我还在学习如何将SAS代码翻译成R，然后收到警告。我需要了解我犯错误的地方。我想要做的是创建一个变量来总结和区分人口的3种状态：大陆，海外，外国人。 我有一个包含2个变量的数据库：
idnat（法国人，外国人），如果idnat是法国人，那么：
idbp（大陆，殖民地，海外）我想将idnat和idbp中的信息汇总成一个名为idnat2的新变量：
所有这些变量都使用“字符类型”。
列idnat2中的预期结果：
idnat idbp idnat2
1 french mainland mainland
2 french colony overseas
3 french overseas overseas
4 foreign foreign foreign
这是我要在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;
这是我在R中的尝试：
if(idnat=="french"){
idnat2 <- "mainland"
} else if(idbp=="overseas"|idbp=="colony"){
idnat2 <- "overseas"
} else {
idnat2 <- "foreigner"
}
我收到此警告：
Warning message:
In if (idnat=="french") { :
the condition has length > 1 and only the first element will be used
我被建议使用“嵌套ifelse”代替其容易，但会收到更多警告：
idnat2 <- ifelse (idnat=="french", "mainland",
ifelse (idbp=="overseas"|idbp=="colony", "overseas")
)
else (idnat2 <- "foreigner")
根据警告信息，长度大于1，因此只考虑第一个括号之间的长度。对不起，但我不明白这个长度与这里有什么关系？谁知道我哪里错了？
答案 0 :(得分：95)
如果您使用的是任何电子表格应用程序，则基本函数if()的语法为：
if(<condition>, <yes>, <no>)
R中的ifelse()语法完全相同：
ifelse(<condition>, <yes>, <no>)
电子表格应用程序中if()的唯一区别是R ifelse()被向量化（将向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，我们希望比较一个＆gt; b如果是则返回1，否则返回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idnat2，您可以：
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")
)
什么是the condition has length > 1 and only the first element will be used？我们来看看：
> # 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
我还可以使用if()吗？是的，你可以，但语法不是很酷：）
test <- function(x) {
if(x=="french") {
"french"
} else{
"not really french"
}
}
apply(array(df[["idnat"]]),MARGIN=1, FUN=test)
答案 1 :(得分：10)
尝试以下内容：
# 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结构。在R中，if和else没有矢量化，这意味着它们会检查单个条件是否为真（即if("french"=="french")有效）并且无法处理多个逻辑（即if(c("french","foreigner")=="french")不起作用）R会给你你收到的警告。
相比之下，ifelse是矢量化的，所以它可以采用你的矢量（也就是输入变量）并测试每个元素的逻辑条件，就像你在SAS中习惯的那样。另一种包围你的方法是使用if和else语句构建一个循环（正如你在这里开始做的那样）但是矢量化ifelse方法会更多效率高，通常涉及的代码较少。
答案 2 :(得分：8)
如果数据集包含许多行，则使用data.table而不是嵌套ifelse()加入查找表可能更有效。
提供下面的查找表
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执行更新：
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
答案 3 :(得分：7)
您可以创建不含idnat2和if的向量ifelse。
函数replace可用于将所有"colony"替换为"overseas"：
idnat2 <- replace(idbp, idbp == "colony", "overseas")
答案 4 :(得分：3)
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))
<强> sqldf
library(sqldf)
sqldf("SELECT idnat, idbp,
CASE
WHEN idbp IN ('colony', 'overseas') THEN 'overseas'
ELSE idbp
END AS idnat2
FROM df")
<强> dplyr
library(dplyr)
df %>%
mutate(idnat2 = case_when(.$idbp == 'mainland' ~ "mainland",
.$idbp %in% c("colony", "overseas") ~ "overseas",
TRUE ~ "foreign"))
<强>输出
idnat idbp idnat2
1 french mainland mainland
2 french colony overseas
3 french overseas overseas
4 foreign foreign foreign
答案 5 :(得分：1)
使用data.table，解决方案是：
DT[, idnat2 := ifelse(idbp %in% "foreign", "foreign",
ifelse(idbp %in% c("colony", "overseas"), "overseas", "mainland" ))]
ifelse是矢量化的。 if-else不是。{1}}。这里，DT是：
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
答案 6 :(得分：0)
# 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)
答案 7 :(得分：0)
使用示例进行解释是帮助我解决问题的关键，但是我遇到的问题是当我复制它时不起作用，因此我不得不以多种方式使其混乱以使其正常工作。 （我是R的超级新手，由于缺乏知识，我对第三个ifelse遇到了一些问题。）
所以对于那些R新手来说会遇到问题...
ifelse(x < -2,"pretty negative", ifelse(x < 1,"close to zero", ifelse(x < 3,"in [1, 3)","large")##all one line
)#normal tab
)
（我在函数中使用了它，因此将“ ifelse ...”标记为一个，但是最后一个“）”完全在左侧）
答案 8 :(得分：-1)
很抱歉加入聚会太晚了。这是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)