What is the equivalent of Stata function inlist() in R?

I think you want %in%:

statevec <- c("NC","AZ","TX","NY","MA","CA","NJ")state <- c("AZ","VT")state %in% statevec ## TRUE FALSEagevec <- c(16, 24, 45, 54, 67,74, 78, 79, 85) age <- c(34,45)age %in% agevec ## FALSE TRUE

edit: working on updated question.

Copying from @NickCox's link:

inlist(z,a,b,...)      Domain:       all reals or all strings      Range:        0 or 1      Description:  returns 1 if z is a member of the remaining arguments;                        otherwise, returns 0.  All arguments must be reals                        or all must be strings.  The number of arguments is                        between 2 and 255 for reals and between 2 and 10 for                        strings.

However, I'm not quite sure how this matches up with the original question. I don't know Stata well enough to know if z can be a vector or not: it doesn't sound that way, in which case the original question (considering z=state as a vector) doesn't make sense. If we consider that it can be a vector then the answer would be as.numeric(state %in% statevec) -- I think.

Edit: Update by Ananda

Using your updated data, here's one approach, again using %in%:

data <- data.frame(age=0:10)within(data, {    m <- as.numeric(!age %in% c(1, 7, 9))})   age m1    0 12    1 03    2 14    3 15    4 16    5 17    6 18    7 09    8 110   9 011  10 1

This matches your expected output, by using ! (NOT) to invert the sense of %in%. It seems to be a little backwards from the way I would think about it (normally, 0=FALSE="is not in the list" and 1=TRUE="is in the list") and my reading of Stata's definition, but if it's what you want ...

Or one can use ifelse for more potential flexibility (i.e. values other than 0/1): substitute within(data, { m <- ifelse(age %in% c(1, 7, 9),0,1)}) in the code above.