使用Ruby，使哈希值通过它们的值找到彼此
我在这个数组中有几个time_tables 。 有四个time_tables通过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。 感谢您对此工作的见解。
你真的不是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。 此外，输入中的B（和start_location B）的start_date值与输出之间存在差异，因此我认为这是一个错误。
考虑到你想要找到最长的线性后续时间表，我已经编写了解决方案。
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"} ]
这将按连续结束和开始位置“加入”您的时间序列。
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，我明白了
arr.map { |x| [x[:start_location], x[:end_location] } => [["A", "B"], ["B", "C"], ["C", "D"], ["D", "E"]]