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。 我需要将这些结果的得分进行比较，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，则其分数应该是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11分。由于下面的查询似乎忽略了内部函数得分并计算：
（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）。
我尝试过不嵌套函数得分并使用简单的应该查询以及尝试 boost_factor 而不是'weight'，并给匹配的字段一个恒定的分数，所有这些都具有相同的结果。
而不是应用恒定权重乘以我想使用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中我是否可以使用除“_score”之外的字段来获取此字段？
提前致谢！
"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"
}
}
"_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"
}
]
}
答案 0 :(得分：2)
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。这意味着它们要么匹配，要么不匹配，因此无法传递嵌套的function_score的分数。
我确实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。