一、需求

A、B、C代表3个用户，第二列代表各自的得分，求A、B、C的最好成绩以及A、B、C最好成绩的均值

A 10
A 11
A 13
B 11
B 11
B 12
C 10
C 10
C 11
C 15

二、思路

先terms分组，求最大值，最后加一个pipeline均值。一开始想用bucket_script解决，实验发现走不通，但是bucket_script在聚合结果之上操作很有用

三、测试数据

PUT sport 
{
  "mappings": {
    "grade": {
      "properties": {
        "user": {
          "type": "keyword"
        },
        "grade":{
          "type": "integer"
        }
      }
    }
  }
}

PUT sport/grade/1
{
  "user":"A",
  "grade":10
}

PUT sport/grade/2
{
  "user":"A",
  "grade":11
}

PUT sport/grade/3
{
  "user":"A",
  "grade":13
}

PUT sport/grade/4
{
  "user":"B",
  "grade":11
}
PUT sport/grade/5
{
  "user":"B",
  "grade":11
}

PUT sport/grade/6
{
  "user":"B",
  "grade":12
}


PUT sport/grade/7
{
  "user":"C",
  "grade":10
}

PUT sport/grade/8
{
  "user":"C",
  "grade":10
}

PUT sport/grade/9
{
  "user":"C",
  "grade":11
}

PUT sport/grade/10
{
  "user":"C",
  "grade":15
}

四、聚合

GET sport/_search
{
  "size": 0,
  "aggs": {
    "avg_score": {
      "terms": {
        "field": "user"
      },
      "aggs": {
        "max_score": {
          "max": {
            "field": "grade"
          }
        }
      }
    },
    "avg_max_score": {
      "avg_bucket": {
        "buckets_path": "avg_score>max_score"
      }
    }
  }
}

结果：

{
  "took": 4,
  "timed_out": false,
  "_shards": {
    "total": 5,
    "successful": 5,
    "failed": 0
  },
  "hits": {
    "total": 10,
    "max_score": 0,
    "hits": []
  },
  "aggregations": {
    "avg_score": {
      "doc_count_error_upper_bound": 0,
      "sum_other_doc_count": 0,
      "buckets": [
        {
          "key": "C",
          "doc_count": 4,
          "max_score": {
            "value": 15 }
        },
        {
          "key": "A",
          "doc_count": 3,
          "max_score": {
            "value": 13 }
        },
        {
          "key": "B",
          "doc_count": 3,
          "max_score": {
            "value": 12 }
        }
      ]
    },
    "avg_max_score": {
      "value": 13.333333333333334
    }
  }
}