How to Generate a Random Number Between a and B in Ruby

How to generate a random number between a and b in Ruby?

UPDATE: Ruby 1.9.3 Kernel#rand also accepts ranges

rand(a..b)

http://www.rubyinside.com/ruby-1-9-3-introduction-and-changes-5428.html

Converting to array may be too expensive, and it's unnecessary.

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(a..b).to_a.sample

Or

[*a..b].sample

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Array#sample

Standard in Ruby 1.8.7+.

Note: was named #choice in 1.8.7 and renamed in later versions.

But anyway, generating array need resources, and solution you already wrote is the best, you can do.

How to get a random number in Ruby

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.

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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.

Trying to generate random number between a range based on an input, returns number outside range

if example_range = 11 || 12

is an assigning, resulting in example_range variable assigned with value 11.
Since this condition is truthy your program does not go further and returns

rand(212.502..255)

Double equal sign (==) is used for comparison.

In your case you want to use something like

if [11, 12].include?(example_range)

P.S.

example_range = gets.chomp

returns a String object, not Integer - you need to convert the example_range (using to_i) before comparing with integers.

Generate a random number with a seed between a range in ruby

You can create a special/designated random number generator with any seed value you like:

special = Random.new 42 # create a new instance of Random seeded with 42
20.times { p special.rand(5..10) } # generate 20 random ints in range 5 to 10

Your special instance of Random is independent of kernel#rand unless you use srand to initialize it with the same seed value.

Ruby on Rails - Generating random number that does not exist in database

To generate a random number and ensuring that the number doesn't exist in the database I would do something like this:

before_create :assign_unique_case_number

validates! :case_number, uniqueness: true

private

CASE_NUMBER_RANGE = (10_000..99_999)

def assign_unique_case_number
  self.case_number = loop do
    number = rand(CASE_NUMBER_RANGE)
    break number unless Case.exists?(case_number: number)
  end
end

Please note that the more case there are in the database the longer it might take to find an unused number. Therefore I suggest using greater numbers right from the start. Greater numbers have another advantage: They are harder to guess what might or might not be important in your application.

Furthermore: Rails cannot guarantee that uniqueness in the database. There might be race conditions that lead to duplicates. The only way to avoid that is to add a unique index to the database column in a migration like this:

add_index :cases, :case_number, unique: true

Ruby: Generate random integers that add up to another number

Edit: see this answer

Just for fun (I'm sure it's not 100% for negatives etc...) :

class Fixnum
  def rand_sum(n = 2)
    arr = (n - 1).times.reduce([]) do |a, _|
      curr_max = self - a.reduce(0, :+)
      a << rand(0..curr_max)
    end

    arr << self - arr.reduce(0, :+)
  end
end

 11.rand_sum
 => [6, 5] 
 23.rand_sum 3
 => [10, 6, 7] 
 11.rand_sum 3
 => [6, 2, 3] 

Get rand number from range

The to_a method takes a lot of time to generate the array, which you don't need.

Just use:

rand(1..1000000000000000)

Negative random numbers

Let's say you want to generate a number between a and b you can always do that using this formula:

randomNum = (b-a)*prng.rand + a

So if you want a number between -8 and +7 for example then a=-8 and b=7 and your code would be

randomNum = (7-(-8))*prng.rand + (-8)

which is equivalent to

randomNum=15*prng.rand - 8

Generate Random Numbers Until a certain

The problem for this code is that you generate an infinite loop because num is always "", you need to update the variable num inside the while loop like this:

num = ""
while num != 7
  num = rand(1..10)
  puts num
end
puts "over"

in this case, you do not know when the loop will stop, but in each loop it can stop with probability 1/10, this is not a good practice, you nedd to know when the loop stops, so will be better to add a variable for maximum number of tries like this. And also you do not need to transform dthe number to string to print it:

num = 0
max_tries = 100
try = 1
guess = 7

while (try < max_tries && num != guess)
  num = rand(1..10)
  try = try + 1
  puts num
end

if (try >= max_tries)
  puts "you do not get the guess"
else
  puts "over"
end

and you get this:

[4] pry(main)> 9
9
3
4
9
6
9
8
4
5
9
10
4
3
7
over
=> true

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