Question

我只是想创建一个用对象填充数组的函数，但是出了点问题：

row1 = []

class Tile
    def initialize(type)
        @type = type
    end
end

def FillRow1

    [1..10].each {
        random = rand(1..3)
        if random == 1 row1.(Tile.new("land") end 
        else if random == 2 row1.(Tile.new("Water") end
        else ifrandom == 3 row1.(Tile.new("empty") end
    }
    row1
end

Answer 1

您的语法错误

    [1..10].each {
        random = rand(1..3)
        if random == 1 then row1.push(Tile.new("land")) end 
        else if random == 2 then row1.push(Tile.new("Water")) end
        else ifrandom == 3 then row1.push(Tile.new("empty") end
    }

这样可行。

但更清洁的解决方案可能是：

types = ["Land","Water","Empty"]
10.times{ row1 << Tile.new(types[rand(0..2)]) }

Answer 2

如果你的第二个错误，那么随机和if之间必须有一个空格。

Answer 3

a= ["land", "water", "empty"]
data= (1..10).map{ a[rand(3)] }
p data

Answer 4

单行选项：

10.times.map{ Tile.new(["land","water","empty"][rand(3)]) }

Answer 5

Ruby的最新版本（＆gt; = 1.9.3）附带方法#sample。它是Array的一部分。您可以使用它来从数组中获取随机元素，甚至不知道数组甚至是多大。

class Tile
    TILE_TYPES = %w(land water empty)

    def initialize(type)
        @type = type
    end

    def to_s
        "Tile of type: #{@type}"
    end
end

# Generate 10 tiles
rows = (1..10).map do 
    Tile.new Tile::TILE_TYPES.sample
end

puts rows
#=> Tile of type: empty
#   Tile of type: land
#   ...

# If you want to pick more then 1 random element you can also do
puts Tile::TILE_TYPES.sample(3)

Answer 6

class Tile
  def initialize(type)
    @type = type
  end
end

types = %w(land water empty)

row = Array.new(10){ Tile.new(types.sample) }

简单阵列分配

6 个答案: