Syntax in R, Overtime Pay - r

Question

I am an intro into computer science student and have learned more on how to use python and am now learning R. I'm not used to R, and I've figured out how to calculate overtime pay, but I am not sure what is wrong with my syntax:
computePay <- function(pay,hours){
}if (hours)>=40{
newpay = 40-hours
total=pay*1.5
return(pay*40)+newpay*total
}else{
return (pay * hours)
}
How would I code this correctly?

Answer 1

Without looking at things like vectorization, a direct correction of your function would look something like:
computePay <- function(pay,hours) {
if (hours >= 40) {
newpay = hours - 40
total = pay * 1.5
return(pay*40 + newpay*total)
} else {
return(pay * hours)
}
}
This supports calling the function with a single pay and a single hours. You mis-calculated newpay (which really should be named something overhours), I corrected it.
You may hear people talk about "avoiding magic constants". A "magic constant" is a hard-coded number within code that is not perfectly clear and/or might be useful to allow the caller to modify. For instance, in some contracts it might be that overtime starts at a number other than 40, so that might be configurable. You can do that by changing the formals to:
computePay <- function(pay, hours, overtime_hours = 40, overtime_factor = 1.5)
and using those variables instead of hard-coded numbers. This allows the user to specify other values, but if not provided then they resort to sane defaults.
Furthermore, it might be useful to call it with a vector of one or the other, in which case the current function will fail because if (hours >= 40) needs a single logical value, but (e.g.) c(40,50) >= 40 returns a logical vector of length 2. We do this by introducing the ifelse function. Though it has some gotchas in advanced usage, it should work just fine here:
computePay1 <- function(pay, hours, overtime_hours = 40, overtime_factor = 1.5) {
ifelse(hours >= overtime_hours,
overtime_hours * pay + (hours - overtime_hours) * overtime_factor * pay,
pay * hours)
}
Because of some gotchas and deep-nested readability (I've seen ifelse stacked 12 levels deep), some people prefer other solutions. If you look at it closer, you may find that you can take further advantage of vectorization and pmax which is max applied piece-wise over each element. (Note the difference of max(c(1,3,5), c(2,4,4)) versus pmax(c(1,3,5), c(2,4,4)).)
Try something like this:
computePay2 <- function(pay, hours, overtime_hours = 40, overtime_factor = 1.5) {
pmax(0, hours - overtime_hours) * overtime_factor * pay +
pmin(hours, overtime_hours) * pay
}
To show how this works, I'll expand the pmax and pmin components:
hours <- c(20, 39, 41, 50)
overtime_hours <- 40
pmax(0, hours - overtime_hours)
# [1] 0 0 1 10
pmin(hours, overtime_hours)
# [1] 20 39 40 40
The rest sorts itself out.

Answer 2

Your "newpay*total" expression is outside the return command. You need put it inside the parentheses. The end bracket at the beginning of the second line should be moved to the last line. You also should have "(hours>=40)" rather than "(hours)>=40". Stylistically, the variable names are poorly chosen and there's no indentation (this might have helped you notice the misplaced bracket). Also, the calculation can be simplified:
total_pay = hourly_wage*(hours+max(0,hours-40)/2))
For every hour you work, you get your hourly wage. For every hour over 40 hours, you get your hourly wage plus half your hourly wage. So the total pay is wage*(total hours + (hours over 40)/2). Hours over 40 is either going to be total hours minus 40, or zero, whichever is larger.