什么是按多列拆分df的** tidyverse **方法？

时间：2017-03-09 20:36:02

标签： r tidyverse purrr

我想将数据框拆分多个列，以便我可以看到每个数据子集的summary()输出。

以下是使用split()中的base执行此操作的方法： 

library(tidyverse)
#> Loading tidyverse: ggplot2
#> Loading tidyverse: tibble
#> Loading tidyverse: tidyr
#> Loading tidyverse: readr
#> Loading tidyverse: purrr
#> Loading tidyverse: dplyr
#> Conflicts with tidy packages ----------------------------------------------
#> filter(): dplyr, stats
#> lag():    dplyr, stats

mtcars %>% 
  select(1:3) %>% 
  mutate(GRP_A = sample(LETTERS[1:2], n(), replace = TRUE),
         GRP_B = sample(c(1:2), n(), replace = TRUE)) %>% 
  split(list(.$GRP_A, .$GRP_B)) %>% 
  map(summary)
#> $A.1
#>       mpg             cyl           disp          GRP_A          
#>  Min.   :10.40   Min.   :4.0   Min.   :108.0   Length:10         
#>  1st Qu.:14.97   1st Qu.:4.5   1st Qu.:151.9   Class :character  
#>  Median :18.50   Median :7.0   Median :259.3   Mode  :character  
#>  Mean   :17.61   Mean   :6.4   Mean   :283.4                     
#>  3rd Qu.:20.85   3rd Qu.:8.0   3rd Qu.:430.0                     
#>  Max.   :24.40   Max.   :8.0   Max.   :472.0                     
#>      GRP_B  
#>  Min.   :1  
#>  1st Qu.:1  
#>  Median :1  
#>  Mean   :1  
#>  3rd Qu.:1  
#>  Max.   :1  
#> 
#> $B.1
#>       mpg             cyl           disp          GRP_A          
#>  Min.   :15.00   Min.   :4.0   Min.   : 75.7   Length:5          
#>  1st Qu.:21.00   1st Qu.:4.0   1st Qu.: 78.7   Class :character  
#>  Median :21.50   Median :4.0   Median :120.1   Mode  :character  
#>  Mean   :24.06   Mean   :5.2   Mean   :147.1                     
#>  3rd Qu.:30.40   3rd Qu.:6.0   3rd Qu.:160.0                     
#>  Max.   :32.40   Max.   :8.0   Max.   :301.0                     
#>      GRP_B  
#>  Min.   :1  
#>  1st Qu.:1  
#>  Median :1  
#>  Mean   :1  
#>  3rd Qu.:1  
#>  Max.   :1  
#> 
#> $A.2
#>       mpg             cyl             disp          GRP_A          
#>  Min.   :15.20   Min.   :4.000   Min.   : 95.1   Length:9          
#>  1st Qu.:16.40   1st Qu.:6.000   1st Qu.:160.0   Class :character  
#>  Median :18.10   Median :8.000   Median :275.8   Mode  :character  
#>  Mean   :19.84   Mean   :6.667   Mean   :234.0                     
#>  3rd Qu.:21.00   3rd Qu.:8.000   3rd Qu.:275.8                     
#>  Max.   :30.40   Max.   :8.000   Max.   :360.0                     
#>      GRP_B  
#>  Min.   :2  
#>  1st Qu.:2  
#>  Median :2  
#>  Mean   :2  
#>  3rd Qu.:2  
#>  Max.   :2  
#> 
#> $B.2
#>       mpg             cyl         disp          GRP_A          
#>  Min.   :13.30   Min.   :4   Min.   : 71.1   Length:8          
#>  1st Qu.:14.97   1st Qu.:4   1st Qu.:125.3   Class :character  
#>  Median :20.55   Median :6   Median :201.5   Mode  :character  
#>  Mean   :20.99   Mean   :6   Mean   :213.5                     
#>  3rd Qu.:23.93   3rd Qu.:8   3rd Qu.:315.5                     
#>  Max.   :33.90   Max.   :8   Max.   :360.0                     
#>      GRP_B  
#>  Min.   :2  
#>  1st Qu.:2  
#>  Median :2  
#>  Mean   :2  
#>  3rd Qu.:2  
#>  Max.   :2

如何使用tidyverse动词获得相同的结果？我最初的想法是使用purrr::by_slice()，但显然已被弃用。

2 个答案:

