Rank a vector based on order and replace ties with their average

Question

I'm new to R, and i find it quite interesting.

I have MATLAB code to rank a vector based on order which works fine. Now I want to convert it to R code, a typical spearman ranking with ties:

# MATLAB CODE

function r=drank(x)

u = unique(x);
[xs,z1] = sort(x);
[z1,z2] = sort(z1);
r = (1:length(x))';
r=r(z2);

for i=1:length(u)

    s=find(u(i)==x);

    r(s,1) = mean(r(s));

end

This is what i tried:

# R CODE

x = c(10.5, 8.2, 11.3, 9.1, 13.0, 11.3, 8.2, 10.1)

drank <- function(x){

    u = unique(x)
    xs = order(x)
    r=r[xs]

    for(i in 1:length(u)){
        s=which(u[i]==x)
        r[i] = mean(r[s])  
    }

    return(r)
}

r <- drank(x)

Results:

r = 5, 1.5, 6.5, 3, 8, 6.5, 1.5, 4

1.5 is average of 8.2 occurring twice ie. tie

6.5 is average of 11.3 occurring twice

Can anyone help me check it?

Thanks,

Answer 1

R has a built-in function for ranking, called rank() and it gives precisely what you are looking for. rank has the argument ties.method, "a character string specifying how ties are treated", which defaults to "average", i.e. replaces ties by their mean.

x = c(10.5, 8.2, 11.3, 9.1, 13.0, 11.3, 8.2, 10.1)
expected <- c(5, 1.5, 6.5, 3, 8, 6.5, 1.5, 4)

rank(x)
# [1] 5.0 1.5 6.5 3.0 8.0 6.5 1.5 4.0

identical(expected, rank(x))
# [1] TRUE