How to get a random number in Ruby

Question

How do I generate a random number between 0 and n?

Answer 1

Use rand(range)

From Ruby Random Numbers:

If you needed a random integer to simulate a roll of a six-sided die, you'd use: 1 + rand(6). A roll in craps could be simulated with 2 + rand(6) + rand(6).

Finally, if you just need a random float, just call rand with no arguments.

---

As Marc-André Lafortune mentions in his answer below (go upvote it), Ruby 1.9.2 has its own Random class (that Marc-André himself helped to debug, hence the 1.9.2 target for that feature).

For instance, in this game where you need to guess 10 numbers, you can initialize them with:

10.times.map{ 20 + Random.rand(11) } 
#=> [26, 26, 22, 20, 30, 26, 23, 23, 25, 22]

Note:

• Using Random.new.rand(20..30) (using Random.new) generally would not be a good idea, as explained in detail (again) by Marc-André Lafortune, in his answer (again).

• But if you don't use Random.new, then the class method rand only takes a max value, not a Range, as banister (energetically) points out in the comment (and as documented in the docs for Random). Only the instance method can take a Range, as illustrated by generate a random number with 7 digits.

This is why the equivalent of Random.new.rand(20..30) would be 20 + Random.rand(11), since Random.rand(int) returns “a random integer greater than or equal to zero and less than the argument.” 20..30 includes 30, I need to come up with a random number between 0 and 11, excluding 11.

Answer 2

While you can use rand(42-10) + 10 to get a random number between 10 and 42 (where 10 is inclusive and 42 exclusive), there's a better way since Ruby 1.9.3, where you are able to call:

rand(10...42) # => 13

Available for all versions of Ruby by requiring my backports gem.

Ruby 1.9.2 also introduced the Random class so you can create your own random number generator objects and has a nice API:

r = Random.new
r.rand(10...42) # => 22
r.bytes(3) # => "rnd"

The Random class itself acts as a random generator, so you call directly:

Random.rand(10...42) # => same as rand(10...42)

Notes on Random.new

In most cases, the simplest is to use rand or Random.rand. Creating a new random generator each time you want a random number is a really bad idea. If you do this, you will get the random properties of the initial seeding algorithm which are atrocious compared to the properties of the random generator itself.

If you use Random.new, you should thus call it as rarely as possible, for example once as MyApp::Random = Random.new and use it everywhere else.

The cases where Random.new is helpful are the following:

• you are writing a gem and don't want to interfere with the sequence of rand/Random.rand that the main programs might be relying on
• you want separate reproducible sequences of random numbers (say one per thread)
• you want to be able to save and resume a reproducible sequence of random numbers (easy as Random objects can marshalled)

Answer 3

If you're not only seeking for a number but also hex or uuid it's worth mentioning that the SecureRandom module found its way from ActiveSupport to the ruby core in 1.9.2+. So without the need for a full blown framework:

require 'securerandom'

p SecureRandom.random_number(100) #=> 15
p SecureRandom.random_number(100) #=> 88

p SecureRandom.random_number #=> 0.596506046187744
p SecureRandom.random_number #=> 0.350621695741409

p SecureRandom.hex #=> "eb693ec8252cd630102fd0d0fb7c3485"

It's documented here: Ruby 1.9.3 - Module: SecureRandom (lib/securerandom.rb)

Answer 4

You can generate a random number with the rand method. The argument passed to the rand method should be an integer or a range, and returns a corresponding random number within the range:

rand(9)       # this generates a number between 0 to 8
rand(0 .. 9)  # this generates a number between 0 to 9
rand(1 .. 50) # this generates a number between 1 to 50
#rand(m .. n) # m is the start of the number range, n is the end of number range

Answer 5

Well, I figured it out. Apparently there is a builtin (?) function called rand:

rand(n + 1)

If someone answers with a more detailed answer, I'll mark that as the correct answer.