答案 0 :(得分：2)

＆＃34; tidy＆＃34;解决方案似乎是＆＃34; mutate + list-cols + purrr＆＃34;的组合。 according to Hadley 


library(tidyverse) 
library(magrittr) 

# group, nest, create a new col leveraging purrr::map()
mt_summary <- 
    mtcars %>% 
    select(1:3) %>% 
    mutate(GRP_A = sample(LETTERS[1:2],  n(), replace = TRUE), 
           GRP_B = sample(c(1:2), n(), replace = TRUE)) %>% 
    group_by(GRP_A, GRP_B) %>% 
    nest() %>% 
    mutate(SUMMARY = map(data, .f = summary))

# check the structure
mt_summary
#> # A tibble: 4 × 4
#>   GRP_A GRP_B              data     SUMMARY
#>   <chr> <int>            <list>      <list>
#> 1     A     1 <tibble [11 × 3]> <S3: table>
#> 2     B     2  <tibble [9 × 3]> <S3: table>
#> 3     A     2  <tibble [7 × 3]> <S3: table>
#> 4     B     1  <tibble [5 × 3]> <S3: table>

# extract the summaries
extract2(mt_summary, "SUMMARY") %>% 
    set_names(paste0(extract2(mt_summary, "GRP_A"), 
                     extract2(mt_summary, "GRP_B")))
#> $A1
#>       mpg             cyl             disp      
#>  Min.   :10.40   Min.   :4.000   Min.   : 75.7  
#>  1st Qu.:15.25   1st Qu.:4.000   1st Qu.:120.9  
#>  Median :19.20   Median :6.000   Median :167.6  
#>  Mean   :20.43   Mean   :6.182   Mean   :229.0  
#>  3rd Qu.:25.85   3rd Qu.:8.000   3rd Qu.:309.5  
#>  Max.   :30.40   Max.   :8.000   Max.   :460.0  
#> 
#> $B2
#>       mpg             cyl             disp      
#>  Min.   :15.20   Min.   :4.000   Min.   : 78.7  
#>  1st Qu.:17.80   1st Qu.:4.000   1st Qu.:120.3  
#>  Median :19.20   Median :6.000   Median :167.6  
#>  Mean   :20.84   Mean   :6.222   Mean   :225.9  
#>  3rd Qu.:21.50   3rd Qu.:8.000   3rd Qu.:351.0  
#>  Max.   :32.40   Max.   :8.000   Max.   :400.0  
#> 
#> $A2
#>       mpg             cyl             disp      
#>  Min.   :15.20   Min.   :4.000   Min.   : 71.1  
#>  1st Qu.:18.90   1st Qu.:4.000   1st Qu.:114.5  
#>  Median :21.40   Median :6.000   Median :145.0  
#>  Mean   :21.79   Mean   :5.429   Mean   :176.0  
#>  3rd Qu.:22.10   3rd Qu.:6.000   3rd Qu.:241.5  
#>  Max.   :33.90   Max.   :8.000   Max.   :304.0  
#> 
#> $B1
#>       mpg             cyl           disp      
#>  Min.   :10.40   Min.   :4.0   Min.   :140.8  
#>  1st Qu.:13.30   1st Qu.:8.0   1st Qu.:275.8  
#>  Median :14.30   Median :8.0   Median :350.0  
#>  Mean   :15.62   Mean   :7.2   Mean   :319.7  
#>  3rd Qu.:17.30   3rd Qu.:8.0   3rd Qu.:360.0  
#>  Max.   :22.80   Max.   :8.0   Max.   :472.0

答案 1 :(得分：1)

dplyr 0.8.0引入了您要搜索的动词：group_split()

从文档中：

  

group_split（）的工作方式类似于base::split()，但

     
      

  • 它使用group_by（）中的分组结构，因此要遵守数据掩码

    •   

  • 它不会基于分组来命名列表的元素，因为这通常会丢失信息并且令人困惑。

    •   
     

group_keys()通过返回数据来说明分组结构   框架，每组一行，每组变量一行一列。

例如：

mtcars %>% 
  select(1:3) %>% 
  mutate(GRP_A = sample(LETTERS[1:2], n(), replace = TRUE),
         GRP_B = sample(c(1:2), n(), replace = TRUE)) %>% 
  group_split(GRP_A, GRP_B) %>% 
  map(summary)
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