Nested ifelse statement

Question

I'm still learning how to translate a SAS code into R and I get warnings. I need to understand where I'm making mistakes. What I want to do is create a variable which summarizes and differentiates 3 status of a population: mainland, overseas, foreigner. I have a database with 2 variables:

• id nationality: idnat (french, foreigner),

If idnat is french then:

• id birthplace: idbp (mainland, colony, overseas)

I want to summarize the info from idnat and idbp into a new variable called idnat2:

• status: k (mainland, overseas, foreigner)

All these variables use "character type".

Results expected in column idnat2 :

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

---

Here is my SAS code I want to translate in R:

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

---

Here is my attempt in R:

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

I receive this warning:

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

I was advised to use a "nested ifelse" instead for its easiness but get more warnings:

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

According to the Warning message, the length is greater than 1 so only what's between the first brackets will be taken into account. Sorry but I don't understand what this length has to do with here? Anybody know where I'm wrong?

Answer 1

Using the SQL CASE statement with the dplyr and sqldf packages:

Data

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")

dplyr

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

Output

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

Answer 2

If the data set contains many rows it might be more efficient to join with a lookup table using data.table instead of nested ifelse().

Provided the lookup table below

lookup

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

and a sample data set

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

then we can do an update while joining:

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

Answer 3

# 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) 

Answer 4

If you are using any spreadsheet application there is a basic function if() with syntax:

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

Syntax is exactly the same for ifelse() in R:

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

The only difference to if() in spreadsheet application is that R ifelse() is vectorized (takes vectors as input and return vector on output). Consider the following comparison of formulas in spreadsheet application and in R for an example where we would like to compare if a > b and return 1 if yes and 0 if not.

In spreadsheet:

  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)

In R:

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

ifelse() can be nested in many ways:

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>)
             )
       )

To calculate column idnat2 you can:

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 Documentation

What is the condition has length > 1 and only the first element will be used? Let's see:

> # 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

Can I still use if()? Yes, you can, but the syntax is not so cool :)

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

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

If you are familiar with SQL, you can also use CASE statement in sqldf package.

Answer 5

You can create the vector idnat2 without if and ifelse.

The function replace can be used to replace all occurrences of "colony" with "overseas":

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

Answer 6

With data.table, the solutions is:

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

The ifelse is vectorized. The if-else is not. Here, DT is:

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

This gives:

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