我想要一个查询,它根据另一个字段的唯一值返回一个字段重复的次数,我有这个 json:
"name" : james,
"city" : "chicago" <----------- same
},
{
"name" : james,
"city" : "san francisco"
},
{
"name" : james,
"city" : "chicago" <-----------same
},
{
"name" : Mike,
"city" : "chicago"
},
{
"name" : Mike,
"city" : "texas"<-----------same
},
{
"name" : Mike,
"city" : "texas"<-----------same
},
{
"name" : Peter,
"city" : "chicago"
},
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我想进行一个查询,根据两个字段的唯一值进行计数。例如,james 等于 2,因为有两个相等的字段(名称:james、city、chicago)和一个不同的字段(名称:james、city:san francisco),那么输出将如下所示:
{
"key" : "james",
"doc_count" : 2
},
{
"key" : "Mike",
"doc_count" : 2
},
{
"key" : "Peter",
"doc_count" : 1
},
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可以对两个字段进行单值计数吗?
您可以进行两级术语聚合:
{
"size": 0,
"aggs": {
"names": {
"terms": {
"field": "name.keyword",
"size": 10
},
"aggs": {
"citys_by_name": {
"terms": {
"field": "city.keyword",
"size": 10,
"min_doc_count": 2
}
}
}
}
}
}
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响应将如下所示:
"aggregations" : {
"names" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "james",
"doc_count" : 15,
"citys_by_name" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "chicago",
"doc_count" : 14
}
]
}
},
{
"key" : "Peter",
"doc_count" : 2,
"citys_by_name" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "chicago",
"doc_count" : 2
}
]
}
},
{
"key" : "mike",
"doc_count" : 2,
"citys_by_name" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [ ]
}
}
]
}
}
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或者您可以连接字段:
GET test/_search
{
"size": 0,
"aggs": {
"names": {
"terms": {
"script": {
"source": "return doc['name.keyword'].value + ' ' + doc['city.keyword'].value",
"lang": "painless"
},
"field": "name.keyword",
"size": 10,
"min_doc_count": 2
}
}
}
}
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响应看起来是这样的:
"aggregations" : {
"names" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "james chicago",
"doc_count" : 14
},
{
"key" : "Peter chicago",
"doc_count" : 2
}
]
}
}
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如果您想要有关存储桶的更多统计信息,请使用 stats_buckets 聚合:
{
"size": 0,
"aggs": {
"names": {
"terms": {
"script": {
"source": "return doc['name.keyword'].value + ' ' + doc['city.keyword'].value",
"lang": "painless"
},
"field": "name.keyword",
"size": 10,
"min_doc_count": 2
}
},
"names_stats":{
"stats_bucket": {
"buckets_path":"names._count"
}
}
}
}
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将导致:
"aggregations" : {
"names" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : "james PARIS",
"doc_count" : 15
},
{
"key" : "james chicago",
"doc_count" : 13
},
{
"key" : "samuel PARIS",
"doc_count" : 11
},
{
"key" : "fred PARIS",
"doc_count" : 2
}
]
},
"names_stats" : {
"count" : 4,
"min" : 2.0,
"max" : 15.0,
"avg" : 10.25,
"sum" : 41.0
}
}
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