嵌套ifelse语句

Question

我仍在学习如何将SAS代码转换为R，并收到警告。我需要弄清楚我在哪里犯错误。我想要做的是创建一个变量来总结和区分一个人口的三个状态:大陆，海外，外国人。我有一个包含两个变量的数据库：

• id国籍：idnat (法语、外国人)

如果idnat是法语，那么：

• 出生地：idbp (内地、殖民地、海外)

我想将来自idnat和idbp的信息归纳为一个名为idnat2的新变量

• 地位:K(内地、海外、外国人)

所有这些变量都使用“字符类型”。

预期结果在idnat2栏中：

   idnat     idbp   idnat2
1  french mainland mainland
2  french   colony overseas
3  french overseas overseas
4 foreign  foreign  foreign

下面是我想在R中翻译的SAS代码：

if idnat = "french" then do;
   if idbp in ("overseas","colony") then idnat2 = "overseas";
   else idnat2 = "mainland";
end;
else idnat2 = "foreigner";
run;

以下是我在R中的尝试：

if(idnat=="french"){
    idnat2 <- "mainland"
} else if(idbp=="overseas"|idbp=="colony"){
    idnat2 <- "overseas"
} else {
    idnat2 <- "foreigner"
}

我收到这样的警告：

Warning message:
In if (idnat=="french") { :
  the condition has length > 1 and only the first element will be used

有人建议我使用“嵌套ifelse”来代替它的简单性，但会收到更多警告：

idnat2 <- ifelse (idnat=="french", "mainland",
        ifelse (idbp=="overseas"|idbp=="colony", "overseas")
      )
            else (idnat2 <- "foreigner")

根据警告消息，长度大于1，因此将只考虑第一个括号之间的内容。对不起，我不明白这个长度和这里有什么关系？有人知道我哪里错了吗？

Answer 1

如果您正在使用任何电子表格应用程序，则有一个具有语法的基本函数if()：

if(<condition>, <yes>, <no>)

在R中，ifelse()的语法完全相同：

ifelse(<condition>, <yes>, <no>)

在电子表格应用程序中，与if()的唯一区别是R ifelse()是矢量化的(将向量作为输入，在输出时返回向量)。考虑下面对电子表格应用程序中公式和R中公式的比较，例如，如果a>b，则返回1，如果不是，则返回0。

在电子表格中：

  A  B C
1 3  1 =if(A1 > B1, 1, 0)
2 2  2 =if(A2 > B2, 1, 0)
3 1  3 =if(A3 > B3, 1, 0)

在R：

> a <- 3:1; b <- 1:3
> ifelse(a > b, 1, 0)
[1] 1 0 0

ifelse()可以以多种方式嵌套：

ifelse(<condition>, <yes>, ifelse(<condition>, <yes>, <no>))

ifelse(<condition>, ifelse(<condition>, <yes>, <no>), <no>)

ifelse(<condition>, 
       ifelse(<condition>, <yes>, <no>), 
       ifelse(<condition>, <yes>, <no>)
      )

ifelse(<condition>, <yes>, 
       ifelse(<condition>, <yes>, 
              ifelse(<condition>, <yes>, <no>)
             )
       )

要计算列idnat2，可以：

df <- read.table(header=TRUE, text="
idnat idbp idnat2
french mainland mainland
french colony overseas
french overseas overseas
foreign foreign foreign"
)

with(df, 
     ifelse(idnat=="french",
       ifelse(idbp %in% c("overseas","colony"),"overseas","mainland"),"foreign")
     )

R文件

the condition has length > 1 and only the first element will be used是什么？让我们看看：

> # What is first condition really testing?
> with(df, idnat=="french")
[1]  TRUE  TRUE  TRUE FALSE
> # This is result of vectorized function - equality of all elements in idnat and 
> # string "french" is tested.
> # Vector of logical values is returned (has the same length as idnat)
> df$idnat2 <- with(df,
+   if(idnat=="french"){
+   idnat2 <- "xxx"
+   }
+   )
Warning message:
In if (idnat == "french") { :
  the condition has length > 1 and only the first element will be used
> # Note that the first element of comparison is TRUE and that's whay we get:
> df
    idnat     idbp idnat2
1  french mainland    xxx
2  french   colony    xxx
3  french overseas    xxx
4 foreign  foreign    xxx
> # There is really logic in it, you have to get used to it

我还能用if()吗？是的，你可以，但语法并不那么酷:)

test <- function(x) {
  if(x=="french") {
    "french"
  } else{
    "not really french"
  }
}

apply(array(df[["idnat"]]),MARGIN=1, FUN=test)

如果您熟悉SQL，也可以在CASE 套餐中使用sqldf 套餐。

Answer 2

尝试如下所示：

# some sample data
idnat <- sample(c("french","foreigner"),100,TRUE)
idbp <- rep(NA,100)
idbp[idnat=="french"] <- sample(c("mainland","overseas","colony"),sum(idnat=="french"),TRUE)

# recoding
out <- ifelse(idnat=="french" & !idbp %in% c("overseas","colony"), "mainland",
              ifelse(idbp %in% c("overseas","colony"),"overseas",
                     "foreigner"))
cbind(idnat,idbp,out) # check result

您的困惑来自SAS和R如何处理if-else结构。在R中，if和else不是向量化的，这意味着它们检查单个条件是否为真(即if("french"=="french")工作)，并且不能处理多个逻辑(即if(c("french","foreigner")=="french")不工作)，并且R会给出您收到的警告。

相反，ifelse是向量化的，因此它可以接受向量(也称为输入变量)，并测试每个元素的逻辑条件，就像您在SAS中所习惯的那样。解决这个问题的另一种方法是使用if和else语句构建一个循环(就像您在这里开始做的那样)，但是矢量化的ifelse方法将更高效，并且通常只涉及较少的代码。

Answer 3

如果数据集包含许多行，则使用data.table而不是嵌套的ifelse()连接查找表可能更有效。

提供了下面的查找表

lookup

idnat idbp idnat2 1: french mainland mainland 2: french colony overseas 3: french overseas overseas 4: foreign foreign foreign

和一个样本数据集

library(data.table)
n_row <- 10L
set.seed(1L)
DT <- data.table(idnat = "french",
                 idbp = sample(c("mainland", "colony", "overseas", "foreign"), n_row, replace = TRUE))
DT[idbp == "foreign", idnat := "foreign"][]

idnat idbp 1: french colony 2: french colony 3: french overseas 4: foreign foreign 5: french mainland 6: foreign foreign 7: foreign foreign 8: french overseas 9: french overseas 10: french mainland

然后我们可以在加入时进行更新。

DT[lookup, on = .(idnat, idbp), idnat2 := i.idnat2][]

idnat idbp idnat2 1: french colony overseas 2: french colony overseas 3: french overseas overseas 4: foreign foreign foreign 5: french mainland mainland 6: foreign foreign foreign 7: foreign foreign foreign 8: french overseas overseas 9: french overseas overseas 10: french mainland mainland