arrays - 在 ruby​​ 数组中查找并展平重叠的时间间隔

Question

我正在尝试对应用程序进行一些停机时间计算。我拥有的是一个巨大的散列数组，显示应用程序的不同部分何时关闭。散列包括开始时间和结束时间。问题是其中一些停机时间可能会重叠。我怎样才能遍历哈希数组并找到重叠的时间间隔。

times = [{"timefrom"=>1461693247, "timeto"=>1461693307},
         {"timefrom"=>1462363987, "timeto"=>1462364607},
         {"timefrom"=>1462364037, "timeto"=>1462366037}]

例如，给定上面的数组，times[1] 和 times[2] 重叠。所以理想情况下，我想做的是合并它们，以便它们形成一个长期中断。 IE。

times[1] = { "timefrom" => times[1]["timefrom"], "timeto" => times[2]["timeto"] }

Answer 1

我理解 times 的元素 g(散列)按 g["timefrom"] 排序。

def combine_times(times)
  times[1..-1].each_with_object([times.first]) do |g,a|
    if g["timefrom"] < a.last["timeto"]
      a[-1]["timeto"] = [ a[-1]["timeto"], g["timeto"] ].max
    else
      a << g
    end
  end
end

times = [{"timefrom"=>1461693247, "timeto"=>1461693307},

         {"timefrom"=>1462363987, "timeto"=>1462364607},
         {"timefrom"=>1462364037, "timeto"=>1462366037}]

行间距显示了times的元素应该如何分组。

combine_times(times)
  #=> [{"timefrom"=>1461693247, "timeto"=>1461693307}, (times[0])
  #    {"timefrom"=>1462363987, "timeto"=>1462366037}] (combines times[1..2])

另一个例子:

times = [{"timefrom"=>10, "timeto"=>20},
         {"timefrom"=>12, "timeto"=>15},

         {"timefrom"=>22, "timeto"=>30},
         {"timefrom"=>28, "timeto"=>32},
         {"timefrom"=>29, "timeto"=>29},

         {"timefrom"=>32, "timeto"=>40},

         {"timefrom"=>42, "timeto"=>50},
         {"timefrom"=>43, "timeto"=>46}]

combine_times(times)
  #=> [{"timefrom"=>10, "timeto"=>20}, (combines times[0..1])
  #    {"timefrom"=>22, "timeto"=>32}, (combines times[2..4])
  #    {"timefrom"=>32, "timeto"=>40}, (times[5])
  #    {"timefrom"=>42, "timeto"=>50}] (combines times[6..7])