How to include NA in ifelse?

Question

I am trying to create a column ID based on logical statements for values of other columns. For example, in the following dataframe

test <- structure(list(time = c(10L, 20L, NA, 30L), type = structure(c(1L,  2L, 3L, NA), .Label = c("A", "B", "C"), class = "factor"), ID = c(NA,  "1", NA, NA)), .Names = c("time", "type", "ID"), row.names = c(NA,  -4L), class = "data.frame") 

which looks like

    time    type 1   10      A 2   20      B 3   NA      C 4   30      NA 

I want to make a new column ID containing a value of 1 for all time that are not NA and all type that are not A. I am using the following code for this:

test$ID <- ifelse(is.na(test$time) | test$type == "A", NA, "1") 

This gives the result as

    time    type    ID 1   10      A       NA 2   20      B       1 3   NA      C       NA 4   30      NA      NA 

However, this code ignores the NA in column type, resulting in a value of NA in column ID. I need this to be a value of 1, so my needed solution should give:

    time    type    ID 1   10      A       NA 2   20      B       1 3   NA      C       NA 4   30      NA      1 

Can anyone tell me how I might do this? I could get this to work with my existing code if I could somehow change the result of is.na(test$type) to return FALSE instead of TRUE, but I'm not sure how to do that. Or, maybe the structure of my existing code needs to be entirely changed? I appreciate any help!

Answer 1

You can't really compare NA with another value, so using == would not work. Consider the following:

NA == NA # [1] NA 

You can just change your comparison from == to %in%:

ifelse(is.na(test$time) | test$type %in% "A", NA, "1") # [1] NA  "1" NA  "1" 

---

Regarding your other question,

I could get this to work with my existing code if I could somehow change the result of is.na(test$type) to return FALSE instead of TRUE, but I'm not sure how to do that.

just use ! to negate the results:

!is.na(test$time) # [1]  TRUE  TRUE FALSE  TRUE