Elasticsearch，根据同级嵌套字段进行合计 [英] Elasticsearch, terms aggs according to sibling nested fields
问题描述
Elasticsearch v7.5
Elasticsearch v7.5
你好，美好的一天！
我们有两个名为 socialmedia 和 Influencers
样本内容:
社交媒体:
{
'_id' : 1001,
'title' : "Title 1",
'smp_id' : 1,
"latest" : [
{
"soc_mm_score" : "5",
}
]
},
{
'_id' : 1002,
'title' : "Title 2",
'smp_id' : 2,
"latest" : [
{
"soc_mm_score" : "10",
}
]
},
{
'_id' : 1003,
'title' : "Title 3",
'smp_id' : 3,
"latest" : [
{
"soc_mm_score" : "35",
}
]
},
{
'_id' : 1004,
'title' : "Title 4",
'smp_id' : 2,
"latest" : [
{
"soc_mm_score" : "30",
}
]
}
///省略了其他一些字段
//omitted some other fields
影响者:
{
'_id' : 1,
'name' : "John",
'smp_id' : 1
},
{
'_id' : 2,
'name' : "Peter",
'smp_id' : 2
},
{
'_id' : 3,
'name' : "Mark",
'smp_id' : 3
}
现在，我有一个简单的查询，可确定 socialmedia 索引中的哪个文档具有最大的 latest.soc_mm_score 值，并显示它们 smp_id
Now I have this simple query that determines which documents in the socialmedia index has the most latest.soc_mm_score value, and also displaying their corresponding influencers determined by the smp_id
GET socialmedia/_search
{
"size": 0,
"_source": "latest",
"query": {
"match_all": {}
},
"aggs": {
"LATEST": {
"nested": {
"path": "latest"
},
"aggs": {
"MM_SCORE": {
"terms": {
"field": "latest.soc_mm_score",
"order": {
"_key": "desc"
},
"size": 3
},
"aggs": {
"REVERSE": {
"reverse_nested": {},
"aggs": {
"SMP_ID": {
"top_hits": {
"_source": ["smp_id"],
"size": 1
}
}
}
}
}
}
}
}
}
}
样品输出:
"aggregations" : {
"LATEST" : {
"doc_count" : //omitted,
"MM_SCORE" : {
"doc_count_error_upper_bound" : //omitted,
"sum_other_doc_count" : //omitted,
"buckets" : [
{
"key" : 35,
"doc_count" : 1,
"REVERSE" : {
"doc_count" : 1,
"SMP_ID" : {
"hits" : {
"total" : {
"value" : 1,
"relation" : "eq"
},
"max_score" : 1.0,
"hits" : [
{
"_index" : "socialmedia",
"_type" : "index",
"_id" : "1003",
"_score" : 1.0,
"_source" : {
"smp_id" : "3"
}
}
]
}
}
}
},
{
"key" : 30,
"doc_count" : 1,
"REVERSE" : {
"doc_count" : 1,
"SMP_ID" : {
"hits" : {
"total" : {
"value" : 1,
"relation" : "eq"
},
"max_score" : 1.0,
"hits" : [
{
"_index" : "socialmedia",
"_type" : "index",
"_id" : "1004",
"_score" : 1.0,
"_source" : {
"smp_id" : "2"
}
}
]
}
}
}
},
{
"key" : 10,
"doc_count" : 1,
"REVERSE" : {
"doc_count" : 1,
"SMP_ID" : {
"hits" : {
"total" : {
"value" : 1,
"relation" : "eq"
},
"max_score" : 1.0,
"hits" : [
{
"_index" : "socialmedia",
"_type" : "index",
"_id" : "1002",
"_score" : 1.0,
"_source" : {
"smp_id" : "2"
}
}
]
}
}
}
}
]
}
}
}
通过上面的查询，我能够成功显示哪些文档具有最高的 latest.soc_mm_score 值
with the query above, I was able to successfully display which documents have the highest latest.soc_mm_score values
上面的示例输出仅显示 DOCUMENTS ，并根据 latest.soc_mm_score 告诉与他们相关的影响者(aka smp_id)是TOP INFLUENCERS.strong>
The sample output above only displays DOCUMENTS, telling that the influencers (a.k.a smp_id) related to them are the TOP INFLUENCERS according to latest.soc_mm_score
理想情况下，只需使用此aggs查询
Ideally just by using this aggs query,
"terms" : {
"field" : "smp_id"
}
根据 doc_count
现在，根据 latest.soc_mm_score 显示字词查询会显示 TOP DOCUMENTS
Now, displaying the terms query according to latest.soc_mm_score displays TOP DOCUMENTS
"terms" : {
"field" : "latest.soc_mm_score"
}
目标目标:
我想根据 socialmedia 索引中的 latest.soc_mm_count 显示 TOP INFLUENCERS .如果Elasticsearch可以根据唯一的smp_id计算所有文档的位置，那么ES是否有办法汇总所有 latest.soc_mm_score 值并将其用作条款?
I want to display the TOP INFLUENCERS according to the latest.soc_mm_count in the socialmedia index. If Elasticsearch can count all the documents where according to unique smp_id, is there a way for ES to sum all latest.soc_mm_score values and use it as terms?
我上面的目标应该输出以下内容:
My objective above should output these:
- smp_id 2成为最有影响力的人，因为他有2个帖子(soc_mm_score分别为30和10)，加上这些帖子后他的排名为40 soc_mm_score
- smp_id 3作为第二位杰出影响者，他拥有1个职位，得分为35 soc_mm_score
- smp_id 1作为第三大影响者，他有1个帖子，还有5个soc_mm_score
是否有适当的查询来实现此目标?
Is there a proper query to meet this objective?
推荐答案
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FINALLY! FOUND AN ANSWER!!!
"aggs": {
"INFS": {
"terms": {
"field": "smp_id.keyword",
"order": {
"LATEST > SUM_SVALUE": "desc"
}
},
"aggs": {
"LATEST": {
"nested": {
"path": "latest"
},
"aggs": {
"SUM_SVALUE": {
"sum" : {
"field": "latest.soc_mm_score"
}
}
}
}
}
}
}
显示以下示例:
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