_sep

library(tidyr)
library(dplyr)
pivot_longer(df, 
           cols = matches('Public|Private'), 
      names_to = c(".value", 'grp'), names_sep ="\\.") %>%
   select(-grp)

-输出

# A tibble: 18 x 4
   plant1 plant2 Public Private
   <chr>  <chr>  <chr>  <chr>  
 1 1      0      1      0      
 2 1      0      0      0      
 3 0      1      0      0      
 4 0      1      0      1      
 5 1      0      1      0      
 6 1      0      0      0      
 7 0      1      0      0      
 8 0      1      1      0      
 9 0      1      0      0      
10 0      1      1      0      
11 1      0      0      1      
12 1      0      0      0      
13 0      1      0      0      
14 0      1      0      1      
15 0      1      0      0      
16 0      1      1      0      
17 1      0      1      0      
18 1      0      0      0