Rails基于优先级对散列数组进行排序

Question

请帮助我根据优先级找到解决方案排序，我没有得到解决方案

基于以下方面的优先事项：

• 第一截最早槽
• 第二最高评级
• 第3 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，通过使用Ruby排序法，您可以减少代码中的开销。在每个块迭代中，只需提供一个正整数(如果a> b )、0(如果它们相等)和一个负整数(b >a )。对于前两种情况，星舰运算符<=>是一个很好的快捷方式，但对布尔函数不起作用，所以我们在这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