红宝石选择与兰特方法随机几个选项

Question

我的代码：

alea = ["x", " "] 
num = alea.length 
choice = rand(num) 
veinte = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 

for i in veinte 
    puts alea[choice] 
end 

我希望我的代码中随机选择几个选项Y没有唯一的一个。例如：

x 

x 
x 

x 
x 
x 

我该怎么做？

Answer 1

Array#sample被发明出来用于此目的：

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10].map do |elem| 
    [elem, [true, false].sample] 
end.to_h 
#⇒ { 
# 1 => true, 
# 10 => false, 
# 2 => true, 
# 3 => true, 
# 4 => false, 
# 5 => true, 
# 6 => false, 
# 7 => true, 
# 8 => false, 
# 9 => false 
#} 

Answer 2

只需编辑代码：

alea = ["x", " "] 
num = alea.length 
choice = rand(num) 
veinte = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 
veinte.map { puts alea[rand(alea.length)] } 

Answer 3

你所得到的输出根据你所编写的代码是正确的：

alea = ["x", " "] # => Array on only two elements 
num = alea.length # => 2 
choice = rand(num) # => rand(2) will only output either 0 or 1 which will always output either alea[0] or alea[1] which is " " or "x" 

如果你真的想要一个数组的随机值，然后我们e @mudasobwa建议的Array类的示例方法

Answer 4

看起来您只需要一定数量的随机选择的项目。在这种情况下，只需使用块形式的Array构造函数：

Array.new(10){ ["x", " "].sample } 

#=>[" ", " ", "x", " ", "x", " ", "x", " ", "x", "x"]