如何按组进行汇总并使用R中的dplyr获取总体数据集的摘要

Posted 2023-02-28

[我想计算不同组的摘要，并同时为整个（未分组的）数据集计算摘要，最好使用dplyr（或非常适合dplyr管道的东西）。

期望的结果可以通过分别计算组摘要，然后是总体摘要，然后合并结果来实现。但是，这似乎效率不高，我希望有一个更简单的解决方案，它需要更少的代码重复。我在文档中或其他问题中都没有找到与此相关的任何内容。

可复制数据：

library(tidyverse)
set.seed(500)
dat <- 
    data_frame(treatment = sample(c("Group1", "Group2", "Group3"), 100, replace = TRUE),
               recruitment_strategy = sample(c("Strategy 1", "Strategy 2", "Strategy 3", "Strategy 4", "Strategy 5"), 100, replace = TRUE),
               Variable_A = rnorm(100),
               Variable_B = rnorm(100),
               Variable_C = rnorm(100))

用于按组和整体数据集中的均值计算多个变量均值的代码：

count_by_group <- dat %>% 
    group_by(treatment) %>% 
    count(recruitment_strategy) %>%
    mutate(`n (%)` = paste0(n, " (", round(n / sum(n)*100,0), "%)")) %>%
    select(-n) %>%
    spread(treatment, `n (%)`)

count_overall <- dat %>% 
    count(recruitment_strategy) %>%
    mutate(`n (%)` = paste0(n, " (", round(n / sum(n)*100,0), "%)")) %>%
    select(-n) %>%
    rename(Overall_dataset = `n (%)`)

left_join(count_by_group, count_overall)

通过上面的代码获得所需的输出：每个组的均值表，紧随整体均值之后：

  variable   Group1  Group2  Group3 Overall_dataset
  <chr>       <dbl>   <dbl>   <dbl>           <dbl>
1 Variable_A -0.154  0.0385  0.263           0.0351
2 Variable_B  0.212 -0.232  -0.124          -0.0671
3 Variable_C -0.195  0.194   0.0508          0.0376

对分类变量进行类似的处理，以获取每个组以及整个数据集的计数和百分比：

count_by_group <- dat %>% 
    group_by(treatment) %>% 
    count(recruitment_strategy) %>%
    mutate(`n (%)` = paste0(n, " (", round(n / sum(n)*100,0), "%)")) %>% # calculate percentage in the desired format for table
    select(-n) %>%
    spread(treatment, `n (%)`)

count_overall <- dat %>% 
    count(recruitment_strategy) %>%
    mutate(`n (%)` = paste0(n, " (", round(n / sum(n)*100,0), "%)")) %>% # calculate percentage in the desired format for table
    select(-n) %>%
    rename(Overall_dataset = `n (%)`)

left_join(count_by_group, count_overall)

  recruitment_strategy Group1  Group2   Group3  Overall_dataset
  <chr>                <chr>   <chr>    <chr>   <chr>          
1 Strategy 1           2 (6%)  13 (30%) 4 (16%) 19 (19%)       
2 Strategy 2           8 (26%) 6 (14%)  6 (24%) 20 (20%)       
3 Strategy 3           6 (19%) 12 (27%) 3 (12%) 21 (21%)       
4 Strategy 4           9 (29%) 4 (9%)   5 (20%) 18 (18%)       
5 Strategy 5           6 (19%) 9 (20%)  7 (28%) 22 (22%) 

是否有一种解决方案可以在一个步骤中获得分组摘要和整体摘要，而不需要分配两个单独的对象，然后将它们合并为第三个对象？

答案

这是我将如何重写您的代码。

使用管道的技巧是使用.将LHS放在RHS的多个位置。这样，您就可以进行联接而无需分配中间对象。我还使用了一些其他步骤来达到不同的清晰度平衡，而不是重复我自己，例如在count()中进行所有分组并使用其name参数，在联接后使用mutate_at进行所有格式化，以及使用str_glue和scales::percent使字符串格式更具可读性。

所有这些都在某种程度上是一个优先事项，但是我认为通过以下方法可以避免中间分配（以及必须为上述对象命名的负担）。

library(tidyverse)
set.seed(500)
dat <- tibble(
  treatment = sample(c("Group1", "Group2", "Group3"), 100, replace = TRUE),
  recruitment_strategy = sample(c("Strategy 1", "Strategy 2", "Strategy 3", "Strategy 4", "Strategy 5"), 100, replace = TRUE),
  Variable_A = rnorm(100),
  Variable_B = rnorm(100),
  Variable_C = rnorm(100)
)

dat %>%
  inner_join(
      x = count(., treatment, recruitment_strategy) %>% spread(treatment, n),
      y = count(., recruitment_strategy, name = "Overall_dataset"),
      by = "recruitment_strategy"
  ) %>%
  mutate_at(
    .vars = vars(-recruitment_strategy),
    .funs = ~ str_glue(". (scales::percent(. / sum(.), accuracy = 1))")
  )
#> # A tibble: 5 x 5
#>   recruitment_strategy Group1  Group2   Group3  Overall_dataset
#>   <chr>                <glue>  <glue>   <glue>  <glue>         
#> 1 Strategy 1           2 (6%)  13 (30%) 4 (16%) 19 (19%)       
#> 2 Strategy 2           8 (26%) 6 (14%)  6 (24%) 20 (20%)       
#> 3 Strategy 3           6 (19%) 12 (27%) 3 (12%) 21 (21%)       
#> 4 Strategy 4           9 (29%) 4 (9%)   5 (20%) 18 (18%)       
#> 5 Strategy 5           6 (19%) 9 (20%)  7 (28%) 22 (22%)

^{由reprex package（v0.3.0）在2019-11-10创建}