Question

我在这个数组中有一对time_tables。有四个time_tables start_location - end_location和start_date - end_date以线性方式相互关联。

当第一个time_table结束时，另一个time_table开始，依此类推。

我的代码：

arr =   [
      { name: 01, start_date: '2014-04-24 22:03:00', start_location: 'A', end_date: '2014-04-24 22:10:00', end_location: 'B' },
      { name: 05, start_date: '2014-04-24 22:10:00', start_location: 'C', end_date: '2014-04-24 23:10:00', end_location: 'D' },
      { name: 01, start_date: '2014-04-24 17:10:00', start_location: 'X', end_date: '2014-04-24 20:10:00', end_location: 'B' },
      { name: 01, start_date: '2014-04-24 17:10:00', start_location: 'Z', end_date: '2014-04-24 20:10:00', end_location: 'B' },
      { name: 06, start_date: '2014-04-24 20:15:00', start_location: 'B', end_date: '2014-04-24 22:10:00', end_location: 'C' },
      { name: 03, start_date: '2014-04-24 23:15:00', start_location: 'D', end_date: '2014-04-24 00:10:00', end_location: 'E' }
    ]
new_array = []

i = 0
while i <= 5 do
  if arr[i][:end_location] == arr[i+1][:start_location] && arr[i][:start_date] <= arr[i+1][:start_date]
    new_array << arr[i+1]
  end
  i = i + 1
end

这是我想要的结果：

  # My expexpected result will be this:
  #     [
  #     { name: 01, start_date: '2014-04-24 22:03:00', start_location: 'A', end_date: '2014-04-24 22:10:00', end_location: 'B' },
  #    { name: 06, start_date: '2014-04-24 22:15:00', start_location: 'B', end_date: '2014-04-24 22:20:00', end_location: 'C' },
  #    { name: 05, start_date: '2014-04-24 22:20:00', start_location: 'C', end_date: '2014-04-24 23:10:00', end_location: 'D' },
  #    { name: 03, start_date: '2014-04-24 23:15:00', start_location: 'D', end_date: '2014-04-24 00:10:00', end_location: 'E' }
  # 

  ]

但我的算法似乎很糟糕。感谢您提供有关此工作的见解。

Answer 1

这将＆＃34;加入＆＃34;你的时间序列是连续的结束和开始的位置。

 def span x; x[:end_location].ord - x[:start_location].ord; end
 def diff x, y; x[:start_location].ord - y[:start_location].ord; end
 arr = arr.sort_by { |x| [x[:start_location], span(x)] }
 prev = arr[0]
 arr = arr.slice_before { |e|
   prev, prev2 = e, prev
   diff(prev, prev2) != 0
 }.to_a.map(&:first).chunk(&method(:span)).first[1]

例如，我得到了

 arr.map { |x| [x[:start_location], x[:end_location] }
 => [["A", "B"], ["B", "C"], ["C", "D"], ["D", "E"]]

Answer 2

Aren你真的在寻找最长的线性后续时间表吗？我想通过评论澄清一下，但我没有得到评论的许可。输入中的B（和start_location B）的start_date值与输出之间也存在差异，因此我假设它是错误的。

我已经编写了解决方案，考虑到你想要找到最长的线性后续时间表。

require 'date'

def getLLTT(time_tables)

    longest = []
    time_tables.sort_by! do |time_table|
        DateTime.parse(time_table[:start_date]).to_time
    end

    0.upto(time_tables.size-1) do |i|

        long_for_i = [time_tables[i]]
        0.upto(i-1) do |j|
            j_end_date = DateTime.parse(longest[j][-1][:end_date]).to_time
            i_start_date = DateTime.parse(time_tables[i][:start_date]).to_time
            if j_end_date <= i_start_date 
                if longest[j][-1][:end_location].eql? time_tables[i][:start_location]
                    if longest[j].size + 1 > long_for_i.size
                        long_for_i = longest[j] + [time_tables[i]]
                    end
                end
            end
        end

        longest[i] = long_for_i
    end
    return longest[-1]
end

puts getLLTT(arr)

所以给出了输入：

arr =   [
{ name: 01, start_date: '2014-04-24 22:03:00', start_location: 'A', end_date: '2014-04-24 22:10:00', end_location: 'B' },
  { name: 05, start_date: '2014-04-24 22:10:00', start_location: 'C', end_date: '2014-04-24 23:10:00', end_location: 'D' },
  { name: 01, start_date: '2014-04-24 17:10:00', start_location: 'X', end_date: '2014-04-24 20:10:00', end_location: 'B' },
  { name: 01, start_date: '2014-04-24 17:10:00', start_location: 'Z', end_date: '2014-04-24 20:10:00', end_location: 'B' },
  { name: 06, start_date: '2014-04-24 20:15:00', start_location: 'B', end_date: '2014-04-24 22:10:00', end_location: 'C' },
  { name: 03, start_date: '2014-04-24 23:15:00', start_location: 'D', end_date: '2014-04-24 00:10:00', end_location: 'E' }
]

输出将是：

[
{:name=>1, :start_date=>"2014-04-24 17:10:00", :start_location=>"Z", :end_date=>"2014-04-24 20:10:00", :end_location=>"B"}
{:name=>6, :start_date=>"2014-04-24 20:15:00", :start_location=>"B", :end_date=>"2014-04-24 22:10:00", :end_location=>"C"}
{:name=>5, :start_date=>"2014-04-24 22:10:00", :start_location=>"C", :end_date=>"2014-04-24 23:10:00", :end_location=>"D"}
{:name=>3, :start_date=>"2014-04-24 23:15:00", :start_location=>"D", :end_date=>"2014-04-24 00:10:00", :end_location=>"E"}
]

使用Ruby通过它们的值使哈希找到彼此

2 个答案: