From: "Jesús Gabriel y Galán" <jgabrielygalan@...>
Date: 2012-11-27T02:30:05+09:00
Subject: Re: Algorithm for choosing object based on priority

On Mon, Nov 26, 2012 at 6:10 PM, Damjan Rems <lists@ruby-forum.com> wrote:
> Thanks so much. I guess I was over-complicating.

There could be one complication depending on the requirements. If you
have to guarantee that for every N ads you show there is exactly one
time ad1, 5 times ad2, etc summing up to N. Then the best way would be
to have a hash of counts, choose a random key and reduce the count,
removing the entry if the count is 0:

ad_count  = {"ad1" => 1, "ad2" => 5, "ad3" => 3}
N = ad_count.inject(0) {|total, (k,v)| total + v}
N.times do
  key = ad_count.keys[rand(ad_count.size)]
  puts key
  ad_count[key] -= 1
  ad_count.delete(key) if ad_count[key] <= 0
end

ad2
ad2
ad3
ad1
ad3
ad3
ad2
ad2
ad2

You could clone the hash and reassing it every time it's empty or
something like that.

Jesus.