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 <-  c(10, 10, 10, 5, 5, 20, 20)
rank(x)
# [1] 4.0 4.0 4.0 1.5 1.5 6.5 6.5

In this case the desired result would be:

my_rank(x)
[1] 2 2 2 1 1 3 3

I've played with all the options for ties.method option (average, max, min, random), none of which are designed to provide the desired result.

Is it possible to acheive this with the rank() function?

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

The "loopless" way to do it is to simply treat the vector as an ordered factor, then convert it to numeric:

> as.numeric( ordered( c( 10,10,10,10, 5,5,5, 10, 10 ) ) )
[1] 2 2 2 2 1 1 1 2 2
> as.numeric( ordered( c(0.5,0.56,0.76,0.23,0.33,0.4) ))
[1] 4 5 6 1 2 3
> as.numeric( ordered( c(1,1,2,3,4,5,8,8) ))
[1] 1 1 2 3 4 5 6 6

Update: Another way, that seems faster is to use findInterval and sort(unique()):

> x <- c( 10, 10, 10, 10, 5,5,5, 10, 10)
> findInterval( x, sort(unique(x)))
[1] 2 2 2 2 1 1 1 2 2

> x <- round( abs( rnorm(1000000)*10))
> system.time( z <- as.numeric( ordered( x )))
   user  system elapsed 
  0.996   0.025   1.021 
> system.time( z <- findInterval( x, sort(unique(x))))
   user  system elapsed 
  0.077   0.003   0.080