如何将多选项分类变量转换为数值变量？以BusinessTravel为例

针对你要把BusinessTravel变量（取值为"Non-Travel"、"Rarely_Travel"、"Frequently_Travel"）转换为0、1、2的需求，这里提供几种适配你现有dplyr代码的实现方式：

方法1：用case_when明确映射（最直观） #

这种方法可以清晰指定每个分类对应的数值，还能处理异常值：

new_df <- new_df %>%
  mutate(Attrition = ifelse(Attrition == "No", 0, 1),
         OverTime = ifelse(OverTime == "No", 0, 1),
         BusinessTravel = case_when(
           BusinessTravel == "Non-Travel" ~ 0,
           BusinessTravel == "Rarely_Travel" ~ 1,
           BusinessTravel == "Frequently_Travel" ~ 2,
           TRUE ~ NA_real_ # 对未匹配到的类别返回NA，避免出错
         ))

方法2：因子转数值（适合有序分类） #

先将变量转为有序因子，再转成数值，注意要指定levels的顺序（默认因子从1开始，所以减1得到0起始的数值）：

new_df <- new_df %>%
  mutate(Attrition = ifelse(Attrition == "No", 0, 1),
         OverTime = ifelse(OverTime == "No", 0, 1),
         BusinessTravel = as.numeric(factor(BusinessTravel, 
                                            levels = c("Non-Travel", "Rarely_Travel", "Frequently_Travel"))) - 1)

方法3：用recode函数快速映射 #

dplyr的recode函数可以简洁完成一对一映射：

new_df <- new_df %>%
  mutate(Attrition = ifelse(Attrition == "No", 0, 1),
         OverTime = ifelse(OverTime == "No", 0, 1),
         BusinessTravel = recode(BusinessTravel,
                                "Non-Travel" = 0,
                                "Rarely_Travel" = 1,
                                "Frequently_Travel" = 2))

注意事项 #

• 你的代码里存在一处变量名错误：读取的数据集是Dataset，但后续select用的是df，需要改成select(Dataset, ...)才能正常运行。
• 转换前建议先检查BusinessTravel的所有取值，确保没有遗漏或拼写错误的类别：table(new_df$BusinessTravel)

内容的提问来源于stack exchange，提问作者Pepe Cobos