Ruby on rails 按特定值匹配数据集中的实例记录_Ruby On Rails_Ruby_Algorithm_Activerecord

Ruby on rails 按特定值匹配数据集中的实例记录

ruby-on-rails ruby algorithm activerecord

Ruby on rails 按特定值匹配数据集中的实例记录,ruby-on-rails,ruby,algorithm,activerecord,Ruby On Rails,Ruby,Algorithm,Activerecord,我有一个解决方案，它工作得很好，但性能不高，需要一些时间才能运行。让我们从最初的两个查询（都是双连接）返回的内容开始：第一组数据如下所示-我们将其称为line\u项。正如您将看到的，line\u项目没有dh\u名字key/value [ [ { pb_id: "133599.0", pbbname: "CUSTOMER", opl_amount: "101.0", ops_type:

我有一个解决方案，它工作得很好，但性能不高，需要一些时间才能运行。让我们从最初的两个查询（都是双连接）返回的内容开始：

第一组数据如下所示-我们将其称为

line\u项

。正如您将看到的，

line\u项目

没有

dh\u名字

key/value


 [
   [
    {
    pb_id: "133599.0",
    pbbname: "CUSTOMER",
    opl_amount: "101.0",
    ops_type: "P",
    ops_stop_id: 269802,
    ops_order_id: 133599,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133599.0",
    pbbname: "CUSTOMER",
    opl_amount: "11.62",
    ops_type: "P",
    ops_stop_id: 269802,
    ops_order_id: 133599,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133590.0",
    pbbname: "CUSTOMER",
    opl_amount: "79.0",
    ops_type: "P",
    ops_stop_id: 269780,
    ops_order_id: 133590,
    ops_driver1: 104,
    ops_delivered_time: null
    },
    {
    pb_id: "133220.0",
    pbbname: "CUSTOMER",
    opl_amount: "625.0",
    ops_type: "D",
    ops_stop_id: 269011,
    ops_order_id: 133220,
    ops_driver1: 62,
    ops_delivered_time: "2021-04-01T12:35:00.000-05:00"
    },
    {
    pb_id: "133357.0",
    pbbname: "CUSTOMER",
    opl_amount: "550.0",
    ops_type: "D",
    ops_stop_id: 269290,
    ops_order_id: 133357,
    ops_driver1: 92,
    ops_delivered_time: "2021-04-01T09:38:00.000-05:00"
    },
    {
    pb_id: "133219.0",
    pbbname: "CUSTOMER",
    opl_amount: "1267.06",
    ops_type: "P",
    ops_stop_id: 269008,
    ops_order_id: 133219,
    ops_driver1: 43,
    ops_delivered_time: null
    },
    {
    pb_id: "133577.0",
    pbbname: "CUSTOMER",
    opl_amount: "150.0",
    ops_type: "P",
    ops_stop_id: 269754,
    ops_order_id: 133577,
    ops_driver1: 94,
    ops_delivered_time: null
    },
    {
    pb_id: "133503.0",
    pbbname: "CUSTOMER",
    opl_amount: "79.0",
    ops_type: "P",
    ops_stop_id: 269592,
    ops_order_id: 133503,
    ops_driver1: 104,
    ops_delivered_time: null
    },
    {
    pb_id: "133643.0",
    pbbname: "HALLMARK CARDS BERMAN BLAKE",
    opl_amount: "79.0",
    ops_type: "P",
    ops_stop_id: 269895,
    ops_order_id: 133643,
    ops_driver1: 104,
    ops_delivered_time: null
    }
   ]
 ]

现在，让我们看看第二个双连接中的下一组数据，这是<代码> LyInCurrest。它看起来像这样：


    [
    {
    pb_id: "133633.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "250.0",
    ops_driver1: 59,
    ops_stop_id: 269869,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-02T13:07:00.000-05:00"
    },
    {
    pb_id: "133127.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "1147.0",
    ops_driver1: 102,
    ops_stop_id: 268801,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: null
    },
    {
    pb_id: "133144.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "650.0",
    ops_driver1: 71,
    ops_stop_id: 268836,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-01T14:38:00.000-05:00"
    },
    {
    pb_id: "133144.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "650.0",
    ops_driver1: 71,
    ops_stop_id: 268837,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: null
    },
    {
    pb_id: "133188.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "700.0",
    ops_driver1: 71,
    ops_stop_id: 268924,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-01T08:04:00.000-05:00"
    },
    ]

    {
    FIRST LAST: [
    {
    pb_id: "133599.0",
    pbbname: "CUSTOMER",
    opl_amount: "101.0",
    ops_type: "P",
    ops_stop_id: 269802,
    ops_order_id: 133599,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133599.0",
    pbbname: "CUSTOMER",
    opl_amount: "11.62",
    ops_type: "P",
    ops_stop_id: 269802,
    ops_order_id: 133599,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133536.0",
    pbbname: "CUSTOMER",
    opl_amount: "45.0",
    ops_type: "P",
    ops_stop_id: 269665,
    ops_order_id: 133536,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133536.0",
    pbbname: "CUSTOMER",
    opl_amount: "5.18",
    ops_type: "P",
    ops_stop_id: 269665,
    ops_order_id: 133536,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133522.0",
    pbbname: "CUSTOMER",
    opl_amount: "150.0",
    ops_type: "P",
    ops_stop_id: 269637,
    ops_order_id: 133522,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133619.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "550.0",
    ops_driver1: 11,
    ops_stop_id: 269841,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-02T11:41:00.000-05:00"
    }
    ],

