wos*_*tom 6 distance aggregation elasticsearch
我的映射如下:
PUT places
{
"mappings": {
"test": {
"properties": {
"id_product": { "type": "keyword" },
"id_product_unique": { "type": "integer" },
"location": { "type": "geo_point" },
"suggest": {
"type": "text"
},
"active": {"type": "boolean"}
}
}
}
}
POST places/test
{
"id_product" : "A",
"id_product_unique": 1,
"location": {
"lat": 1.378446,
"lon": 103.763427
},
"suggest": ["coke","zero"],
"active": true
}
POST places/test
{
"id_product" : "A",
"id_product_unique": 2,
"location": {
"lat": 1.878446,
"lon": 108.763427
},
"suggest": ["coke","zero"],
"active": true
}
POST places/test
{
"id_product" : "B",
"id_product_unique": 3,
"location": {
"lat": 1.478446,
"lon": 104.763427
},
"suggest": ["coke"],
"active": true
}
POST places/test
{
"id_product" : "C",
"id_product_unique": 4,
"location": {
"lat": 1.218446,
"lon": 102.763427
},
"suggest": ["coke","light"],
"active": true
}
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在我的例子中,有2罐可乐零("id_product_unique" = 1和2),1罐可乐("id_product_unique" = 3)和一罐可乐灯("id_product_unique" = 4)
所有这些罐都在不同的位置.
" id_product"不是唯一的,因为完全相同的"焦炭罐"可以在两个不同的位置(ex "id_product_unique" = 1和2)出售.
只有" id_product_unique"和"位置"从"可乐罐"变为另一个(2个相同的"焦炭罐"具有相同的字段"建议"和" id_product"但不是相同的" id_product_unique"和" location").
我的目标是从给定的GPS位置搜索产品,并通过id_product(最近的一个)显示唯一的结果:
POST /places/_search?size=0
{
"aggs" : {
"group-by-type" : {
"terms" : { "field" : "id_product"},
"aggs": {
"min-distance": {
"top_hits": {
"sort": {
"_script": {
"type": "number",
"script": {
"source": "def x = doc['location'].lat; def y = doc['location'].lon; return Math.abs(x-1.178446) + Math.abs(y-101.763427)",
"lang": "painless"
},
"order": "asc"
}
},
"size" : 1
}
}
}
}
}
}
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从这个结果列表中我现在想应用一个应该查询并按计算得分重新排序我的结果列表.我尝试了以下方法:
POST /places/_search?size=0
{
"query" : {
"bool": {
"filter": {"term" : { "active" : "true" }},
"should": [
{"match" : { "suggest" : "coke" }},
{"match" : { "suggest" : "light" }}
]
}
},
"aggs" : {
"group-by-type" : {
"terms" : { "field" : "id_product"},
"aggs": {
"min-distance": {
"top_hits": {
"sort": {
"_script": {
"type": "number",
"script": {
"source": "def x = doc['location'].lat; def y = doc['location'].lon; return Math.abs(x-1.178446) + Math.abs(y-101.763427)",
"lang": "painless"
},
"order": "asc"
}
},
"size" : 1
}
}
}
}
}
}
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但我无法想象如何用doc分数替换距离排序分数.
任何帮助都会很棒.
我设法通过添加新的聚合“max_score”来做到这一点:
"max_score": {
"max": {
"script": {
"lang": "painless",
"source": "_score"
}
}
}
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并按 max_score.value desc 排序:
"order": {"max_score.value": "desc"}
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我的最终查询如下:
POST /places/_search?size=0
{
"query" : {
"bool": {
"filter": {"term" : { "active" : "true" }},
"should": [
{"match" : { "suggest" : "coke" }},
{"match" : { "suggest" : "light" }}
]
}
},
"aggs" : {
"group-by-type" : {
"terms" : {
"field" : "id_product",
"order": {"max_score.value": "desc"}
},
"aggs": {
"min-distance": {
"top_hits": {
"sort": {
"_script": {
"type": "number",
"script": {
"source": "def x = doc['location'].lat; def y = doc['location'].lon; return Math.abs(x-1.178446) + Math.abs(y-101.763427)",
"lang": "painless"
},
"order": "asc"
}
},
"size" : 1
}
},
"max_score": {
"max": {
"script": {
"lang": "painless",
"inline": "_score"
}
}
}
}
}
}
}
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回答:
{
"took": 3,
"timed_out": false,
"_shards": {
"total": 5,
"successful": 5,
"skipped": 0,
"failed": 0
},
"hits": {
"total": 4,
"max_score": 0,
"hits": []
},
"aggregations": {
"group-by-type": {
"doc_count_error_upper_bound": 0,
"sum_other_doc_count": 0,
"buckets": [
{
"key": "C",
"doc_count": 1,
"max_score": {
"value": 1.0300811529159546
},
"min-distance": {
"hits": {
"total": 1,
"max_score": null,
"hits": [
{
"_index": "places",
"_type": "test",
"_id": "VhJdOmIBKhzTB9xcDvfk",
"_score": null,
"_source": {
"id_product": "C",
"id_product_unique": 4,
"location": {
"lat": 1.218446,
"lon": 102.763427
},
"suggest": [
"coke",
"light"
],
"active": true
},
"sort": [
1.0399999646503995
]
}
]
}
}
},
{
"key": "A",
"doc_count": 2,
"max_score": {
"value": 0.28768208622932434
},
"min-distance": {
"hits": {
"total": 2,
"max_score": null,
"hits": [
{
"_index": "places",
"_type": "test",
"_id": "UhJcOmIBKhzTB9xc6ve-",
"_score": null,
"_source": {
"id_product": "A",
"id_product_unique": 1,
"location": {
"lat": 1.378446,
"lon": 103.763427
},
"suggest": [
"coke",
"zero"
],
"active": true
},
"sort": [
2.1999999592114756
]
}
]
}
}
},
{
"key": "B",
"doc_count": 1,
"max_score": {
"value": 0.1596570909023285
},
"min-distance": {
"hits": {
"total": 1,
"max_score": null,
"hits": [
{
"_index": "places",
"_type": "test",
"_id": "VRJcOmIBKhzTB9xc_vc0",
"_score": null,
"_source": {
"id_product": "B",
"id_product_unique": 3,
"location": {
"lat": 1.478446,
"lon": 104.763427
},
"suggest": [
"coke"
],
"active": true
},
"sort": [
3.2999999020282695
]
}
]
}
}
}
]
}
}
}
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