How do I generate a list of n unique random numbers in Ruby?

Question

This is what I have so far:

myArray.map!{ rand(max) } 

Obviously, however, sometimes the numbers in the list are not unique. How can I make sure my list only contains unique numbers without having to create a bigger list from which I then just pick the n unique numbers?

Edit:
I'd really like to see this done w/o loop - if at all possible.

Answer 1

(0..50).to_a.sort{ rand() - 0.5 }[0..x]  

(0..50).to_a can be replaced with any array. 0 is "minvalue", 50 is "max value" x is "how many values i want out"

of course, its impossible for x to be permitted to be greater than max-min :)

In expansion of how this works

(0..5).to_a  ==> [0,1,2,3,4,5] [0,1,2,3,4,5].sort{ -1 }  ==>  [0, 1, 2, 4, 3, 5]  # constant [0,1,2,3,4,5].sort{  1 }  ==>  [5, 3, 0, 4, 2, 1]  # constant [0,1,2,3,4,5].sort{ rand() - 0.5 }   ==>  [1, 5, 0, 3, 4, 2 ]  # random [1, 5, 0, 3, 4, 2 ][ 0..2 ]   ==>  [1, 5, 0 ] 

Footnotes:

It is worth mentioning that at the time this question was originally answered, September 2008, that Array#shuffle was either not available or not already known to me, hence the approximation in Array#sort

And there's a barrage of suggested edits to this as a result.

So:

.sort{ rand() - 0.5 } 

Can be better, and shorter expressed on modern ruby implementations using

.shuffle 

Additionally,

[0..x] 

Can be more obviously written with Array#take as:

.take(x) 

Thus, the easiest way to produce a sequence of random numbers on a modern ruby is:

(0..50).to_a.shuffle.take(x) 

Answer 2

This uses Set:

require 'set'  def rand_n(n, max)     randoms = Set.new     loop do         randoms << rand(max)         return randoms.to_a if randoms.size >= n     end end