Question

我有这个对象数组：

a = [#<Person id: 9, name: "Bob", bin_id: "114628">, 
   #<Person id: 10, name: "Sally", bin_id: "114626">, 
   #<Person id: 11, name: "Jessie", bin_id: "114627">, 
   #<Person id: 12, name: "Rapheal", bin_id: "114620">, 
   #<Company _id: 55295, name: "X", bin_id: "114619">, 
   #<Company _id: 55295, name: "Y", bin_id: "114629">, 
   #<Company _id: 55295, name: "Z", bin_id: "16074802">, 
   #<Company _id: 55295, name: "W", bin_id: "16074815">]

当我这样做时

a.group_by { |objects| object.calculate_age }

我明白了：

 { 
     33 => 
            [#<Person id: 9, name: "Bob", bin_id: "114628">, 
            #<Person id: 10, name: "Sally", bin_id: "114626">, 
            #<Person id: 11, name: "Jessie", bin_id: "114627">],
     53 => [#<Company _id: 55295, name: "X", bin_id: "114619">],
     45 => [#<Company _id: 55295, name: "Y", bin_id: "114629">,],
     56 => [#<Company _id: 55295, name: "Z", bin_id: "16074802">],
     60 => [#<Company _id: 55295, name: "W", bin_id: "16074815">]
 }

但我如何得到这样的东西：

 {   
     33 => [#<Person id: 9, name: "Bob", bin_id: "114628">], 
     33 => [#<Person id: 10, name: "Sally", bin_id: "114626">], 
     33 => [#<Person id: 11, name: "Jessie", bin_id: "114627">],
     53 => [#<Company _id: 55295, name: "X", bin_id: "114619">],
     45 => [#<Company _id: 55295, name: "Y", bin_id: "114629">,],
     56 => [#<Company _id: 55295, name: "Z", bin_id: "16074802">],
     60 => [#<Company _id: 55295, name: "W", bin_id: "16074815">]
    }

每个键是否明确地设置为每个值？也许group_by不是正确的方法。我想返回一个像你上面看到的哈希。

Answer 1

出于问题的目的，您可以编写数组：

arr = [obj0, obj1,...,obj7]

因为每个对象的实例变量的值是无关紧要的。您无法将其转换为具有重复键（age）的哈希值，但您可以将其转换为哈希数组，每个哈希值都包含一个密钥age，如果这样做有帮助的话：

arr.map { |obj| { obj.calculate_age=>obj } }

要了解这是如何工作的，假设我们有：

class Friends
  attr_reader :calculate_age
  def initialize name, age
    @name, @calculate_age = name, age
  end
end

a = [["Amy", 21], ["Billy-Bob", 53], ["Wilber", 21], ["Trixi", 34], ["Bo", 53]]
arr = a.map { |name,age| Friends.new(name,age) }
  #=> [#<Friends:0x007fc1f28b5518 @name="Amy", @calculate_age=21>,
  #    #<Friends:0x007fc1f28b54a0 @name="Billy-Bob", @calculate_age=53>,
  #    #<Friends:0x007fc1f28b5450 @name="Wilber", @calculate_age=21>,
  #    #<Friends:0x007fc1f28b5400 @name="Trixi", @calculate_age=34>,
  #    #<Friends:0x007fc1f28b5388 @name="Bo", @calculate_age=53>]

我们现在可以将其转换为哈希数组：

a = arr.map { |obj| { obj.calculate_age=>obj } }
  #=> [{21=>#<Friends:0x007fc1f28b5518 @name="Amy",...},
  #    {53=>#<Friends:0x007fc1f28b54a0 @name="Billy-Bob",...},
  #    {21=>#<Friends:0x007fc1f28b5450 @name="Wilber"...},
  #    {34=>#<Friends:0x007fc1f28b5400 @name="Trixi"...},
  #    {53=>#<Friends:0x007fc1f28b5388 @name="Bo",...}]

如果您希望这些数组按年龄排序：

a.sort_by { |h| h.keys.first }
  #=> [{21=>#<Friends:0x007fc1f28b5518 @name="Amy",...},
  #    {21=>#<Friends:0x007fc1f28b5450 @name="Wilber"...},
  #    {34=>#<Friends:0x007fc1f28b5400 @name="Trixi"...},
  #    {53=>#<Friends:0x007fc1f28b54a0 @name="Billy-Bob",...},
  #    {53=>#<Friends:0x007fc1f28b5388 @name="Bo",...}]

枚举通过哈希并保持密钥的单独键值对是相同的

1 个答案: