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

(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)

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    endend

Ruby 1.9 offers the Array#sample method which returns an element, or elements randomly selected from an Array. The results of #sample won't include the same Array element twice.

(1..999).to_a.sample 5 # => [389, 30, 326, 946, 746]

When compared to the to_a.sort_by approach, the sample method appears to be significantly faster. In a simple scenario I compared sort_by to sample, and got the following results.

require 'benchmark'range = 0...1000000how_many = 5Benchmark.realtime do  range.to_a.sample(how_many)end=> 0.081083Benchmark.realtime do  (range).sort_by{rand}[0...how_many]end=> 2.907445