Question

我有一系列哈希。

每个哈希都有3个键值（计划，数字，时间）

所以它会像：

[0]{plan => "my_plan", number=> 1234, time =>"2016-03-02T13:56:02.000-05:00" }
[1]{plan => "my_plan", number=> 1234, time =>"2016-03-02T14:56:02.000-05:00" }
[2]{plan => "my_plan", number=> 1234, time =>"2016-03-02T13:59:02.000-05:00" }
[3]{plan => "my_plan", number=> 1234, time =>"2016-03-02T12:56:02.000-05:00" }
[4]{plan => "my_plan", number=> 4444, time =>"2016-03-02T16:42:02.000-05:00" }
[5]{plan => "my_plan", number=> 4444, time =>"2016-03-02T16:45:02.000-05:00" }
[6]{plan => "my_plan", number=> 4444, time =>"2016-03-02T16:44:02.000-05:00" }
[7]{plan => "my_plan", number=> 4444, time =>"2016-03-02T16:50:02.000-05:00" }

等等

我希望保留最新时间（最长时间）。

所以数字是“1234”，我想保留这一行：

{plan => "my_plan", number=> 1234, time =>"2016-03-02T14:56:02.000-05:00" }

并丢弃其余部分。

对于号码“4444”，我会保留：

{plan => "my_plan", number=> 4444, time =>"2016-03-02T16:50:02.000-05:00" }

因为它是具有该数字最大日期的那个

Answer 1

input = [
  {plan: "my_plan", number: 1234, time: "2016-03-02T13:56:02.000-05:00" },
  {plan: "my_plan", number: 1234, time: "2016-03-02T14:56:02.000-05:00" },
  {plan: "my_plan", number: 1234, time: "2016-03-02T13:59:02.000-05:00" },
  {plan: "my_plan", number: 1234, time: "2016-03-02T12:56:02.000-05:00" },
  {plan: "my_plan", number: 4444, time: "2016-03-02T16:42:02.000-05:00" },
  {plan: "my_plan", number: 4444, time: "2016-03-02T16:45:02.000-05:00" },
  {plan: "my_plan", number: 4444, time: "2016-03-02T16:44:02.000-05:00" },
  {plan: "my_plan", number: 4444, time: "2016-03-02T16:50:02.000-05:00" }
]

input.group_by { |e| e[:number] }
     .values
     .map { |vals| vals.sort_by { |e| e[:time] }.last }

#⇒ [
#  [0] {
#    :number => 1234,
#      :plan => "my_plan",
#      :time => "2016-03-02T14:56:02.000-05:00"
#  },
#  [1] {
#    :number => 4444,
#      :plan => "my_plan",
#      :time => "2016-03-02T16:50:02.000-05:00"
#  }
# ]

无论您是否可以自由使用精彩的Hashie::Mash实施，它都更具表现力：

require 'hashie'
input.map { |h| Hashie::Mash.new(h) }
     .group_by(&:number)
     .values
     .map { |vals| vals.sort_by(&:time).last }

NB 日期时间值是否可能位于不同的时区，在排序前将字符串显式转换为Time实例更安全：

input.group_by { |e| e[:number] }
     .values
     .map { |vals| vals.sort_by { |e| Time.parse e[:time] }.last }

Ruby中的一系列哈希 - 删除不需要的东西

1 个答案: