Ruby On Rails 根据优先级对 hash 的数组进行排序

Question

请帮助我找到基于优先级排序的解决方案我没有得到解决方案

优先级基于：

• 第一个选择最早的插槽
• 第二个最大评级
• 第三个 cronofy_enabled

interview_proposal = [
  { interviewer_id: 3903, rating: 4, cronofy_enabled: false, slot_date: "2022-05-09" },
  { interviewer_id: 10, rating: 3.5, cronofy_enabled: false, slot_date: "2022-05-06" },
  { interviewer_id: 3902, rating: 2, cronofy_enabled: true, slot_date: "2022-05-06" },
{ interviewer_id: 3904, rating: 2.5, cronofy_enabled: false, slot_date: "2022-05-09" },
{ interviewer_id: 3905, rating: 3.5, cronofy_enabled: false, slot_date: "2022-05-09" }
]
    
# First priority picked earliest_slot
@earliest_slot = interview_proposal.find{|proposal| proposal[:slot_date] == "2022-05-06"}
#second priority rating
@max_rating = interview_proposal.max_by{|proposal| proposal[:rating]}
# Third priority cronofy_enabled
@calendar_synced = interview_proposal.find{|proposal| proposal[:cronofy_enabled]}
    
sorted_array = []
interview_proposal.each do |interview_proposal|
  priority = match_count
  sorted_array << { priority: priority, interview_proposal: interview_proposal }
end
sorted_array = sorted_array.
  sort_by { |obj| obj[:priority] }.
  reverse.map { |obj| obj[:interview_proposal] }
sorted_array
    
def match_count
  count = 0
  count += 1 if @earliest_slot[:slot_date].present?
  count += 1 if @max_rating[:rating].present?
  count += 1 if @calendar_synced[:cronofy_enabled]
  count
end

Answer 1

在我看来，您可以通过使用Ruby 的排序方法来减少代码的开销。 如果 a > b，您需要在每个块迭代中提供正数 integer，如果它们相等，则为 0；如果 b >a，则为负数 integer，以便排序工作。 starship 运算符 <=> 是前两种情况的一个很好的快捷方式，但不适用于布尔值，因此我们在那里添加了额外的逻辑。

sorted_array = interview_proposal.sort do |a, b|

   if  (a[:slot_date] <=> b[:slot_date]) != 0
      a[:slot_date] <=> b[:slot_date] 
   elsif (a[:rating] <=> b[:rating]) != 0
      a[:rating] <=> b[:rating]
   elsif a[:cronofy_enabled] == b[:cronofy_enabled]
       0
   elsif a[:cronofy_enabled]
       1
   else 
      -1
   end
end

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