elasticsearch - Elasticsearch 路口查询

Question

我想获取按总数排序的用户列表的常用字词。

例:
我有一个用户使用的单词索引。

docs:

[
  {
    user_id: 1,
    word: 'food',
    count: 2
  },
  {
    user_id: 1,
    word: 'thor',
    count: 1
  },
  {
    user_id: 1,
    word: 'beer',
    count: 7
  },
  {
    user_id: 2,
    word: 'summer',
    count: 12
  },
  {
    user_id: 2,
    word: 'thor',
    count: 4
  },
  {
    user_id: 1,
    word: 'beer',
    count: 2
  },
  ..otheruserdetails..
]

输入:user_ids: [1, 2]
所需的输出:

[
  {
    'word': 'beer',
    'total_count': 9
  },
  {
    'word': 'thor',
    'total_count': 5
  }
]

我到目前为止所拥有的:

在user_id列表中使用user_id获取所有文档( bool(boolean) 应查询)

在应用程序层处理文档。

遍历每个关键字

检查每个user_id是否存在关键字

如果是，请找到计数

else，配置并转到下一个关键字

但是，这是不可行的，因为word文档会变得越来越大，并且应用程序层将无法跟上。任何方式将其移至ES查询？

Answer 1

您可以使用Terms aggregation和Value Count aggregation

可以将“术语聚合”视为“分组依据”。输出将给出一个唯一的userIds列表，该用户下所有单词的列表以及每个单词的最终计数

{
  "from": 0, 
  "size": 10, 
  "query": {
    "terms": {
      "user_id": [
        "1",
        "2"
      ]
    }
  },
  "aggs": {
    "users": {
      "terms": {
        "field": "user_id",
        "size": 10
      },
      "aggs": {
        "words": {
          "terms": {
            "field": "word.keyword",
            "size": 10
          },
          "aggs": {
            "word_count": {
              "value_count": {
                "field": "word.keyword"
              }
            }
          }
        }
      }
    }
  }
}

结果

    "hits" : [
      {
        "_index" : "index89",
        "_type" : "_doc",
        "_id" : "gFRzr3ABAWOsYG7t2tpt",
        "_score" : 1.0,
        "_source" : {
          "user_id" : 1,
          "word" : "thor",
          "count" : 1
        }
      },
      {
        "_index" : "index89",
        "_type" : "_doc",
        "_id" : "flRzr3ABAWOsYG7t0dqI",
        "_score" : 1.0,
        "_source" : {
          "user_id" : 1,
          "word" : "food",
          "count" : 2
        }
      },
      {
        "_index" : "index89",
        "_type" : "_doc",
        "_id" : "f1Rzr3ABAWOsYG7t19ps",
        "_score" : 1.0,
        "_source" : {
          "user_id" : 2,
          "word" : "thor",
          "count" : 4
        }
      },
      {
        "_index" : "index89",
        "_type" : "_doc",
        "_id" : "gVRzr3ABAWOsYG7t8NrR",
        "_score" : 1.0,
        "_source" : {
          "user_id" : 1,
          "word" : "food",
          "count" : 2
        }
      },
      {
        "_index" : "index89",
        "_type" : "_doc",
        "_id" : "glRzr3ABAWOsYG7t-Npj",
        "_score" : 1.0,
        "_source" : {
          "user_id" : 1,
          "word" : "thor",
          "count" : 1
        }
      },
      {
        "_index" : "index89",
        "_type" : "_doc",
        "_id" : "g1Rzr3ABAWOsYG7t_9po",
        "_score" : 1.0,
        "_source" : {
          "user_id" : 2,
          "word" : "thor",
          "count" : 4
        }
      }
    ]
  },
  "aggregations" : {
    "users" : {
      "doc_count_error_upper_bound" : 0,
      "sum_other_doc_count" : 0,
      "buckets" : [
        {
          "key" : 1,
          "doc_count" : 4,
          "words" : {
            "doc_count_error_upper_bound" : 0,
            "sum_other_doc_count" : 0,
            "buckets" : [
              {
                "key" : "food",
                "doc_count" : 2,
                "word_count" : {
                  "value" : 2
                }
              },
              {
                "key" : "thor",
                "doc_count" : 2,
                "word_count" : {
                  "value" : 2
                }
              }
            ]
          }
        },
        {
          "key" : 2,
          "doc_count" : 2,
          "words" : {
            "doc_count_error_upper_bound" : 0,
            "sum_other_doc_count" : 0,
            "buckets" : [
              {
                "key" : "thor",
                "doc_count" : 2,
                "word_count" : {
                  "value" : 2
                }
              }
            ]
          }
        }
      ]
    }
  }