嵌套的 ifelse 语句

请尝试如下操作：

# 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

您的困惑来自于 SAS 和 R 如何处理 if-else 构造。在 R 中，并且没有向量化，这意味着它们检查单个条件是否为真（即有效）并且不能处理多个逻辑（即不起作用），并且 R 会向您发出您收到的警告。ifelseif("french"=="french")if(c("french","foreigner")=="french")

相比之下，它是矢量化的，因此它可以获取您的向量（又名输入变量）并测试其每个元素的逻辑条件，就像您在 SAS 中习惯的那样。另一种方法是使用 and 语句构建一个循环（正如您在这里开始做的那样），但矢量化方法将更有效，并且通常涉及更少的代码。ifelseifelseifelse

如果您使用的是任何电子表格应用程序，则有一个带有语法的基本功能：if()

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

R 中的语法完全相同：ifelse()

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

与电子表格应用程序的唯一区别是 R 是矢量化的（将矢量作为输入并在输出时返回向量）。考虑以下电子表格应用程序和 R 中的公式比较，例如，我们想比较 a 是否> b，如果是则返回 1，如果不是则返回 0。if()ifelse()

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 中：

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

ifelse()可以通过多种方式嵌套：

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，您可以：idnat2

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 文档

什么？我看看：the condition has length > 1 and only the first element will be used

> # 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

我还能使用吗？是的，你可以，但语法不是那么酷:)if()

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

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

如果你熟悉SQL，也可以在package中使用语句。CASEsqldf

您可以创建不带 和 的向量。idnat2ififelse

该函数可用于将所有出现的 ：replace"colony""overseas"

idnat2 <- replace(idbp, idbp == "colony", "overseas")

使用 data.table，解决方案是：

DT[, idnat2 := ifelse(idbp %in% "foreign", "foreign", 
        ifelse(idbp %in% c("colony", "overseas"), "overseas", "mainland" ))]

是矢量化的。不是。在这里，DT 是：ifelseif-else

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

这给出了：

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

将 SQL CASE 语句与 dplyr 和 sqldf 包一起使用：

数据

df <-structure(list(idnat = structure(c(2L, 2L, 2L, 1L), .Label = c("foreign", 
"french"), class = "factor"), idbp = structure(c(3L, 1L, 4L, 
2L), .Label = c("colony", "foreign", "mainland", "overseas"), class = "factor")), .Names = c("idnat", 
"idbp"), class = "data.frame", row.names = c(NA, -4L))

SQLDF

library(sqldf)
sqldf("SELECT idnat, idbp,
        CASE 
          WHEN idbp IN ('colony', 'overseas') THEN 'overseas' 
          ELSE idbp 
        END AS idnat2
       FROM df")

德普利尔

library(dplyr)
df %>% 
mutate(idnat2 = case_when(idbp == 'mainland' ~ "mainland", 
                          idbp %in% c("colony", "overseas") ~ "overseas", 
                         TRUE ~ "foreign"))

输出

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

如果数据集包含许多行，则使用 而不是 nested 来联接查找表可能会更有效。data.tableifelse()

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

和示例数据集

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

# Read in the data.

idnat=c("french","french","french","foreign")
idbp=c("mainland","colony","overseas","foreign")

# Initialize the new variable.

idnat2=as.character(vector())

# Logically evaluate "idnat" and "idbp" for each case, assigning the appropriate level to "idnat2".

for(i in 1:length(idnat)) {
  if(idnat[i] == "french" & idbp[i] == "mainland") {
    idnat2[i] = "mainland"
} else if (idnat[i] == "french" & (idbp[i] == "colony" | idbp[i] == "overseas")) {
  idnat2[i] = "overseas"
} else {
  idnat2[i] = "foreign"
} 
}

# Create a data frame with the two old variables and the new variable.

data.frame(idnat,idbp,idnat2) 

对不起，参加聚会太晚了。这里有一个简单的解决方案。

#building up your initial table
idnat <- c(1,1,1,2) #1 is french, 2 is foreign

idbp <- c(1,2,3,4) #1 is mainland, 2 is colony, 3 is overseas, 4 is foreign

t <- cbind(idnat, idbp)

#the last column will be a vector of row length = row length of your matrix
idnat2 <- vector()

#.. and we will populate that vector with a cursor

for(i in 1:length(idnat))

     #*check that we selected the cursor to for the length of one of the vectors*

{  

  if (t[i,1] == 2) #*this says: if idnat = foreign, then it's foreign*

    {

      idnat2[i] <- 3 #3 is foreign

    }

  else if (t[i,2] == 1) #*this says: if not foreign and idbp = mainland then it's mainland*

    {

      idnat2[i] <- 2 # 2 is mainland  

    }

  else #*this says: anything else will be classified as colony or overseas*

    {

      idnat2[i] <- 1 # 1 is colony or overseas 

    }

}


cbind(t,idnat2)

示例的解释是帮助我的关键，但是我遇到的问题是当我复制它不起作用时，所以我不得不以多种方式弄乱它才能使其正常工作。（我是 R 的新手，由于缺乏知识，我对第三个 if 有一些问题）。

所以对于那些对 R 超级陌生的人遇到问题......

   ifelse(x < -2,"pretty negative", ifelse(x < 1,"close to zero", ifelse(x < 3,"in [1, 3)","large")##all one line
     )#normal tab
)

（我在一个函数中使用了它，所以它“ifelse...”被标记在一个上面，但最后一个“）”完全在左边）

我整理了一个用于嵌套 if-else 语句的函数。未针对速度进行优化。认为它可能对其他人有用。

ifelse_nested <- function(...) {
  args <- list(...)
  nargs <- length(args)
  
  default_ind <- nargs
  condition_inds <- which(seq_len(nargs) %% 2 == 1)
  condition_inds <- condition_inds[-length(condition_inds)] # remove default_ind
  value_inds <- which(seq_len(nargs) %% 2 == 0)
  
  .init <- args[[default_ind]]
  .x <- mapply(
    function(icond_ind, ivalue_ind) {
      return(list(condition=args[[icond_ind]], value=args[[ivalue_ind]]))
    }
    , icond_ind=condition_inds
    , ivalue_ind=value_inds
    , SIMPLIFY = FALSE
  ) # generate pairs of conditions & resulting-values
  
  out <- Reduce(
    function(x, y) ifelse(x$condition, x$value, y)
    , x = .x
    , init=.init
    , right=TRUE
  )
  
  return(out)
}

例如：

x <- seq_len(10)
ifelse_nested(x%%2==0, 2,x%%3==0, x^2, 0)