问题
How do I generate a random number between 0
and n
?
回答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 with2 + 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)
(usingRandom.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 amax
value, not aRange
, as banister (energetically) points out in the comment (and as documented in the docs for Random). Only the instance method can take aRange
, 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.
回答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)
回答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)
回答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
回答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.
回答6:
What about this?
n = 3
(0..n).to_a.sample
回答7:
Simplest answer to the question:
rand(0..n)
回答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.
回答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.
回答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)
回答11:
you can do rand(range)
x = rand(1..5)
回答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)}
回答13:
Easy way to get random number in ruby is,
def random
(1..10).to_a.sample.to_s
end
回答14:
Try array#shuffle
method for randomization
array = (1..10).to_a
array.shuffle.first
回答15:
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
回答16:
How about this one?
num = Random.new
num.rand(1..n)
回答17:
Don't forget to seed the RNG with srand() first.
回答18:
use 'rand' function Just Like thisrand(10)
来源:https://stackoverflow.com/questions/198460/how-to-get-a-random-number-in-ruby