Answer 6

What about this?

n = 3
(0..n).to_a.sample

Answer 7

Simplest answer to the question:

rand(0..n)

Answer 8

You can simply use random_number.

If a positive integer is given as n, random_number returns an integer: 0 <= random_number < n.

Use it like this:

any_number = SecureRandom.random_number(100) 

The output will be any number between 0 and 100.

Answer 9

rand(6)    #=> gives a random number between 0 and 6 inclusively 
rand(1..6) #=> gives a random number between 1 and 6 inclusively

Note that the range option is only available in newer(1.9+ I believe) versions of ruby.

Answer 10

range = 10..50

rand(range)

or

range.to_a.sample

or

range.to_a.shuffle(this will shuffle whole array and you can pick a random number by first or last or any from this array to pick random one)

Answer 11

you can do rand(range)

x = rand(1..5)

Answer 12

This link is going to be helpful regarding this;

http://ruby-doc.org/core-1.9.3/Random.html

And some more clarity below over the random numbers in ruby;

Generate an integer from 0 to 10

puts (rand() * 10).to_i

Generate a number from 0 to 10 In a more readable way

puts rand(10)

Generate a number from 10 to 15 Including 15

puts rand(10..15)

Non-Random Random Numbers

Generate the same sequence of numbers every time the program is run

srand(5)

Generate 10 random numbers

puts (0..10).map{rand(0..10)}

Answer 13

Easy way to get random number in ruby is,

def random    
  (1..10).to_a.sample.to_s
end

Answer 14

Maybe it help you. I use this in my app

https://github.com/rubyworks/facets
class String

  # Create a random String of given length, using given character set
  #
  # Character set is an Array which can contain Ranges, Arrays, Characters
  #
  # Examples
  #
  #     String.random
  #     => "D9DxFIaqR3dr8Ct1AfmFxHxqGsmA4Oz3"
  #
  #     String.random(10)
  #     => "t8BIna341S"
  #
  #     String.random(10, ['a'..'z'])
  #     => "nstpvixfri"
  #
  #     String.random(10, ['0'..'9'] )
  #     => "0982541042"
  #
  #     String.random(10, ['0'..'9','A'..'F'] )
  #     => "3EBF48AD3D"
  #
  #     BASE64_CHAR_SET =  ["A".."Z", "a".."z", "0".."9", '_', '-']
  #     String.random(10, BASE64_CHAR_SET)
  #     => "xM_1t3qcNn"
  #
  #     SPECIAL_CHARS = ["!", "@", "#", "$", "%", "^", "&", "*", "(", ")", "-", "_", "=", "+", "|", "/", "?", ".", ",", ";", ":", "~", "`", "[", "]", "{", "}", "<", ">"]
  #     BASE91_CHAR_SET =  ["A".."Z", "a".."z", "0".."9", SPECIAL_CHARS]
  #     String.random(10, BASE91_CHAR_SET)
  #      => "S(Z]z,J{v;"
  #
  # CREDIT: Tilo Sloboda
  #
  # SEE: https://gist.github.com/tilo/3ee8d94871d30416feba
  #
  # TODO: Move to random.rb in standard library?

  def self.random(len=32, character_set = ["A".."Z", "a".."z", "0".."9"])
    chars = character_set.map{|x| x.is_a?(Range) ? x.to_a : x }.flatten
    Array.new(len){ chars.sample }.join
  end

end

https://github.com/rubyworks/facets/blob/5569b03b4c6fd25897444a266ffe25872284be2b/lib/core/facets/string/random.rb

It works fine for me

Answer 15

How about this one?

num = Random.new
num.rand(1..n)

Answer 16

Don't forget to seed the RNG with srand() first.

Answer 17

Try array#shuffle method for randomization

array = (1..10).to_a
array.shuffle.first

Answer 18

You can use ruby rand method for this like below:

rand(n+1)

You need to use n+1 as the rand method returns any random number greater than or equal to 0 but less than the passed parameter value.