r - 如何使用 dplyr 来总结两个函数

Question

我尝试将此数据集作为示例进行总结，并尝试使用多个函数 n() 和 mean()。如何将两者结合在同一个工作流程中？

这是一个反射(reflect)我的较大数据的玩具数据集:

library(tidyverse)
df <- structure(list(group_var = c(70, 72, 73, 70, 70, 71, 70, 71, 
                                   71, 70), var1_scr = c(50.5, 25.75, 50.5, 50.5, 50.5, 50.5, 75.25, 
                                                         75.25, 50.5, 75.25), var2_scr = c(50.5, 50.5, NA, 75.25, 50.5, 
                                                                                           50.5, 75.25, 75.25, 100, 75.25), var3_scr = c(NA, NA, 75.25, 
                                                                                                                                         NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA, -10L), class = c("tbl_df", 
                                                                                                                                                                                                          "tbl", "data.frame"))
df
#> # A tibble: 10 x 4
#>    group_var var1_scr var2_scr var3_scr
#>        <dbl>    <dbl>    <dbl>    <dbl>
#>  1        70     50.5     50.5     NA  
#>  2        72     25.8     50.5     NA  
#>  3        73     50.5     NA       75.2
#>  4        70     50.5     75.2     NA  
#>  5        70     50.5     50.5     NA  
#>  6        71     50.5     50.5     NA  
#>  7        70     75.2     75.2     NA  
#>  8        71     75.2     75.2     NA  
#>  9        71     50.5    100       NA  
#> 10        70     75.2     75.2     NA

# summarize the scores
df %>%  group_by(group_var) %>% 
  summarise_at(vars(ends_with("_scr")), funs(mean(., na.rm = TRUE)))

#> # A tibble: 4 x 4
#>   group_var var1_scr var2_scr var3_scr
#>       <dbl>    <dbl>    <dbl>    <dbl>
#> 1        70     60.4     65.4    NaN  
#> 2        71     58.8     75.2    NaN  
#> 3        72     25.8     50.5    NaN  
#> 4        73     50.5    NaN       75.2

# count all the oberservations
df %>%  group_by(group_var) %>% 
  summarise(obs = n())
#> # A tibble: 4 x 2
#>   group_var   obs
#>       <dbl> <int>
#> 1        70     5
#> 2        71     3
#> 3        72     1
#> 4        73     1

# my goal is to produce this dataset but using the mutate_at function
df %>%  group_by(group_var) %>% 
  summarise(var1_scr = mean(var1_scr), 
            var2_scr = mean(var2_scr),
            var3_scr = mean(var3_scr), 
            obs = n())
#> # A tibble: 4 x 5
#>   group_var var1_scr var2_scr var3_scr   obs
#>       <dbl>    <dbl>    <dbl>    <dbl> <int>
#> 1        70     60.4     65.4     NA       5
#> 2        71     58.8     75.2     NA       3
#> 3        72     25.8     50.5     NA       1
#> 4        73     50.5     NA       75.2     1

^{由reprex package于2019年8月15日创建(v0.3.0)}

Answer 1

一个选项是在按“group_var”分组后在分组变量中添加“n”，然后执行summarise_at

library(dplyr)
df %>% 
     group_by(group_var)  %>%
     group_by(obs = n(), add = TRUE) %>%
     summarise_at(vars(ends_with("_scr")), list(~mean(., na.rm = TRUE)))
# A tibble: 4 x 5
# Groups:   group_var [4]
#  group_var   obs var1_scr var2_scr var3_scr
#      <dbl> <int>    <dbl>    <dbl>    <dbl>
#1        70     5     60.4     65.4    NaN  
#2        71     3     58.8     75.2    NaN  
#3        72     1     25.8     50.5    NaN  
#4        73     1     50.5    NaN       75.2

<小时/>

另一种选择是使用 mutate 创建频率列，并通过将其也包含在 summarise_at 中来获取平均值(例如 >平均值(rep(3, 5)) -> 3)

df %>% 
     group_by(group_var)  %>%
     mutate(obs = n()) %>%
     summarise_at(vars(ends_with("_scr"), obs), list(~mean(., na.rm = TRUE)))
# A tibble: 4 x 5
#  group_var var1_scr var2_scr var3_scr   obs
#      <dbl>    <dbl>    <dbl>    <dbl> <dbl>
#1        70     60.4     65.4    NaN       5
#2        71     58.8     75.2    NaN       3
#3        72     25.8     50.5    NaN       1
#4        73     50.5    NaN       75.2     1

注意:这两者都为“obs”提供一列

<小时/>

这里，OP 的预期输出是一个汇总输出，对于该输出，summarise/summarise_at/summarise_all/summarise_if 非常有效。但是，如果我们需要使用mutate_at(仅用于演示)

df %>% 
     group_by(group_var)  %>%
     mutate(obs = n()) %>%
     mutate_at(vars(ends_with("_scr"), obs), list(~mean(., na.rm = TRUE))) %>%
     distinct_at(vars(group_var, ends_with("_scr"), obs))
# A tibble: 4 x 5
# Groups:   group_var [4]
#  group_var var1_scr var2_scr var3_scr   obs
#      <dbl>    <dbl>    <dbl>    <dbl> <dbl>
#1        70     60.4     65.4    NaN       5
#2        72     25.8     50.5    NaN       1
#3        73     50.5    NaN       75.2     1
#4        71     58.8     75.2    NaN       3