Continuous Error in argument "x" is missing, with no default - r

Question

I am trying to build a function that takes a numeric vector of homework scores (of length n), and an optional logical argument drop, to compute a single homework value. If drop = TRUE, the lowest HW score must be dropped.
step1 function to get average
get_average <- function(x,na.rm=TRUE) {
if(na.rm==TRUE){
x = remove_missing(x)}
total <- 0
for (n in 1:length(x)) {
total= total + x[n]
}
return(total/length(x))
}
put it all together
score_homework <- function(x,drop=TRUE)
{
if(drop==TRUE)
x = drop_lowest(x)
{get_average(x)}}
However I keep getting the error Error in score_homework() : argument "x" is missing, with no default

Answer 1

I'm not sure this is what you want, but here goes.
First generate some dummy data:
# Set seed
set.seed(1234)
# Generate dummy homework data with <NA> values
homework <- c(rep(NA, 20), rnorm(n = 100, mean = 50, sd = 10))
# Have a quick look
hist(homework)
Then we write the function:
# Make function
homework_func <- function(data, drop = TRUE) {
# Remove NA
data <- data[!is.na(data)]
# Calculate the average depending on whether 'drop' is T or F
if(drop == TRUE) {
data <- data[data > min(data)]
mean(data)
} else {
mean(data)
}
}
# Use function with 'drop = TRUE'
homework_func(data = homework, drop = TRUE)
#> [1] 48.65349
# Use function with 'drop = FALSE'
homework_func(data = homework, drop = FALSE)
#> [1] 48.43238

Answer 2

Here is a function to eliminate the lowest missing score that's less complicated than the version in the original post. I sort the scores in descending order in case the there is a tie for the lowest score. In that case, we should only remove one instance of the lowest score. Also, you're really better off using R's mean() function than writing your own.
scores <- c(78,93,61,NA,61,83,92,95,NA,100)
removeMinScore <- function(x) {
x <- x[order(-x)] # order descending
x <- x[!is.na(x)] # remove NAs
x[1:length(x)-1] # return all but lowest score, removes only 1 tied value
}
That said, if you must write your own version of mean(), here is a simpler approach that takes advantage of existing R functions.
TIP: Since is.na() returns a vector of TRUE and FALSE values, you can sum these to count the number of non-missing values in a vector.
mymean <- function(x) {sum(x, na.rm=TRUE) / sum(!is.na(x))}
The results look like this.
The modified version of score_homework() would be:
score_homework <- function(x,drop=TRUE){
if(drop == TRUE) return mean(removeMinScore(x),na.rm=TRUE)
else mean(x,na.rm=TRUE)
}
The results from testing the function are as follows.