How to get ranks with no gaps when there are ties among values?

Question

When there are ties in the original data, is there a way to create a ranking without gaps in the ranks (consecutive, integer rank values)? Suppose:

x <-           


        

Answer 1

Modified crayola solution but using match instead of merge:

x_unique <- unique(x)
x_ranks <- rank(x_unique)
x_ranks[match(x,x_unique)]

edit

or in a one-liner, as per @hadley 's comment:

match(x, sort(unique(x)))

Answer 2

Another function that does this, but it seems inefficient. There is no for loop, but I doubt it is more efficient than Sacha's suggestion!

x=c(1,1,2,3,4,5,8,8)
fancy.rank <- function(x) {
    x.unique <- unique(x)
    d1 <- data.frame(x=x)
    d2 <- data.frame(x=x.unique, rank(x.unique))
    merge(d1, d2, by="x")[,2]
}

fancy.rank(x)

[1] 1 1 2 3 4 5 6 6

Answer 3

try to think about another way

x <-  c(10,10,10,5,5,20,20)
as.numeric(as.factor(x))
[1] 2 2 2 1 1 3 3

Answer 4

If you don't mind leaving base-R:

library(data.table)
frank(x, ties.method = "dense")
[1] 2 2 2 1 1 3 3

data:

x <-  c(10, 10, 10, 5, 5, 20, 20)

Answer 5

What about sort()?

x <- c(1,1,2,3,4,5)
sort(x)

> sort(x) 
[1] 1 1 2 3 4 5

Answer 6

I can think of a quick function to do this. It's not optimal with a for loop but it works:)

x=c(1,1,2,3,4,5,8,8)

foo <- function(x){
    su=sort(unique(x))
    for (i in 1:length(su)) x[x==su[i]] = i
    return(x)
}

foo(x)

[1] 1 1 2 3 4 5 6 6