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 10A 11A 13B 11B 11B 12C 10C 10C 11C 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    }  }}