r - 是否有更优雅的方法将 88 个级别的变量折叠为 5 个级别的变量？

Question

我有一个包含 88 个级别(县)的分类变量，我想将它们聚合到五个更大的地理区域中。有没有比大量 ifelse 语句(如下所示)更优雅的方法来做到这一点？

survey.responses$admin<-ifelse(survey.responses$CNTY=="Lake","Northeast",
                         ifelse(survey.responses$CNTY=="Traverse","Northwest", 
                         ifelse(survey.responses$CNTY=="Ramsey","Central",
                         ifelse(survey.responses$CNTY=="Cottonwood","South","out of state")

除了想象 CNTY 有 88 个级别!有什么想法吗？

Answer 1

两个快速的方法，我推荐merge一种用于较大的套装。

数据

dat <- data.frame(cnty = c("Lake", "Traverse", "Ramsey", "Cottonwood"),
                  stringsAsFactors = FALSE)

合并/加入 .我更喜欢这个有几个原因，最重要的是维护匹配和 read.csv 的 CSV 文件很容易。将 CSV 导入 ref查找表。我会故意将“湖”排除在外，以显示不匹配时会发生什么。

ref <- data.frame(cnty = c("Cottonwood", "Ramsey", "Traverse", "SomeOther"),
                  admin = c("South", "Central", "Northwest", "NeverNeverLand"),
                  stringsAsFactors = FALSE)
out <- merge(dat, ref, by = "cnty", all.x = TRUE)
out
#         cnty     admin
# 1 Cottonwood     South
# 2       Lake      <NA>
# 3     Ramsey   Central
# 4   Traverse Northwest

默认值以这种方式分配:

out$admin[is.na(out$admin)] <- "out of state"
out
#         cnty        admin
# 1 Cottonwood        South
# 2       Lake out of state
# 3     Ramsey      Central
# 4   Traverse    Northwest

如果您正在使用 tidyverse 的其他组件，这可以用

library(dplyr)
left_join(dat, ref, by = "cnty") %>%
  mutate(admin = if_else(is.na(admin), "out of state", admin))

查找 .这适用于小事，但可能不适合您。 (同样，我已经将“Lake”注释掉以显示不匹配。)

c(Cottonwood="South", # Lake="Northeast",
  Ramsey="Central", Traverse="Northwest")[dat$cnty]
#        <NA>    Traverse      Ramsey  Cottonwood 
#          NA "Northwest"   "Central"     "South"