使用 dplyr 创建因子变量？

假设我有一个看起来像这样的数据框：

df1=structure(list(Name = structure(1:6, .Label = c("N1", "N2", "N3", 
                                                    "N4", "N5", "N6", "N7"), class = "factor"), sector = structure(c(4L, 
                                                                                                                     4L, 4L, 3L, 3L, 2L), .Label = c("other stuff", "Private for-profit, 4-year or above", 
                                                                                                                                                     "Private not-for-profit, 4-year or above", "Public, 4-year or above"
                                                                                                                     ), class = "factor"), flagship = c(1, 0, 0, 0, 0, 0)), .Names = c("Name", 
                                                                                                                                                                                       "sector", "flagship"), row.names = c(NA, 6L), class = "data.frame")

我想创建一个新的因子变量“Sector”。我可以用很多行代码来完成它，但我确信有一种更有效的方法。

现在这就是我正在做的事情：

df1$PublicFlag=0
df1$PublicFlag[df1$sector=="Public, 4-year or above" & df1$flagship==1]=1
df1$Public=0
df1$Public[df1$sector=="Public, 4-year or above" & df1$flagship==0]=1
df1$PrivateNP=0
df1$PrivateNP[df1$sector=="Private not-for-profit"]=1
df1$Private4P=0
df1$Private4P[df1$sector=="Private for-profit, 4-year or above"]=1

library(reshape)
df2 = melt(df1, id=c("Name", "sector", "flagship"))
df2 = df2[df2$value==1,c("Name", "sector", "flagship", "variable")]
library(plyr)
df2 = rename(df2, c("variable"="Sector"))

谢谢您的帮助！

这是一篇老帖子，但我经常偶然发现它。这就是为什么我想给出一个最新的答案。dplyr 0.5.0 版本 https://blog.rstudio.com/2016/06/27/dplyr-0-5-0/引入了很多有用的向量函数来解决这个问题。

使用 case_when() 避免 ifelse 嵌套（从而让很多很多小猫活下来）：

df1 %>% 
  mutate(Sector = case_when(
        sector=="Public, 4-year or above" & flagship==1 ~ "PublicFlag",
        sector=="Public, 4-year or above" & flagship==0 ~ "Public",
        sector=="Private not-for-profit" ~ "PrivateNP",
        sector=="Private for-profit, 4-year or above" ~ "Private4P"),
    Sector = factor(Sector, levels=c("Public","PublicFlag","PrivateNP","Private4P"))
  )

使用 recode_factor() 从字符（或数字）变量生成因子：

df1 %>%
    mutate(Sector = recode_factor(sector,
                               "Public, 4-year or above" = "Public",
                               "Private not-for-profit" = "PrivateNP",
                               "Private for-profit, 4-year or above" = "Private4P"))