多字段聚合
elasticsearch同时对多个字段聚合(两个字段值类型一致)
Elasticsearch • God_lockin 回复了问题 • 3 人关注 • 1 个回复 • 4024 次浏览 • 2019-08-03 18:20
es 一次进行多个metric聚合,性能很差,什么原因
Elasticsearch • kennywu76 回复了问题 • 3 人关注 • 1 个回复 • 4538 次浏览 • 2017-07-20 10:40
ElasticSearch java API - 聚合查询
Elasticsearch • carlislelee 发表了文章 • 3 个评论 • 54040 次浏览 • 2016-09-20 17:16
"mappings": {
"player": {
"properties": {
"name": {
"index": "not_analyzed",
"type": "string"
},
"age": {
"type": "integer"
},
"salary": {
"type": "integer"
},
"team": {
"index": "not_analyzed",
"type": "string"
},
"position": {
"index": "not_analyzed",
"type": "string"
}
},
"_all": {
"enabled": false
}
}
}
索引中的全部数据:
首先,初始化Builder:SearchRequestBuilder sbuilder = client.prepareSearch("player").setTypes("player");
接下来举例说明各种聚合操作的实现方法,因为在es的api中,多字段上的聚合操作需要用到子聚合(subAggregation),初学者可能找不到方法(网上资料比较少,笔者在这个问题上折腾了两天,最后度了源码才彻底搞清楚T_T),后边会特意说明多字段聚合的实现方法。另外,聚合后的排序也会单独说明。
- group by/count
select team, count(*) as player_count from player group by team;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("player_count ").field("team");
sbuilder.addAggregation(teamAgg);
SearchResponse response = sbuilder.execute().actionGet();
- group by多个field
select team, position, count(*) as pos_count from player group by team, position;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("player_count ").field("team");
TermsBuilder posAgg= AggregationBuilders.terms("pos_count").field("position");
sbuilder.addAggregation(teamAgg.subAggregation(posAgg));
SearchResponse response = sbuilder.execute().actionGet();
- max/min/sum/avg
select team, max(age) as max_age from player group by team;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("player_count ").field("team");
MaxBuilder ageAgg= AggregationBuilders.max("max_age").field("age");
sbuilder.addAggregation(teamAgg.subAggregation(ageAgg));
SearchResponse response = sbuilder.execute().actionGet();
- 对多个field求max/min/sum/avg
select team, avg(age)as avg_age, sum(salary) as total_salary from player group by team;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("team");
AvgBuilder ageAgg= AggregationBuilders.avg("avg_age").field("age");
SumBuilder salaryAgg= AggregationBuilders.avg("total_salary ").field("salary");
sbuilder.addAggregation(teamAgg.subAggregation(ageAgg).subAggregation(salaryAgg));
SearchResponse response = sbuilder.execute().actionGet();
- 聚合后对Aggregation结果排序
select team, sum(salary) as total_salary from player group by team order by total_salary desc;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("team").order(Order.aggregation("total_salary ", false);
SumBuilder salaryAgg= AggregationBuilders.avg("total_salary ").field("salary");
sbuilder.addAggregation(teamAgg.subAggregation(salaryAgg));
SearchResponse response = sbuilder.execute().actionGet();
需要特别注意的是,排序是在TermAggregation处执行的,Order.aggregation函数的第一个参数是aggregation的名字,第二个参数是boolean型,true表示正序,false表示倒序。
- Aggregation结果条数的问题
TermsBuilder teamAgg= AggregationBuilders.terms("team").size(15);
- Aggregation结果的解析/输出
Map<String, Aggregation> aggMap = response.getAggregations().asMap();
StringTerms teamAgg= (StringTerms) aggMap.get("keywordAgg");
Iterator<Bucket> teamBucketIt = teamAgg.getBuckets().iterator();
while (teamBucketIt .hasNext()) {
Bucket buck = teamBucketIt .next();
//球队名
String team = buck.getKey();
//记录数
long count = buck.getDocCount();
//得到所有子聚合
Map subaggmap = buck.getAggregations().asMap();
//avg值获取方法
double avg_age= ((InternalAvg) subaggmap.get("avg_age")).getValue();
//sum值获取方法
double total_salary = ((InternalSum) subaggmap.get("total_salary")).getValue();
//...