我目前正在做的是循环遍历这两个变量，并根据这些

值匹配它们
ops_stop_id, ops_driver_1, pb_id

如果这三个匹配，那么我需要在特定驱动程序的名称下构造它们，该名称只能来自具有dh_first_name
的实例。此数据结构完成后如下所示：

    [
    {
    pb_id: "133633.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "250.0",
    ops_driver1: 59,
    ops_stop_id: 269869,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-02T13:07:00.000-05:00"
    },
    {
    pb_id: "133127.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "1147.0",
    ops_driver1: 102,
    ops_stop_id: 268801,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: null
    },
    {
    pb_id: "133144.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "650.0",
    ops_driver1: 71,
    ops_stop_id: 268836,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-01T14:38:00.000-05:00"
    },
    {
    pb_id: "133144.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "650.0",
    ops_driver1: 71,
    ops_stop_id: 268837,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: null
    },
    {
    pb_id: "133188.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "700.0",
    ops_driver1: 71,
    ops_stop_id: 268924,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-01T08:04:00.000-05:00"
    },
    ]


    {
    FIRST LAST: [
    {
    pb_id: "133599.0",
    pbbname: "CUSTOMER",
    opl_amount: "101.0",
    ops_type: "P",
    ops_stop_id: 269802,
    ops_order_id: 133599,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133599.0",
    pbbname: "CUSTOMER",
    opl_amount: "11.62",
    ops_type: "P",
    ops_stop_id: 269802,
    ops_order_id: 133599,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133536.0",
    pbbname: "CUSTOMER",
    opl_amount: "45.0",
    ops_type: "P",
    ops_stop_id: 269665,
    ops_order_id: 133536,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133536.0",
    pbbname: "CUSTOMER",
    opl_amount: "5.18",
    ops_type: "P",
    ops_stop_id: 269665,
    ops_order_id: 133536,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133522.0",
    pbbname: "CUSTOMER",
    opl_amount: "150.0",
    ops_type: "P",
    ops_stop_id: 269637,
    ops_order_id: 133522,
    ops_driver1: 11,
    ops_delivered_time: null
    },
    {
    pb_id: "133619.0",
    pbbname: "CUSTOMER",
    pb_net_rev: "550.0",
    ops_driver1: 11,
    ops_stop_id: 269841,
    dh_first_name: "FIRST",
    dh_last_name: "LAST",
    ops_delivered_time: "2021-04-02T11:41:00.000-05:00"
    }
    ],


您将看到两个记录的混合，匹配的参数组织正确
这就是我目前正在解决的问题

    merger = {}
      line_items.each do |lines, i|
        line_stops.each do |stops|
            if (lines.ops_stop_id == stops.ops_stop_id && lines.ops_driver1 == stops.ops_driver1 && lines.pb_id == stops.pb_id)
              stops_arr.push(stops)
              merger[stops.dh_first_name + ' ' + stops.dh_last_name] = (merger[stops.dh_first_name + ' ' + stops.dh_last_name] ||= []) << lines
            end
         end
      end

      line_stops.each do |stops|
            if (!stops_arr.include?(stops))
              stops_arr.push(stops)
              merger[stops.dh_first_name + ' ' + stops.dh_last_name] = (merger[stops.dh_first_name + ' ' + stops.dh_last_name] ||= []) << stops
            end
        end


合并={}
行|项。每个do |行，i|
线路停止。每个do停止|
如果（lines.ops\u stop\u id==stops.ops\u stop\u id&&lines.ops\u driver1==stops.ops\u driver1&&lines.pb\u id==stops.pb\u id）
停止推送（停止）
合并[stops.dh_first_name+“”+stops.dh_last_name]=（合并[stops.dh_first_name+“”+stops.dh_last_name]|=[]）代码的时间复杂度是O（lines.size*stops.size）

这里是我的提案，它的时间复杂度约为O（lines.size+stops.size）

def merge_键（停止）
stops.dh_first_name+“”+stops.dh_last_name
结束
#请注意，下面的散列密钥代码可能不够好
def哈希_键（行）
“{lines.ops#u stop_id}{lines.ops#u driver1}{lines.pb_id}”
结束
merge=Hash.new{| Hash，key | Hash[key]=[]}
停止\u hash=hash.new
#O（线_停止。大小）
线路停止。每个do停止|
合并密钥=合并密钥（停止）
下一个if merge.hash_key？（merge_key）#因为在代码中没有添加dup停止，对吗？
合并[merge_key]如果你从数据库中取出这些数据，那么你可能应该在那里进行，而不是在Ruby中进行。我担心这就是答案，但我只是更愿意编写脚本来完成它。。。但我认为除了通过查询进行优化之外，没有其他真正的优化方法。我甚至不知道从哪里开始，通过SQL查询实现逻辑。我会尝试一下，如果不顺利，然后问一个关于模式、模型、数据示例和预期输出的问题。你能为JSON变量提供一个可运行的示例吗？它可以是抽象的，不需要拥有所有那些不能告诉我们太多的属性（显然它们对您的业务很重要，但我不关心pb_net_rev:-），或者提供模式和预期结果？所以我们可以帮助您在SQL中实现这一点。我想stops\u arr
是一个数组，对吗？如果是真的，那么我猜罪魁祸首是逻辑：！停止\u arr.包括？（停止）
。把changestops\u arr
作为散列值怎么样。