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按数据对JSON进行分组？

ruby-on-railsrubyarraysjson
4

我有以下JSON数据：

{
  "events":
  {
      "event":
      [
          {
              "city":"Birmingham",
              "state":"AL",
              "country":"US",
              "lat":"33.5206608",
              "lng":"-86.80249",
              "status":"Delivered",
              "occured_at":"2012-04-06
14:17:00 UTC"
          },
          {
              "city":"Birmingham",
              "state":"AL",
              "country":"US",
              "lat":"33.5206608",
              "lng":"-86.80249",
              "status":"Out
For Delivery",
              "occured_at":"2012-04-06 04:44:00 UTC"
          },
          {
              "city":"Birmingham",
              "state":"AL",
              "country":"US",
              "lat":"33.5206608",
              "lng":"-86.80249",
              "status":"Arrival
Scan",
              "occured_at":"2012-04-05 19:07:00 UTC"
          },
          {
              "city":"Doraville",
              "state":"GA",
              "country":"US",
              "lat":"33.8981579",
              "lng":"-84.2832564",
              "status":"Departure
Scan",
              "occured_at":"2012-04-05 17:08:00 UTC"
          },
          {
              "city":"Doraville",
              "state":"GA",
              "country":"US",
              "lat":"33.8981579",
              "lng":"-84.2832564",
              "status":"Arrival
Scan",
              "occured_at":"2012-04-05 11:15:00 UTC"
          },
          {
              "city":"Spartanburg",
              "state":"SC",
              "country":"US",
              "lat":"34.9495672",
              "lng":"-81.9320482",
              "status":"Departure
Scan",
              "occured_at":"2012-04-05 08:42:00 UTC"
          },
          {
              "city":"Spartanburg",
              "state":"SC",
              "country":"US",
              "lat":"34.9495672",
              "lng":"-81.9320482",
              "status":"Arrival
Scan",
              "occured_at":"2012-04-05 08:21:00 UTC"
          },
          {
              "city":"Greensboro",
              "state":"NC",
              "country":"US",
              "lat":"36.0726354",
              "lng":"-79.7919754",
              "status":"Departure
Scan",
              "occured_at":"2012-04-05 04:45:00 UTC"
          },
          {
              "city":"Greensboro",
              "state":"NC",
              "country":"US",
              "lat":"36.0726354",
              "lng":"-79.7919754",
              "status":"Origin
Scan",
              "occured_at":"2012-04-05 00:11:00 UTC"
          },
          {
              "city":null,
              "state":null,
              "country":"US",
              "status":"Billing
Information Received",
              "occured_at":"2012-04-04 18:20:27 UTC"
          }
      ]
  }
}

我需要做的是将数据按城市、州和国家的组合进行分组，但仍然返回每个项目的数据。
例如，按“Birmingham，AL，US”进行分组，但仍能遍历每个事件的状态（即已交付、派送中、到达扫描）。
这是JSON格式并不重要，对吧？你可以使用JSON.parse将其转换为Ruby哈希表，然后直接使用它（如果需要，最后再将其转换回JSON）。 - Phrogz
2个回答
6
你需要的魔法是 Enumerable#group_by ：
require 'json'
all = JSON.parse(DATA.read)['events']['event']
all.group_by{ |h| [h['city'],h['state'],h['country']] }.each do |loc,events|
  puts "'#{loc.join(',')}': #{events.length} event#{:s if events.length!=1}"
  print "--> "
  puts events.map{ |e| e['status'] }.join(', ')
end

#=> 'Birmingham,AL,US': 3 events
#=> --> Delivered, Out For Delivery, Arrival Scan
#=> 'Doraville,GA,US': 2 events
#=> --> Departure Scan, Arrival Scan
#=> 'Spartanburg,SC,US': 2 events
#=> --> Departure Scan, Arrival Scan
#=> 'Greensboro,NC,US': 2 events
#=> --> Departure Scan, Origin Scan
#=> ',,US': 1 event
#=> --> Billing Information Received

请注意，在上面的代码中，loc是由group_by评估的块返回的三元素数组，而events是同一组中所有项的数组。
0
你可以使用关联数组，这样你就可以使用字符串作为键，并使用它来访问你想要分组的记录。例如，所有伯明翰的事件都可以通过以下方式访问：
events.event.Birmingham[i]

所有的斯巴达堡记录都存储在通过以下方式访问的数组中：

events.event.Spartanburg[i]

以下是一个部分JSON示例，展示了您序列化对象的样子：
{
"events": {
    "event": {
        "Birmingham": [
            {
                "city": "Birmingham",
                "state": "AL",
                "country": "US",
                "lat": "33.5206608",
                "lng": "-86.80249",
                "status": "Delivered",
                "occured_at": "2012-04-06 14:17:00 UTC"
            },
            {
                "city": "Birmingham",
                "state": "AL",
                "country": "US",
                "lat": "33.5206608",
                "lng": "-86.80249",
                "status": "Out For Delivery",
                "occured_at": "2012-04-06 04:44:00 UTC"
            }
        ],
        "Spartanburg": [
            {
                "city":"Spartanburg",
                "state":"SC",
                "country":"US",
                "lat":"34.9495672",
                "lng":"-81.9320482",
                "status":"Departure Scan",
                "occured_at":"2012-04-05 08:42:00 UTC"
            },
            {
                "city": "Spartanburg",
                "state": "SC",
                "country": "US",
                "lat": "34.9495672",
                "lng": "-81.9320482",
                "status": "Arrival Scan",
                "occured_at": "2012-04-05 08:21:00 UTC"
            }
        ]
    }
}
}

这里是一个在 Ruby 中生成 JSON 的示例。该示例来自于 Ruby 的 JSON 实现：

puts JSON.pretty_generate([1, 2, {"a"=>3.141}, false, true, nil, 4..10])

以下是该命令生成的JSON结果：

[
    1,
    2,
    {
      "a": 3.141
    },
    false,
    true,
    null,
    {
      "json_class": "Range",
      "data": [
         4,
        10,
        false
      ]
    }
]

以下是一个部分示例，它将向你展示如何从一开始就使用修改后的示例启动:

puts JSON.pretty_generate({"events"=>{"event=>{"Birmingham"=>[{"city"=>"Birmingham","state"=>"AL"},{"city"=>"Birmingham","state"=>"AL"}]},{"Spartanburg"=>[{"city"=>"Spartanburg","state"=>"GA"}]}}})
那看起来就是我需要的。我该如何将我的当前JSON代码转换为您的示例？ - Shpigford
请查看https://dev59.com/M2025IYBdhLWcg3wr4B6。它描述了如何在Ruby中构建JSON对象。如果没有看到您的代码，我无法更具体地说明。如果有帮助的话，“[]”符号表示“数组”，而JSON由每个城市的单独数组组成。 - jamesmortensen
另外，也可以查看这个链接。它包含更复杂的JSON字符串示例：http://flori.github.com/json/doc/index.html - jamesmortensen
JSON对象直接来自第三方API，没有ActiveRecord对象或类似的东西。就像我的示例代码一样直截了当。 - Shpigford
@Shpigford - 我更新了我的答案，并提供了一个示例。虽然它尚未经过测试，但应该足以让您开始并演示概念。 - jamesmortensen
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