Question

我有一个简单的分配多个数组的要求。现在我正在做类似的事情。

   employees = Array.new
      projects = Array.new
      practices = Array.new
      entities = Array.new
      categories = Array.new
      groups = Array.new
      external_inputs = Array.new
      payrolls = Array.new

我希望employees, projects, practices, entities ...payrolls所有内容都应该在一行中分配给Array.new 请建议一些更好，更干净的方法。也许通过Metaprogramming。

Answer 1

employees, projects, practices, entities, categories, groups, external_inputs, payrolls = 
  8.times.map { [] }

Answer 2

丑陋但满足要求：

employees, projects, practices, entities, categories, groups, external_inputs, payrolls = Array.new(8) { [] }

Answer 3

我希望employees，projects，practices，entities，...，payrolls所有内容都应分配到Array.new线

这里你去：

employees = Array.new; projects = Array.new; practices = Array.new; entities = Array.new; categories = Array.new; groups = Array.new; external_inputs = Array.new; payrolls = Array.new

Voilà，就像您要求的那样，单行。

现在，为什么你有一个如此庞大，复杂的方法，它需要8个局部变量是另一个问题。

Answer 4

嗯，这不是一条线，但使用起来要干净得多：

employees        = []
projects         = []
practices        = []
entities         = []
categories       = []
groups           = []
external_inputs  = []
payrolls         = []

但是，如果你想“巧妙地”做到这一点，你可以尝试：

employees = projects = practices = entities = categories = groups = external_inputs = payrolls = nil
%w(employees projects practices entities categories groups external_inputs payrolls).each {|v| eval "#{v} = []" }

也就是说，将值集合到哈希中可能更容易。

Hash[*%w(employees projects practices entities categories groups external_inputs payrolls).map {|k| [k, []] }.flatten(1)]

# Result
# {"employees"=>[], "projects"=>[], "practices"=>[], "entities"=>[], "categories"=>[], "groups"=>[], "external_inputs"=>[], "payrolls"=>[]}

将数组分配给不同的变量

4 个答案: