用 dplyr 创建一个因子变量？答案

【问题标题】：creating a factor variable with dplyr?用 dplyr 创建一个因子变量？
【发布时间】：2014-09-30 14:37:17
【问题描述】：

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"))

感谢您的帮助！

【问题讨论】：

标签： r dplyr

【解决方案1】：

这是一个旧帖子，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。 Version 0.5.0 of dplyr 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（从而使许多小猫存活）：

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"))

【讨论】：

【解决方案2】：

试试：

df1$Sector <-  with(df1, c("Private4P", NA, "Public",
                 "PublicFlag")[as.numeric(factor(1+2*as.numeric(sector)+4*flagship))])



 subset(df1, !is.na(Sector))
 #  Name                              sector flagship     Sector 
 #1   N1             Public, 4-year or above        1 PublicFlag
 #2   N2             Public, 4-year or above        0     Public
 #3   N3             Public, 4-year or above        0     Public
 #6   N6 Private for-profit, 4-year or above        0  Private4P

【讨论】：

【解决方案3】：

你甚至不需要dplyr：

df1$Sector <- factor(ifelse(df1$sector=="Public, 4-year or above" & df1$flagship==1, "PublicFlag",
                       ifelse(df1$sector=="Public, 4-year or above" & df1$flagship==0, "Public",
                         ifelse(df1$sector=="Private not-for-profit", "PrivateNP", 
                           ifelse(df1$sector=="Private for-profit, 4-year or above", "Private4P", NA)))))


df1

##   Name                                  sector flagship     Sector
## 1   N1                 Public, 4-year or above        1 PublicFlag
## 2   N2                 Public, 4-year or above        0     Public
## 3   N3                 Public, 4-year or above        0     Public
## 4   N4 Private not-for-profit, 4-year or above        0       <NA>
## 5   N5 Private not-for-profit, 4-year or above        0       <NA>
## 6   N6     Private for-profit, 4-year or above        0  Private4P

如果需要，您可以将NA 替换为最终可能的因子水平

【讨论】：

每当你把嵌套的 if-else 嵌套得那么深，我就杀了一只小猫。

【解决方案4】：

所选答案不适用于我正在处理的特定问题，因为我在 case_when() 中分配了数值并尝试为其赋予字符级别。我想添加我为解决我的特定问题所做的事情作为替代方案，以防将来有人发现它有用。

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") %>%
  as.factor() %>%
  structure(levels = c("Public","PublicFlag","PrivateNP","Private4P"))
  )

【讨论】：