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理想情况下，我需要计算城市名称中使用“伦敦”的次数。但是查询返回的“london”、“London”和“LoNdOn”等值不同。我曾尝试使用不区分大小写作为选项，但它没有给我所需的结果。

这是我的查询，

 {
"queryType": "topN",
"dataSource": "wikiticker",
"dimension":"cityName",
"granularity": "ALL",
"metric": "count",
"threshold": 10,
"filter":
        { 
            "type": "search",
            "dimension": "cityName",
            "query": {
                "type": "insensitive_contains",
                "value": "london",                  
                 }
             }, 
"aggregations": [
        {
             "type": "longSum",
              "name": "count",
             "fieldName": "count"
         }  
       ],
 "intervals": ["2014-10-01T00:00:00.000Z/2016-10-07T00:00:00.000Z"]
 }

这是我的结果：

     [ {
     "timestamp" : "2015-09-12T00:46:58.771Z",
      "result" : [ {
         "count" : 21,
         "cityName" : "London"
        }, 
        {
          "count" : 10,
          "cityName" : "New London"
        }, 
        {
           "count" : 3,
           "cityName" : "london"
        }, 
        {
            "count" : 1,
            "cityName" : "LoNdon"
        }, 
        {
             "count" : 1,
             "cityName" : "LondOn"
         } ]
      } ]

我应该得到类似的东西：

        [ {
            "timestamp" : "2015-09-12T00:46:58.771Z",
            "result" : [ {
            "count" : 26,
            "cityName" : "London"
           }, 
           {
             "count" : 10,
             "cityName" : "New London"
           }  ]
       } ]
4

1 回答 1

1

使用过滤聚合器：

过滤聚合器包装任何给定的聚合器，但只聚合给定维度过滤器匹配的值。

{
  "type" : "filtered",
  "filter" : {
             "type" : "search",
             "dimension" : cityName,
             "query": {
                      "type":"contains",
                      "value":"london"
                      }          
             },
  "aggregator" : {
                 "type": "count",
                 "name": "Total Count of the Name London"
                 }
}

参考

于 2017-04-14T06:00:29.613 回答