我正在尝试根据字段的重要性来实现自定义分数.
但是,我需要比较不同文档类型的多个索引.这些文件有不同的字段,具有不同的重要性.
我需要从这些结果中得出的分数具有可比性,因此我们希望忽略TF / IDF和得分归一化.
因此,如果搜索查询匹配2个重要字段和1个不太重要的字段,则其分数应为重要分数的两倍加上不太重要的分数：
(8* (1+1)) + (3*(1)) = 19
我得到的结果是得分11.由于下面的查询似乎忽略内部函数得分和计算：
(8*1) + (3*1).
分数解释也在下面,这似乎表明它忽略了内部的function_score,只是给它一个1的常数分数(这就是我想要停止发生的事情).
我试过没有嵌套函数得分和使用简单的应该查询以及尝试使用boost_factor而不是’weight’并给匹配的字段一个恒定的分数,所有这些都有相同的结果.
而不是应用恒定的权重乘以我想使用script_score来计算外部结果.然而,传递的’_score’不是我刚刚计算的分数,而是原始搜索分数.
在script_score中我是否可以使用“_score”以外的字段来获取此字段？
提前致谢！
询问
"query": {
"function_score": {
"functions": [
{
"weight": 8.0,
"filter": {
"fquery": {
"query": {
"function_score": {
"functions": [
{
"weight": 1.0,
"filter": {
"fquery": {
"query": {
"query_string": {
"query": "match*",
"fields": [
"ImportantField1"
],
"default_operator": "and",
"analyzer": "english",
"analyze_wildcard": true
}
}
}
}
},
{
"weight": 1.0,
"filter": {
"fquery": {
"query": {
"query_string": {
"query": "match*",
"fields": [
"ImportantField2"
],
"default_operator": "and",
"analyzer": "english",
"analyze_wildcard": true
}
}
}
} // More field queries that don't match omitted for clarity
}
],
"score_mode": "sum",
"boost_mode": "replace"
}
}
}
}
},
{
"weight": 3.0,
"filter": {
"fquery": {
"query": {
"function_score": {
"functions": [
{
"weight": 1.0,
"filter": {
"fquery": {
"query": {
"query_string": {
"query": "match*",
"fields": [
"LessImportantField"
],
"default_operator": "and",
"analyzer": "english",
"analyze_wildcard": true
}
}
}
}
}// More field queries that don't match omitted for clarity
],
"query": {
"match_all": {}
},
"score_mode": "sum",
"boost_mode": "replace"
}
}
}
}
}
],
"query": {
"match_all": {} // Filtering done here, omitted for clarity
}
},
"score_mode": "sum",
"boost_mode": "replace"
}
}
分数说明
"_explanation": {
"value": 11,
"description": "function score, product of:",
"details": [
{
"value": 11,
"description": "Math.min of",
"details": [
{
"value": 11,
"description": "function score, score mode [sum]",
"details": [
{
"value": 8,
"description": "function score, product of:",
"details": [
{
"value": 1,
"description": "match filter: QueryWrapperFilter(function score (ConstantScore(*:*), functions: [{filter(QueryWrapperFilter(ImportantField1:match*)), function [org.elasticsearch.common.lucene.search.function.WeightFactorFunction@64b3fd0e]}{filter(QueryWrapperFilter(ImportantField2:match*)), function [org.elasticsearch.common.lucene.search.function.WeightFactorFunction@38ed4b5c]}]))"
},
{
"value": 8,
"description": "product of:",
"details": [
{
"value": 1,
"description": "constant score 1.0 - no function provided"
},
{
"value": 8,
"description": "weight"
}
]
}
]
},
{
"value": 3,
"description": "function score, product of:",
"details": [
{
"value": 1,
"description": "match filter: QueryWrapperFilter(function score (ConstantScore(*:*), functions: [{filter(QueryWrapperFilter(LessImportantField:match*)), function [org.elasticsearch.common.lucene.search.function.WeightFactorFunction@3ce99ebf]}]))"
},
{
"value": 3,
"description": "product of:",
"details": [
{
"value": 1,
"description": "constant score 1.0 - no function provided"
},
{
"value": 3,
"description": "weight"
}
]
}
]
}
]
},
{
"value": 3.4028235e+38,
"description": "maxBoost"
}
]
},
{
"value": 1,
"description": "queryBoost"
}
]
}
最佳答案 所以这是不可能的. Function_score仅在其函数中使用过滤器来应用分数.这意味着它们匹配或不匹配,因此嵌套的function_score的分数无法传递.
我设法使用以下方法禁用查询规范化：
"similarity": {
"default": {
"queryNorm": "1",
"type": //whatever type you want
}
}
然而这意味着TF / IDF对我来说成了问题,因为我的每个索引的这些值都不同,所以我最终使用编写自定义相似度类并将这些值设置为1的常量.
