重命名数据框中的观察结果

Question

我在 R 中有一个数据帧，其中一列称为状态。在本专栏中，我们所有的州均采用缩写形式，例如NY、ME 等。现在我需要用它们的全名重命名这些值，例如纽约、缅因州等。我尝试了多种函数，例如 rename() 或 revalue()，但不知何故它们要么给我一个错误通知，要么创建一个新值，这不是我想要的。您可以在附件中找到我的多次失败之一。我非常感谢各种帮助，因为几天来我一直在处理这个问题。祝大家初临快乐!

str(rename(zipcode_new,
            "AL"="alaska","AK"="alaska","AR"="arizona","AZ"="arizona","CA"="california","CO"="colorado","CT"="conneticut","DE"="delaware","FL"="florida","GA"="georgia","HI"="hawaii","IA"="iowa","ID"="idaho","IL"="illinois","IN"="indiana","KS"="kansas","KY"="kentucky","LA"="louisiana","MA"="massachussets","MD"="maryland","ME"="maine","MN"="minnesota","MI"="michigan","MS"="mississippi","MS"="mississippi","MO"="missouri","MT"="montana","NE"="north carolina","ND"="north dakota","NH"="new hampshire","NJ"="new jersey","NM"="new mexico","NV"="nevada","NY"="new york","OH"="ohio","OK"="oklahoma","OR"="oregon","PA"="pennsylvania","SC"="south carolina","SD"="south dakota","TN"="tennessee","TX"="texas","UT"="utah","VA"="virginia","WA"="washington","WI"="wisconsin","WY"="wyoming"))

Answer 1

如果我们可以使用内置的 state.name 和 state.abb 会更容易

df1$state <- with(df1, tolower(setNames(state.name, state.abb)[state]))

---

使用OP的数据集

zipcode_new$state <- with(zipcode_new, tolower(setNames(state.name, 
             state.abb)[state]))

数据

set.seed(24)
df1 <- data.frame(state = sample(state.abb, 100, replace = TRUE),
         stringsAsFactors = FALSE)