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 summ

Answer 1

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

if(, , )

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

ifelse(, , )

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(, , ifelse(, , ))

ifelse(, ifelse(, , ), )

ifelse(, 
       ifelse(, , ), 
       ifelse(, , )
      )

ifelse(, , 
       ifelse(, , 
              ifelse(, , )
             )
       )

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.