将一个变量条件赋值给另外两个变量之一的值

Question

我想创建一个新变量,它等于其他两个变量之一的值,以其他变量的值为条件.这是一个假数据的玩具示例.

数据框的每一行代表一名学生.每个学生可以学习最多两个科目(subj1和subj2),并且可以在每个科目中攻读学位("BA")或未成年人("MN").我的真实数据包括数千名学生,几种类型的学位,约50个科目,学生可以有多达五个专业/未成年人.

   ID  subj1 degree1  subj2 degree2
1   1    BUS      BA   <NA>    <NA>
2   2    SCI      BA    ENG      BA
3   3    BUS      MN    ENG      BA
4   4    SCI      MN    BUS      BA
5   5    ENG      BA    BUS      MN
6   6    SCI      MN   <NA>    <NA>
7   7    ENG      MN    SCI      BA
8   8    BUS      BA    ENG      MN
...

现在我想创建一个第六个变量,df$major它等于subj1if 的值subj1是学生的主要专业,或者subj2if 的值subj2是主要专业.主要专业是第一个学位等于"BA"的学科.我尝试了以下代码:

df$major[df$degree1 == "BA"] = df$subj1
df$major[df$degree1 != "BA" & df$degree2 == "BA"] = df$subj2

不幸的是,我收到一条错误消息:

> df$major[df$degree1 == "BA"] = df$subj1
Error in df$major[df$degree1 == "BA"] = df$subj1 : 
  NAs are not allowed in subscripted assignments

我假设这意味着如果赋值至少评估一行的NA,则不能使用向量化赋值.

我觉得我必须遗漏一些基本的东西,但上面的代码似乎是显而易见的事情,我无法想出一个替代方案.

如果它有助于编写答案,这里的示例数据是使用dput()与上面列出的假数据相同的格式创建的:

structure(list(ID = 1:20, subj1 = structure(c(3L, NA, 1L, 2L, 
2L, 3L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 3L, 3L, 1L, 2L, 1L
), .Label = c("BUS", "ENG", "SCI"), class = "factor"), degree1 = structure(c(2L, 
NA, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L), .Label = c("BA", "MN"), class = "factor"), subj2 = structure(c(1L, 
2L, NA, NA, 1L, NA, 3L, 2L, NA, 2L, 2L, 1L, 3L, NA, 2L, 1L, 1L, 
NA, 2L, 2L), .Label = c("BUS", "ENG", "SCI"), class = "factor"), 
    degree2 = structure(c(2L, 2L, NA, NA, 2L, NA, 1L, 2L, NA, 
    2L, 1L, 1L, 2L, NA, 1L, 2L, 2L, NA, 1L, 2L), .Label = c("BA", 
    "MN"), class = "factor")), .Names = c("ID", "subj1", "degree1", 
"subj2", "degree2"), row.names = c(NA, -20L), class = "data.frame")

Answer 1

您的原始分配方法失败至少有两个原因.

1)下标分配的问题df$major[df$degree1 == "BA"] <-.使用==可以产生NA,这是提示错误的原因.From ?"[<-":"当替换时(即在赋值的lhs上使用索引)NA不会选择任何要替换的元素.由于是否应该使用rhs的元素存在歧义,这仅允许rhs值的长度为1(因此两种解释会产生相同的结果)." 有很多方法可以解决这个问题,但我更喜欢使用which:

df$major[which(df$degree1 == "BA")] <-

不同的是,==返回TRUE,FALSE和NA,而which回报是TRUE的对象的索引

> df$degree1 == "BA"
 [1] FALSE    NA  TRUE  TRUE  TRUE FALSE FALSE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE

> which(df$degree1 == "BA")
 [1]  3  4  5  8  9 10 11 12 13 14 15 16 17 18 19 20

2)当您执行下标任务时,右侧需要明智地适应左侧(这是我想到的方式).这可能意味着左右两边长度相等,这就是你的例子所暗示的.因此,您还需要对赋值的右侧进行子集化:

df$major[which(df$degree1 == "BA")] <- df$subj1[which(df$degree1 == "BA")]

我希望澄清你原来的尝试产生错误的原因.

ifelse正如@DavidRobinson所建议的那样,使用这种类型的赋值是一种很好的方法.我接受它:

df$major2 <- ifelse(df$degree1 == "BA", df$subj1, ifelse(df$degree2 == "BA",
  df$subj2,NA))

这相当于

df$major[which(df$degree1 == "BA")] <- df$subj1[which(df$degree1 == "BA")]
df$major[which(df$degree1 != "BA" & df$degree2 == "BA")] <- 
  df$subj2[which(df$degree1 != "BA" & df$degree2 == "BA")]

根据嵌套ifelse语句的深度,另一种方法可能更适合您的实际数据.

编辑:

我打算写第三个原因导致原始代码失败(即df$major尚未分配),但它对我有用,而不必这样做.不过,这是我记得的一个问题.你在运行什么版本的R？(对我来说是2.15.0.)如果使用这种ifelse()方法,则无需执行此步骤.使用时你的解决方案很好[,尽管我会选择

df$major <- NA

要获取主题的字符值,而不是因子级别索引,请使用as.character()(因子相当于和调用levels(x)[x]):

df$major[which(df$degree1 == "BA")] <- as.character(df$subj1)[which(df$degree1 == "BA")]
df$major[which(df$degree1 != "BA" & df$degree2 == "BA")] <- 
  as.character(df$subj2)[which(df$degree1 != "BA" & df$degree2 == "BA")]

方式相同ifelse():

df$major2 <- ifelse(df$degree1 == "BA", as.character(df$subj1),
  ifelse(df$degree2 == "BA", as.character(df$subj2), NA))

Answer 2

通常,ifelse函数是这些情况的正确选择,例如:

df$major = ifelse((!is.na(df$degree1) & df$degree1 == "BA") & (is.na(df$degree2) | df$degree1 != "BA"), df$subj1, df$subj2)

然而,其精确的使用取决于你做什么,如果这两个df$degree1和df$degree2是"BA".