r - R 中的表操作 - dplyr/tidyverse 解决方案

Question

我正在尝试将一个 df 转换为另一种格式。

开始 df:

df <- data.frame(id = c("1", "2", "3","4", "5", "6", "7", "8", "9", "10"),
    criteria_A = c("present", "present", "absent", "absent", "absent", "present", "absent", "present", "absent", "present"),
    criteria_B =c("absent", "absent", "present", "absent", "absent", "present", "absent", "absent", "present", "present"))

我想按存在/缺席对每个标准进行计数，并按如下方式重新制表:

df2 <- data.frame(criteria = c("criteria_A", "criteria_A", "criteria_B", "criteria_B"),
    count = c("5", "5", "4", "6"),
    status = c("present", "absent", "present", "absent"))

我考虑过按照标准以这种方式获取计数:

library(dplyr)
tmp1 <- df %>% group_by(criteria_A) %>% count() %>% mutate(criteria="criteria_A")
tmp1 <- tmp1 %>% rename(criteria_A=status)
tmp2 <- df %>% group_by(criteria_B) %>% count() %>% mutate(criteria="criteria_B")
tmp2 <- tmp2 %>% rename(criteria_B=status)

我想我可以垂直合并输出。当实际上我有数百个标准时，这不是一种有效或聪明的方法......

我确信有一个优雅的解决方案，但我不够聪明，无法弄清楚!

一如既往地提供任何帮助，我们将不胜感激。

Answer 1

在使用 pivot_longer 将数据转换为长格式后，您可以尝试使用 dplyr::tally。

library(dplyr)

df %>%
  pivot_longer(-id, 
               names_to = 'criteria',
               values_to = 'status') %>%
  group_by(criteria, status) %>%
  tally

#----
  criteria   status      n
  <chr>      <chr>   <int>
1 criteria_A absent      5
2 criteria_A present     5
3 criteria_B absent      6
4 criteria_B present     4