views:

2162

answers:

11

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.

+1  A: 

You could use a hash to track the random numbers you've used so far:

seen = {}
max = 100
(1..10).map { |n|
  x = rand(max)
  while (seen[x]) 
    x = rand(max)
  end
  x
}
Kyle Burton
+1  A: 

Rather than add the items to a list/array, add them to a Set.

jon
+6  A: 

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
Ryan Leavengood
Any way to do this w/o looping? Any way to do it with a map?
Esteban Araya
A: 

Here is one solution:

Suppose you want these random numbers to be between r_min and r_max. For each element in your list, generate a random number r, and make list[i]=list[i-1]+r. This would give you random numbers which are monotonically increasing, guaranteeing uniqueness provided that

  • r+list[i-1] does not over flow
  • r > 0

For the first element, you would use r_min instead of list[i-1]. Once you are done, you can shuffle the list so the elements are not so obviously in order.

The only problem with this method is when you go over r_max and still have more elements to generate. In this case, you can reset r_min and r_max to 2 adjacent element you have already computed, and simply repeat the process. This effectively runs the same algorithm over an interval where there are no numbers already used. You can keep doing this until you have the list populated.

freespace
+3  A: 
(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 ]
Kent Fredric
A: 

If you have a finite list of possible random numbers (i.e. 1 to 100), then Kent's solution is good.

Otherwise there is no other good way to do it without looping. The problem is you MUST do a loop if you get a duplicate. My solution should be efficient and the looping should not be too much more than the size of your array (i.e. if you want 20 unique random numbers, it might take 25 iterations on average.) Though the number of iterations gets worse the more numbers you need and the smaller max is. Here is my above code modified to show how many iterations are needed for the given input:

require 'set'

def rand_n(n, max)
    randoms = Set.new
    i = 0
    loop do
     randoms << rand(max)
     break if randoms.size > n
     i += 1
    end
    puts "Took #{i} iterations for #{n} random numbers to a max of #{max}"
    return randoms.to_a
end

I could write this code to LOOK more like Array.map if you want :)

Ryan Leavengood
A: 

As far as it is nice to know in advance the maxium value, you can do this way:

class NoLoopRand
  def initialize(max)
    @deck = (0..max).to_a
  end

  def getrnd
    return @deck.delete_at(rand(@deck.length - 1))
  end
end

and you can obtain random data in this way:

aRndNum = NoLoopRand.new(10)
puts aRndNum.getrnd

you'll obtain nil when all the values will be exausted from the deck.

TuxmAL
+2  A: 

Just to give you an idea about speed, I ran four versions of this:

  1. Using Sets, like Ryan's suggestion.
  2. Using an Array slightly larger than necessary, then doing uniq! at the end.
  3. Using a Hash, like Kyle suggested.
  4. Creating an Array of the required size, then sorting it randomly, like Kent's suggestion (but without the extraneous "- 0.5", which does nothing).

They're all fast at small scales, so I had them each create a list of 1,000,000 numbers. Here are the times, in seconds:

  1. Sets: 628
  2. Array + uniq: 629
  3. Hash: 645
  4. fixed Array + sort: 8

And no, that last one is not a typo. So if you care about speed, and it's OK for the numbers to be integers from 0 to whatever, then my exact code was:

a = (0...1000000).sort_by{rand}
glenn mcdonald
A: 

How about a play on this? Unique random numbers without needing to use Set or Hash.

x = 0
(1..100).map{|iter| x += rand(100)}.shuffle
Sam Saffron
I somehow feel that these numbers will be significantly less random than by picking 100 unique ones from the range of 0 to 10000.
Claudiu
yerp, it needs to be improved, the higher the number the lower the odds you will get it. But surely there is a way of getting something along these lines to work.
Sam Saffron