Writing a median function in R - r

Question

I have been tasked to write my own median function in R, without using the built-in median function. If the numbers are odd; calculate the two middle values, as is usual concerning the median value.
Something i probably could do in Java, but I struggle with some of the syntax in
R Code:
list1 <- c(7, 24, 9, 42, 12, 88, 91, 131, 47, 71)
sorted=list1[order(list1)]
sorted
n = length(sorted)
n
if(n%2==0) # problem here, implementing mod() and the rest of logic.

Answer 1

Here is a self-written function mymedian:
mymedian <- function(lst) {
n <- length(lst)
s <- sort(lst)
ifelse(n%%2==1,s[(n+1)/2],mean(s[n/2+0:1]))
}
Example
list1 <- c(7, 24, 9, 42, 12, 88, 91, 131, 47, 71)
list2 <- c(7, 24, 9, 42, 12, 88, 91, 131, 47)
mymedian(list1)
mymedian(list2)
such that
> mymedian(list1)
[1] 44.5
> mymedian(list2)
[1] 42

Answer 2

I believe this should get you the median you're looking for:
homemade_median <- function(vec){
sorted <- sort(vec)
n <- length(sorted)
if(n %% 2 == 0){
mid <- sorted[c(floor(n/2),floor(n/2)+1)]
med <- sum(mid)/2
} else {
med <- sorted[ceiling(n/2)]
}
med
}
homemade_median(list1)
median(list1) # for comparison

Answer 3

A short function that does the trick:
my_median <- function(x){
# Order Vector ascending
x <- sort(x)
# For even lenght average the value of the surrounding numbers
if((length(x) %% 2) == 0){
return((x[length(x)/2] + x[length(x)/2 + 1]) / 2)
}
# For uneven lenght just take the value thats right in the center
else{
return(x[(length(x)/2) + 0.5])
}
}
Check to see if it returns desired outcomes:
my_median(list1)
44.5
median(list1)
44.5
#
list2 <- c(1,4,5,90,18)
my_median(list2)
5
median(list2)
5

Answer 4

You don't need to test for evenness, you can just create a sequence from half the length plus one, using floor and ceiling as appriopriate:
x <- rnorm(100)
y <- rnorm(101)
my_median <- function(x)
{
mid <- seq(floor((length(x)+1)/2),ceiling((length(x)+1)/2))
mean(sort(x)[mid])
}
my_median(x)
[1] 0.1682606
median(x)
[1] 0.1682606
my_median(y)
[1] 0.2473015
median(y)
[1] 0.2473015