如何在R语言中生成新列并返回所有符合条件的列标题（行拆分）

原始数据 #

先还原你的原始DataFrame：

DF <- data.frame(
  ID = 1:7,
  Mon = c("Yes", "No", "No", "No", "Yes", "Yes", "No"),
  Tue = c("No", "No", "No", "Yes", "Yes", "Yes", "No"),
  Wed = c("No", "Yes", "No", "No", "No", "Yes", "Yes"),
  Thur = c("No", "No", "No", "Yes", "Yes", "Yes", "Yes"),
  Count_Y = c(1, 1, 0, 2, 3, 4, 2),
  stringsAsFactors = FALSE
)

你原来用case_when的写法只能匹配第一个符合条件的列，没法把所有值为"Yes"的列标题都提取出来并拆分成多行，下面是两种可行的解决方案：

方法1：用tidyr::pivot_longer（推荐） #

这是tidyverse生态里最简洁的实现方式，通过宽表转长表完成需求：

library(dplyr)
library(tidyr)

DF_result <- DF %>%
  # 将Mon到Thur的列转成长表，列名存到New_column，值存到临时列status
  pivot_longer(
    cols = Mon:Thur,
    names_to = "New_column",
    values_to = "status"
  ) %>%
  # 按ID分组，判断当前行是否存在Yes
  group_by(ID) %>%
  mutate(has_yes = any(status == "Yes")) %>%
  ungroup() %>%
  # 保留Yes的行，或者全No的行
  filter(status == "Yes" | !has_yes) %>%
  # 全No的行把New_column设为OFF
  mutate(New_column = ifelse(!has_yes, "OFF", New_column)) %>%
  # 删除临时列，按ID排序
  select(-status, -has_yes) %>%
  arrange(ID)

方法2：用rowwise+purrr::map逐行处理 #

如果习惯逐行逻辑处理，可以用这种方式：

library(dplyr)
library(purrr)
library(tidyr)

DF_result <- DF %>%
  rowwise() %>%
  # 逐行判断：全No则返回"OFF"列表，否则返回所有Yes对应的列名列表
  mutate(
    New_column = list(
      if (all(c(Mon, Tue, Wed, Thur) == "No")) {
        "OFF"
      } else {
        c("Mon", "Tue", "Wed", "Thur")[c(Mon, Tue, Wed, Thur) == "Yes"]
      }
    )
  ) %>%
  # 将列表列拆分成多行
  unnest(New_column) %>%
  ungroup() %>%
  arrange(ID)

运行任意一种方法后，都能得到你期望的输出：

# A tibble: 15 × 6
      ID Mon   Tue   Wed   Thur Count_Y New_column
   <int> <chr> <chr> <chr> <chr>   <dbl> <chr>     
 1     1 Yes   No    No    No         1 Mon       
 2     2 No    No    Yes   No         1 Wed       
 3     3 No    No    No    No         0 OFF       
 4     4 No    Yes   No    Yes        2 Tue       
 5     4 No    Yes   No    Yes        2 Thur      
 6     5 Yes   Yes   No    Yes        3 Mon       
 7     5 Yes   Yes   No    Yes        3 Tue       
 8     5 Yes   Yes   No    Yes        3 Thur      
 9     6 Yes   Yes   Yes   Yes        4 Mon       
10     6 Yes   Yes   Yes   Yes        4 Tue       
11     6 Yes   Yes   Yes   Yes        4 Wed       
12     6 Yes   Yes   Yes   Yes        4 Thur      
13     7 No    No    Yes   Yes        2 Wed       
14     7 No    No    Yes   Yes        2 Thur      

内容的提问来源于stack exchange，提问作者user20835011