Question

我创建了一个包含以下数据的数据框

name <- c("A","B","C","D","E","F","G","H","I","J")
age <- c(22,43,12,17,29,5,51,56,9,44)
sex <- c("M","F","M","M","M","F","F","M","F","F")
rock <- data.frame(name,age,sex,stringsAsFactors = TRUE)
rock

现在我想找出：

如果名称为E至J且性别不等于F，则状态为“1F”，如果名称为A至D且年龄大于15，则状态为“年轻”。其他一切都是“他人”

所以，我正在申请以下代码：

rock$status <- ifelse(rock$name==c("E","F","G","H","I","J")& 
rock$sex!="F","1F",            
ifelse(rock$name==c("E","F","G","H","I","J")&rock$sex=="F","Fenamle",
ifelse(rock$name==c("A","B","C","D") & rock$age>15,"Young","Others")))
rock

但我得到的结果如下：

  name  age    sex    status
1     A   22     M   Young   
2     B   43     F   Young   
3     C   12     M  Others  
4     D   17     M  Young   
5     E   29     M  Others  
6     F    5     F  Others  
7     G   51     F  Others  
8     H   56     M  Others 
9     I    9     F  Others  
10    J   44     F  Others

但是，它必须是E和H上的“1F”，但它显示“其他”

我在代码中做了什么错误？

请纠正我，并就此提出一些有价值的建议。

Answer 1

我们需要使用%in%代替==：

rock$status <- ifelse(rock$name %in% c("E", "F", "G", "H", "I", "J") & 
                        rock$sex != "F", "1F",            
                      ifelse(rock$name %in% c("E", "F", "G", "H", "I", "J") & 
                               rock$sex == "F", "Female",
                             ifelse(rock$name %in% c("A", "B", "C", "D") &
                                      rock$age > 15, "Young", "Others")))
rock

#    name age sex  status
# 1     A  22   M   Young
# 2     B  43   F   Young
# 3     C  12   M  Others
# 4     D  17   M   Young
# 5     E  29   M      1F
# 6     F   5   F  Female
# 7     G  51   F  Female
# 8     H  56   M      1F
# 9     I   9   F  Female
# 10    J  44   F  Female

Answer 2

在这种情况下，我经常更喜欢预先分配索引，然后使用这些索引的总和来索引唯一值。它比嵌套的ifelse（imo）更快，更易读。一个例子：

i1 <- rock$name %in% c("E", "F", "G", "H", "I", "J") & rock$sex != "F"
i2 <- rock$name %in% c("E", "F", "G", "H", "I", "J") & rock$sex == "F"
i3 <- rock$name %in% c("A", "B", "C", "D") & rock$age > 15

rock$status <- c("Other", "1F", "Female", "Young")[1 + i1 + 2*i2 + 3*i3]

给出了期望的结果：

> rock
   name age sex status
1     A  22   M  Young
2     B  43   F  Young
3     C  12   M  Other
4     D  17   M  Young
5     E  29   M     1F
6     F   5   F Female
7     G  51   F Female
8     H  56   M     1F
9     I   9   F Female
10    J  44   F Female

Answer 3

使用data.table，您可以：

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Answer 4

使用dplyr case_when()函数的解决方案：

library(dplyr)

name <- c("A","B","C","D","E","F","G","H","I","J")
age <- c(22,43,12,17,29,5,51,56,9,44)
sex <- c("M","F","M","M","M","F","F","M","F","F")
rock <- data.frame(name,age,sex,stringsAsFactors = TRUE)

name_condition_1 <- c("E","F","G","H","I","J")
name_condition_2 <- c("A","B","C","D")

rock %>% mutate(
  status = case_when(
    name %in% name_condition_1 & sex != "F" ~ "1F",
    name %in% name_condition_1 & sex == "F" ~ "Female",
    name %in% name_condition_2 & age >  15  ~ "Young",
    TRUE ~ "Others"
  )
)

制造

   name age sex status
1     A  22   M  Young
2     B  43   F  Young
3     C  12   M Others
4     D  17   M  Young
5     E  29   M     1F
6     F   5   F Female
7     G  51   F Female
8     H  56   M     1F
9     I   9   F Female
10    J  44   F Female

Answer 5

为了完整起见，这里还有一个解决方案，使用 join 和 non-equi join 来更新status列：

library(data.table)
setDT(rock)[.(name = LETTERS[1:4], age = 15), on = .(name, age > age), status := "Young"][
  .(name = LETTERS[5:10], sex = "F"), on = .(name, sex), status := "Female"][
    .(name = LETTERS[5:10], status = NA_character_), on = .(name, status), status := "1F"][
      .(status = NA_character_), on = .(status), status := "Other"][]

    name age sex status
 1:    A  22   M  Young
 2:    B  43   F  Young
 3:    C  12   M  Other
 4:    D  17   M  Young
 5:    E  29   M     1F
 6:    F   5   F Female
 7:    G  51   F Female
 8:    H  56   M     1F
 9:    I   9   F Female
10:    J  44   F Female

不幸的是，非equi连接不适用于不等运算符!=。所以，

setDT(rock)[.(name = LETTERS[1:4], age = 15), on = .(name, age > age), status := "Young"][
  .(name = LETTERS[5:10], sex = "F"), on = .(name, sex != sex), status := "1F"][]

给出错误消息。相反，我必须首先加入name和sex以将status设置为Female，然后检查NA中的status获得免费套装。

但是，还有另一种使用两种非等连接的解决方法：

setDT(rock)[.(name = LETTERS[1:4], age = 15), on = .(name, age > age), status := "Young"][
  .(name = LETTERS[5:10], sex = "F"), on = .(name, sex < sex), status := "1F"][
    .(name = LETTERS[5:10], sex = "F"), on = .(name, sex > sex), status := "1F"][]

Answer 6

data$status <- ifelse(data$name %in% c("A", "B", "C", "D") & data$age > 15,"Young",ifelse(data$sex != "F" & data$name %in% c("E", "F", "G", "H", "I", "J"),"1F","Others"))
data

我应该如何使用嵌套的ifelse语句？

6 个答案: