在 R 中使用 tidyverse 在单个 tibble 中汇总分组和未分组的数据答案

【问题标题】：Summarizing grouped and ungrouped data in a single tibble with tidyverse in R在 R 中使用 tidyverse 在单个 tibble 中汇总分组和未分组的数据
【发布时间】：2021-09-02 19:07:31
【问题描述】：

为了总结数据中的每个变量，我通常会创建两个表：一个按实验条件分组，另一个显示所有实验组的汇总统计数据。

但是，我想在一个表格中同时显示分组和汇总的描述。

我已经完成了这个，如下面的演示代码所示——但是，我想找到一个更优雅的解决方案，最好使用单链。

有谁知道我如何做到这一点？

感谢您的帮助。


# Data
df <- tibble(Condition = rep(c("Group1",
                           "Group2",
                           "Group3",
                           "Group4"),20),
             comp_fail = rbinom(n = 80,
                                size = 1,
                                prob = .1)
)

# Converting Condition to factor to mimic my actual df
df$Condition <- as.factor(df$Condition)

# Descriptives grouped by condition
tbl_comp_fail_condition <- 
 df %>% 
  group_by(Condition) %>% 
  summarize("# with Zero Fails" = sum(comp_fail == 0, na.rm = T),
            "# with One Fail" = sum(comp_fail == 1, na.rm = T))

# Descriptives aggregated
tbl_comp_fail_aggregate <- 
 df %>% 
  summarize("# with Zero Fails" = sum(comp_fail == 0, na.rm = T),
            "# with One Fail" = sum(comp_fail == 1, na.rm = T))

# Joining grouped and aggregated
tbl_comp_fail_combined <- full_join(
 tbl_comp_fail_aggregate,
 tbl_comp_fail_condition) %>% 
  select(Condition, everything())

# Converting Condition to character to replace NA with All
tbl_comp_fail_combined$Condition <- as.character(tbl_comp_fail_combined$Condition)

# Replace NA with All
tbl_comp_fail_combined %>% replace_na(list(Condition = "All")) ``

【问题讨论】：

标签： r grouping tidyverse summarize

【解决方案1】：

这是一种使用dplyr::bind_rows的方法：

df %>%
  group_by(Condition) %>%
  summarize("# with Zero Fails" = sum(comp_fail == 0, na.rm = T),
            "# with One Fail" = sum(comp_fail == 1, na.rm = T)) %>%
  bind_rows(list(summarise(., across(-Condition, sum))))

# A tibble: 5 x 3
  Condition `# with Zero Fails` `# with One Fail`
  <fct>                   <int>             <int>
1 Group1                     18                 2
2 Group2                     18                 2
3 Group3                     17                 3
4 Group4                     18                 2
5 NA                         71                 9

或者先获取摘要：

df %>%
  group_by(Condition) %>%
  summarize("# with Zero Fails" = sum(comp_fail == 0, na.rm = T),
            "# with One Fail" = sum(comp_fail == 1, na.rm = T)) %>%
  bind_rows(list(summarise(., across(-Condition, sum))), .) %>%
  relocate(Condition)

# A tibble: 5 x 3
  Condition `# with Zero Fails` `# with One Fail`
  <fct>                   <int>             <int>
1 NA                         71                 9
2 Group1                     18                 2
3 Group2                     18                 2
4 Group3                     17                 3
5 Group4                     18                 2

【讨论】：

【解决方案2】：

使用adorn_totals 会更容易

library(dplyr)
library(janitor)
df %>% 
    group_by(Condition) %>% 
    summarise("# with Zero Fails" = sum(!comp_fail, na.rm = TRUE), 
            "# with One Fail" = sum(comp_fail, na.rm = TRUE)) %>% 
    adorn_totals(name = "All")

-输出

   Condition # with Zero Fails # with One Fail
    Group1                17               3
    Group2                14               6
    Group3                18               2
    Group4                14               6
       All                63              17

OP 的输出

out
# A tibble: 5 x 3
  Condition `# with Zero Fails` `# with One Fail`
  <chr>                   <int>             <int>
1 All                        63                17
2 Group1                     17                 3
3 Group2                     14                 6
4 Group3                     18                 2
5 Group4                     14                 6

或者另一个选项是count 和pivot_wider（如果有超过 2 个组，这会更通用）

library(tidyr)
df %>%
     count(Condition, comp_fail) %>% 
     pivot_wider(names_from = comp_fail, values_from = n) %>% 
     adorn_totals(name = 'All')

或者这可以在base R 和table/add_margins 中完成

addmargins(table(df), 1)
         comp_fail
Condition  0  1
   Group1 17  3
   Group2 14  6
   Group3 18  2
   Group4 14  6
   Sum    63 17

【讨论】：