//max/min以此类推
}
- 总结
ElasticSearch怎么同时对两个字段做sum聚合
Elasticsearch • weizijun 回复了问题 • 2 人关注 • 1 个回复 • 14317 次浏览 • 2016-09-20 12:35
elasticsearch同时对多个字段聚合(两个字段值类型一致)
回复Elasticsearch • God_lockin 回复了问题 • 3 人关注 • 1 个回复 • 4024 次浏览 • 2019-08-03 18:20
es 一次进行多个metric聚合,性能很差,什么原因
回复Elasticsearch • kennywu76 回复了问题 • 3 人关注 • 1 个回复 • 4538 次浏览 • 2017-07-20 10:40
ElasticSearch怎么同时对两个字段做sum聚合
回复Elasticsearch • weizijun 回复了问题 • 2 人关注 • 1 个回复 • 14317 次浏览 • 2016-09-20 12:35
ElasticSearch java API - 聚合查询
Elasticsearch • carlislelee 发表了文章 • 3 个评论 • 54040 次浏览 • 2016-09-20 17:16
"mappings": {
"player": {
"properties": {
"name": {
"index": "not_analyzed",
"type": "string"
},
"age": {
"type": "integer"
},
"salary": {
"type": "integer"
},
"team": {
"index": "not_analyzed",
"type": "string"
},
"position": {
"index": "not_analyzed",
"type": "string"
}
},
"_all": {
"enabled": false
}
}
}
索引中的全部数据:
首先,初始化Builder:SearchRequestBuilder sbuilder = client.prepareSearch("player").setTypes("player");
接下来举例说明各种聚合操作的实现方法,因为在es的api中,多字段上的聚合操作需要用到子聚合(subAggregation),初学者可能找不到方法(网上资料比较少,笔者在这个问题上折腾了两天,最后度了源码才彻底搞清楚T_T),后边会特意说明多字段聚合的实现方法。另外,聚合后的排序也会单独说明。
- group by/count
select team, count(*) as player_count from player group by team;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("player_count ").field("team");
sbuilder.addAggregation(teamAgg);
SearchResponse response = sbuilder.execute().actionGet();
- group by多个field
select team, position, count(*) as pos_count from player group by team, position;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("player_count ").field("team");
TermsBuilder posAgg= AggregationBuilders.terms("pos_count").field("position");
sbuilder.addAggregation(teamAgg.subAggregation(posAgg));
SearchResponse response = sbuilder.execute().actionGet();
- max/min/sum/avg
select team, max(age) as max_age from player group by team;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("player_count ").field("team");
MaxBuilder ageAgg= AggregationBuilders.max("max_age").field("age");
sbuilder.addAggregation(teamAgg.subAggregation(ageAgg));
SearchResponse response = sbuilder.execute().actionGet();
- 对多个field求max/min/sum/avg
select team, avg(age)as avg_age, sum(salary) as total_salary from player group by team;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("team");
AvgBuilder ageAgg= AggregationBuilders.avg("avg_age").field("age");
SumBuilder salaryAgg= AggregationBuilders.avg("total_salary ").field("salary");
sbuilder.addAggregation(teamAgg.subAggregation(ageAgg).subAggregation(salaryAgg));
SearchResponse response = sbuilder.execute().actionGet();
- 聚合后对Aggregation结果排序
select team, sum(salary) as total_salary from player group by team order by total_salary desc;
ES的java api:TermsBuilder teamAgg= AggregationBuilders.terms("team").order(Order.aggregation("total_salary ", false);
SumBuilder salaryAgg= AggregationBuilders.avg("total_salary ").field("salary");
sbuilder.addAggregation(teamAgg.subAggregation(salaryAgg));
SearchResponse response = sbuilder.execute().actionGet();
需要特别注意的是,排序是在TermAggregation处执行的,Order.aggregation函数的第一个参数是aggregation的名字,第二个参数是boolean型,true表示正序,false表示倒序。
- Aggregation结果条数的问题
TermsBuilder teamAgg= AggregationBuilders.terms("team").size(15);
- Aggregation结果的解析/输出
Map<String, Aggregation> aggMap = response.getAggregations().asMap();
StringTerms teamAgg= (StringTerms) aggMap.get("keywordAgg");
Iterator<Bucket> teamBucketIt = teamAgg.getBuckets().iterator();
while (teamBucketIt .hasNext()) {
Bucket buck = teamBucketIt .next();
//球队名
String team = buck.getKey();
//记录数
long count = buck.getDocCount();
//得到所有子聚合
Map subaggmap = buck.getAggregations().asMap();
//avg值获取方法
double avg_age= ((InternalAvg) subaggmap.get("avg_age")).getValue();
//sum值获取方法
double total_salary = ((InternalSum) subaggmap.get("total_salary")).getValue();
//...
//max/min以此类推
}
- 总结