Question

我有一个数组“大小”，如下所示：

[#<OPTIONVALUE ID: 5, NAME: "M">,
#<OPTIONVALUE ID: 6, NAME: "M/L">, 
#<OPTIONVALUE ID: 7, NAME: "XS/S">]

考虑属性NAME的值。数组排序：M，M / L，XS / S.

但排序顺序应如下所示：

@sizes_sort_order = ["XS", "XS/S", "S", "S/M", "M", "M/L", "L", "L/XL", "XL"]

应用于前一个数组，元素的顺序应如下所示：

[#<SPREE::OPTIONVALUE ID: 7, NAME: "XS/S">,
#<SPREE::OPTIONVALUE ID: 5, NAME: "M">,
#<SPREE::OPTIONVALUE ID: 6, NAME: "M/L">]

def sizes
  @sizes ||= grouped_option_values_by_option_type[Spree::OptionType.find_by!(name: 'size')]
  @sizes_sort_order = ["XS", "XS/S", "S", "S/M", "M", "M/L", "L", "L/XL", "XL"]
  @sizes.map { # sort after @size_sort_order }
end

如何在@sizes_sort_order之后将数组中的元素排序？

Answer 1

您可以使用Enumerable#sort_by

my_array.sort_by {|x| @sizes_sort_order.index(x.name) }

Answer 2

您可以包含Comparable模块以获得对象的自然排序。

http://ruby-doc.org/core-2.2.3/Comparable.html

Comparable mixin由可以对其对象进行排序的类使用。该类必须定义＆lt; =＆gt;运算符，用于比较接收器对另一个对象，返回-1,0或+1取决于是否接收器小于，等于或大于另一个物体。

class Size
  include Comparable

  SIZES = ["XS", "XS/S", "S", "S/M", "M", "M/L", "L", "L/XL", "XL"]

  attr_reader :name
  def initialize(id, name)
    @id = id
    @name = name
  end

  def <=>(b)
    SIZES.index(name) <=> SIZES.index(b.name)
  end
end

a = Size.new(5, 'M')
b = Size.new(6, 'M/L')
c = Size.new(7, 'XS/S')

print [a, b, c].sort

[#<Size:0x007f8e910458e0 @id=7, @name="XS/S">, #<Size:0x007f8e910459a8 @id=5, @name="M">, #<Size:0x007f8e91045930 @id=6, @name="M/L">]

Answer 3

这种方法涉及的步骤多于使用sort或sort_by的方法，但对于较大的数组，它可能更快，因为不涉及排序 - 这是相对昂贵的。

<强>代码

def reorder_by_size(instances, size_order)
  instances.each_with_object({}) { |inst, h| h.update(inst.name=>inst) }.
    values_at(*(size_order & (instances.map { |s| s.name })))
end

示例

首先让我们创建一个
的实例数组
class Sizes attr_reader :name def initialize(id, name) @id = id @name = name end end
像这样：

instances = [Sizes.new(5,'M'), Sizes.new(6,'M/L'), Sizes.new(7, 'XS/S')] #=> [#<Sizes:0x007fa66a955ac0 @id=5, @name="M">, # #<Sizes:0x007fa66a955a70 @id=6, @name="M/L">, # #<Sizes:0x007fa66a955a20 @id=7, @name="XS/S">]

然后

reorder_by_size(instances, @sizes_sort_order) #=> [#<Sizes:0x007fa66a01dfc0 @id=7, @name="XS/S">, # #<Sizes:0x007fa66a86fdb8 @id=5, @name="M">, # #<Sizes:0x007fa66a8404f0 @id=6, @name="M/L">]

<强>解释

对于示例中定义的instances，首先按所需顺序创建一个大小数组：

names = @sizes_sort_order & (instances.map { |s| s.name }) #=> ["XS/S", "M", "M/L"]

重要：Array#&状态的文档，“订单将从原始数组中保留。”

现在我们可以在不进行排序的情况下创建所需的重新排序，方法是创建一个带有键大小和值的实例的哈希，然后使用Hash#values_at以所需的顺序提取实例。

instances.each_with_object({}) { |inst, h| h.update(inst.name=>inst) }.values_at(*names) #=> [#<Sizes:0x007fa66a01dfc0 @id=7, @name="XS/S">, # #<Sizes:0x007fa66a86fdb8 @id=5, @name="M">, # #<Sizes:0x007fa66a8404f0 @id=6, @name="M/L">]

最后一项操作涉及以下两个步骤。

h = instances.each_with_object({}) { |inst, h| h.update(inst.name=>inst) } #=> {"M" => #<Sizes:0x007fa66a955ac0 @id=5, @name="M">, # "M/L" => #<Sizes:0x007fa66a955a70 @id=6, @name="M/L">, # "XS/S" => #<Sizes:0x007fa66a955a20 @id=7, @name="XS/S">} h.values_at(*names) #=> h.values_at(*["XS/S", "M", "M/L"]) #=> h.values_at("XS/S", "M", "M/L") #=> [#<Sizes:0x007fa66a955a20 @id=7, @name="XS/S">, # #<Sizes:0x007fa66a955ac0 @id=5, @name="M">, # #<Sizes:0x007fa66a955a70 @id=6, @name="M/L">]

Ruby：在名称属性数组之后对对象数组进行排序

3 个答案: